text:
Hamilton Region
Construction beginning on Centennial Parkway GO train station
The province announced the start of the $50-million project Thursday
By Matthew Van DongenSpectator Reporter
MORE ON THE GO STATION
Are GO trains finally on track to east Hamilton?
Construction will begin this fall on a long-delayed east-city GO train station that is now slated to see commuter trains roll in by 2025.
On Thursday, Premier Doug Ford and Transportation Minister Caroline Mulroney announced the start of the $50-million project at the eventual site on Centennial Parkway.
Ford said the new Confederation station will eventually offer "two-way, all-day service" to Stoney Creek residents who right now either have to drive to West Harbour GO in the core to take the train, or hop on a Route 12 bus.
"That's an absolute game-changer for the people of Hamilton," he said, promising faster commutes and more transit options — including the city's on-again, off-again LRT.
That original light-rail transit project was famously cancelled by the Tory government in 2019 over budget woes — but later resurrected in 2021 with shared capital funding of $3.4 billion from the province and federal Liberal government.
When asked if LRT remained on track in the face of inflationary pressures and pandemic supply-chain challenges, Ford said the province is still on board. "We're gonna get it built as soon as possible and we've committed to it," he said.
The new GO station at 397 Centennial Parkway will include an accessible island platform with canopies, a pedestrian tunnel and direct stair access from Centennial Parkway. There will be a kiss-and-ride area as well as 150 parking spots.
The planned new station will be "self-serve" rather than staffed, however.
Once upon a time, the station was supposed to be open in 2019. The former Liberal government even held a ceremonial groundbreaking for presumed construction in 2018.
But a new tender call went out in 2020 after the Tory government announced a "market-driven approach" to future station construction.
Metrolinx chief capital officer Stephanie Davies said the agency must also do some track work through the lower city to allow competing users room to move in the crowded CN-owned railroad corridor that also ushers freight trains and other passenger services through Hamilton.
Matthew Van Dongen is a transportation and environment reporter at The Spectator. mvandongen@thespec.com
Go Transit, Metrolinx


text:
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Boston Tech Watch: Polaris, GE, DataRobot, Facial Recognition Woes
Mark Your Calendars: Startup Week Detroit Kicks Off on June 17
"More to Life Than Profit": Protest at Wayfair for Sales to Migrant Camp
Bill Mortimore
CEO of Foundry
David Broecker
President & CEO, Indiana Biosciences Research Institute
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Weaving Words with Wordle: A Talk with IBM's Jonathan Feinberg
Wade Roush
@soonishpodcast
Xconomy chose to set up shop in Kendall Square because we wanted to be at the epicenter of investment and innovation in the Boston area. But it was just luck that we ended up right across the street from IBM's Cambridge research facility at 1 Rogers Street, which is home to both the Collaborative User Experience (CUE) Research group and the new Center for Social Software. The creative software engineers, user interface designers, and visualization researchers in the group have come up with a whole gallery of fascinating tools for studying the way business people (and average Web surfers) interact with data, including Many Eyes, a community portal for visualization experiments.
But one of the Web-based applications included in Many Eyes has taken on a surprising life of its own outside IBM. It's a free visualization tool called Wordle, where users can dump a bunch of text into a window and then see it automatically yet artfully arranged into a cloud of words, with the size of each word corresponding to its frequency in the original passage. As the Wordle site puts it, it's "a toy for generating 'word clouds' from text that you provide."
These aren't your father's tag clouds: they're playful, colorful, and made from attractive fonts. And they fill up the space around their center points in a clever way that often seems to say something profound about the nature of the ideas in the text. In fact, Wordle achieved minor fame during the 2008 presidential campaign, when several newspapers and other media outlets used the tool to analyze publications and speeches by the major candidates. It's also found its way into popular culture: a Google image search for Wordle diagrams turns up more than 200,000 examples. (The Wordle word cloud on this page is based on the text of this article, and the one on page 2 is based in Lincoln's Gettysburg Address.)
I dropped by the CUE Research group's offices several weeks ago for a conversation with Wordle's creator and shepherd, Jonathan Feinberg. Though Feinberg is a senior software engineer at CUE, he developed Wordle as a personal project, and maintains it on a server outside of IBM. Martin Wattenberg, a mathematician who lead's the CUE Research group's Visual Communications Lab, was also on hand for the talk. Below is an abridged transcription.
Xconomy: Where did Wordle come from?
Jonathan Feinberg: It came from a Lotus project I worked on called Dogear. It was a social bookmarking engine. In every piece of social software there's a tag cloud, and I have implemented some, but I thought they were ugly and boring. I was prompted by Dogear to think about the ways you could fit words together on a page. The core bit of code I wrote was a Java applet within Dogear called TagExplorer, but when Dogear came out as a product and TagExplorer was not part of it, because Java applets are considered too slow to load. Then two years later, while clearing out one of my workspaces, I stumbled onto the code and thought I'd like to do something with it. That's when I built Wordle. So, the sand that formed the pearl was a tag cloud, but it's really divorced from that idea now.
There was no way to put it out there as a branded IBM thing. I didn't even begin working on the real Web application for Wordle until I had gotten permission to use the code externally. The lawyers and the product managers didn't want this code for a product, and I got permission to use it non-commercially—which serviced me well, because I don't like doing business. So Wordle the Web app belongs to me personally, not ManyEyes.
Martin Wattenberg: But ManyEyes has Wordle in it, and it's become one of the most popular visualization types on ManyEyes.
X: What's really wrong with conventional tag clouds?
MW: If you look at how the academic community has viewed things like tag clouds, it's been with a certain amount of suspicion and skepticism about what they convey. And that skepticism has been backed up by studies. If you are going to try to convey a list of word frequencies and measure what people remember about a text, it's not clear that tag clouds give you any benefit over, say, a list of words ordered by frequency.
That said, when Wordle went up, Jon positioned it as a toy. To me there is a fine line between a toy and an experiment. We've had thousands of people from all over the Web experimenting with Wordle and finding all sorts of things to do with it. There is a blog on the top 25 uses for Wordle. For me that is an indication that there is a lot more to it than just being a toy. There are teachers who have found that it's extremely useful as part of their teaching. And maybe if you are analyzing a text very serious, a Wordle diagram has no advantage over a list of word frequencies. But if you are trying to convey something emotionally, Wordle has a huge advantage.
There is this analogy with cameras. A Wordle is almost like a Polaroid camera for text. It lets people take snapshots that mean something to them. We've found more than one blog where guys talk about getting "boyfriend points" by making Wordles out of their love letters to their girlfriends. All of these things point to really interested new uses of visualization.
JF: But how am I ever going to get back all of the husband points I lost by spending all that time making Wordle?
MW: One of the interesting things we've seen is a bunch of fairly high profile media outlets like the Boston Globe and the Washington Post have used Wordles to illustrate articles about speeches. That indicates that it is in some sense more than a toy. Poeple who are professional political analysts find this an interesting way to illustrate their points. Jon looked at tag clouds, thought about what was wrong with them, and created this fantastic alterantive. And by putting it on the Web, we've seen ways that it's useful that I don't think we would have invented ourselves or trusted. It's a little counterintuitive, and it's certianly contrary to the usual paradigm of interface design, were you have a task in mind and an audience in mind before you start.
JF: Calling it a toy was a defensive measure. I don't want to make any claims for Wordle as a visualization tool that gets you an accurate idea of your text. People write to me and say 'Wow, I made a Wordle and now I see wasn't writing about what I thought I was writing about.' But I feel very resistant to those kinds of analyses.
MW: That raises a research question about what people are getting out of this way of visualizing text and what they think they're getting. Wordle really is the starting point of an important line of research. We are looking at the way it's being used, and are in the midst of writing our first paper on it. By putting it out in public without making any strong claims about it, we get to see … Next Page »
Wade Roush is the producer and host of the podcast Soonish and a contributing editor at Xconomy. Follow @soonishpodcast

redpajama_set_name: RedPajamaCommonCrawl


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Flowering of American Folk Art by Jean Lipman and Alice Wincheste, 1776-1816, 1974, Courage Books and Whitney Museum of Art, 288 pages, dust jacket, cloth, more than 400 hundreds of illustrations in color and black and white of outstanding examples of folk art.
Five Star Folk Art: one hundred American masterpieces by Lipman-Bishop-Warren-Eisenstat, Abrams, '90, 176pp, 126 illustrations-including 125 in color.
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Hong Kong International Airport (HKIA) has seen continued growth in all three traffic categories during March 2019.
Passenger volume rose by 0.3 per cent to 6.4 million compared to the same month in 2018, while flight movements increased by 1.3 per cent to 36,735. Cargo throughput recorded year-on-year growth of 0.2 per cent; reaching 432,000 tonnes.
The growth in passenger volume was driven by higher levels of transfer and transit traffic, which together registered a year-on-year upsurge of nine per cent. Passenger traffic to and from India, mainland China and Japan recorded the most significant increases. Because the Easter peak season fell in late March last year, Hong Kong resident travel in March 2019 experienced a 10 per cent year-on-year drop.
Exports saw an 11 per cent increase in March, making the biggest contribution to cargo growth. Amongst key trading regions, cargo exported to North America and Europe posted the largest increases. Imports and transshipments both declined.
Over the first three months of 2019, HKIA handled 18.8 million passengers and 105,695 flight movements, representing year-on-year increases of 2.5 per cent and 1.1 per cent respectively. HKIA handled 1.1 million tonnes of cargo between January and March 2019, a 5.4 per cent decrease compared to the same period in 2018.
On a 12-month rolling basis, HKIA handled 75.1 million passengers and 428,870 flight movements, representing respective increases of two per cent and 1.3 per cent over the same period last year. Cargo throughput experienced a slight decline of 0.6 per cent to 5.06 million tonnes.
In order to manage the expected travel surge during the Easter holidays, additional manpower will be deployed to assist passengers at the terminals.
Meanwhile, HKIA has been working with Travelport, a leading travel commerce platform, on a new arrangement to promote multi-modal connectivity services. Under the arrangement, overseas travel buyers can book multi-modal transfer services conveniently when reserving air tickets, ensuring a hassle-free experience for passengers commuting between HKIA and the cities in the Guangdong-Hong Kong-Macao Greater Bay Area. Passengers will be able to take advantage of this new offering within this year.


text:
\section{Introduction}
Let $S=K[x_1,\dots,x_n]$ be a polynomial ring over a field $K$, $\lambda$  a partition of $n$, and $\tab \lambda$ the set of tableaux of shape $\lambda$.
For a tableau $T \in \tab \lambda$, its {\bf Specht polynomial} $f_T \in S$ is the product of all $x_i-x_j$ such that $i$ and $j$ are in the same column of $T$.
The ideal $I_\lambda$ of $S$ generated by all Specht polynomials of shape $\lambda$ is called the {\bf Specht ideal} of shape $\lambda$.
Specht ideals have been studied by several authors from several points of view (sometimes under other names).
See, for example, \cite{BGS,MW,MRV,SY2}.

A finite subset $G$ of $I_\lambda$ is called a \textbf{universal Gr\"obner basis},
if $G$ is a Gr\"obner basis of $I_\lambda$ with respect to any monomial order.
In their unpublished manuscript, Haiman and Woo found a   universal Gr\"obner basis of $I_\lambda$,
and Murai and the authors of the present paper gave a short proof of this result (\cite{OMY}).
See Theorem~\ref{HW} below.
In the present paper, we study the number of all possible initial ideals of $I_\lambda$.
By the above universal Gr\"obner basis, we see that the initial ideal $\operatorname{in}_\prec(I_\lambda)$ depends only the order of variables
$x_{\sigma(1)} \prec x_{\sigma(2)} \prec \cdots \prec x_{\sigma(n)}$ for $\sigma \in {\mathfrak S}_n$. The following is a main result of this paper.

\medskip

\noindent{\bf Theorem~\ref{n!/(m+1)!}}
{\it For a partition $\lambda=(\lambda_1, \ldots, \lambda_m)$ of $n$ with $\lambda_m>0$, set
$k:=\min \{ \, \lambda_{i-1}-\lambda_i \mid i=2,3, \ldots, m \, \}$.
Then $I_\lambda$ admits exactly $n!/(k+1)!$ distinct initial ideals under all possible monomial orders of $S$
{\rm (}equivalently, all $\sigma \in {\mathfrak S}_n${\rm )}.}

\medskip

The Gr\"obner fan of an ideal is introduced by Mora and Robbiano \cite{MoRo}.
Although there is a good software package {\tt Gfan} \cite{GfanSoft} for computing Gr\"obner fans,
the computation is very difficult in general.
On the other hand, the state polytope of a homogeneous ideal is
introduced by Bayer and Morrison \cite{BaMo}.
It is a dual of a Gr\"obner fan and
a generalization of the Newton polytope of a single homogeneous polynomial.

We give a brief introduction of these concepts.
Given a vector ${\mathbf w} \in {\mathbb R}^n$,
a partial order on the set of monomials in $S$ is
defined by $x_1^{a_1} \dots x_n^{a_n} > x_1^{b_1} \dots x_n^{b_n}$
if $(a_1,\dots, a_n) \cdot {\mathbf w} > (b_1,\dots,b_n) \cdot {\mathbf w}$.
For a homogeneous ideal $I \subset S$, let
$\operatorname{in}_{\mathbf w} (I)$ be the ideal generated by
the initial forms of polynomials in $I$ with respect to ${\mathbf w}$.
Clearly, $\operatorname{in}_{\mathbf w} (I)$ is not a monomial ideal in general.
However, for any monomial order $<$, the set
$
\{ {\mathbf w} \in {\mathbb R}^n \mid  \operatorname{in}_{\mathbf w} (I) = \operatorname{in}_< (I) \}
$
is nonempty, and forms an open convex polyhedral cone.
The \textbf{Gr\"obner fan} ${\rm GF} (I)$ of $I$ is a polyhedral complex generated by the
closures of these cones.
A convex polytope $P \subset {\mathbb R}^n$ is called a \textbf{state polytope} of $I$
if ${\rm GF} (I)$ is the normal fan of $P$. In particular, each initial ideal of $I$ corresponds to each vertex of the state polytope of $I$.
See, for example, \cite{CLO, Stu} for the details.

\medskip

\noindent{\bf Theorem~\ref{genpermuto} and Corollary~\ref{permuto}.}
{\it
With the same notation as the above theorem, the state polytope of $I_\lambda$ is a generalized permutohedron.
In particular,
 the state polytope is a (usual) permutohedron
 if and only if $\lambda_i=\lambda_{i-1}$ for some $i \le m$.
}

\section{Preliminaries}

\ytableausetup{centertableaux,boxsize=0.3em}
A {\bf partition} of a positive integer $n$ is a non-increasing sequence of non-negative  integers  $\lambda=(\lambda_1,\dots,\lambda_m)$ with $\lambda_1+ \cdots+\lambda_m=n$, but we identify $(\lambda_1,\dots,\lambda_m)$ with $(\lambda_1,\dots,\lambda_m,0)$.
Therefore we frequently assume that $\lambda_m >0$.
If $\lambda$ is a partition of $n$, then we write $\lambda \vdash n$.

 A partition $\lambda \vdash n$ is represented by its Young diagram.
 For example, $(4,2,1)$ is represented as $\ydiagram {4,2,1}$.
 A (\textbf{Young}) \textbf{tableau} of shape $\lambda$ is a bijective filling of the Young diagram of  $\lambda$
by the integers in $\{1,2, \ldots, n\}$.
For example,
$$
\ytableausetup{mathmode, boxsize=1.3em}
\begin{ytableau}
3 & 5 & 1& 7   \\
4 & 2    \\
6 \\
\end{ytableau}
$$
is a tableau of shape $(4,2,1)$.
The box in the $i$-th row and the $j$-th column has the coordinates $(i,j)$, as in a matrix. For example, in the above tableau, the box in the $(3,1)$ position is filled by the number 6.

Recall that the Specht polynomial $f_T$ of $T \in \tab \lambda$ is the product of all $x_i-x_j$ such that $i$ and $j$ are in the same column of $T$ (and $j$ is in a lower position than $i$).
For example, if $T$ is the above tableau,
then $f_T=(x_3-x_4)(x_3-x_6)(x_4-x_6)(x_5-x_2).$
In this paper, we study the Specht ideal
$$I_\lambda:=\langle \, f_T \mid T \in \tab \lambda \, \rangle \subset S$$
of $\lambda$.
If $\lambda=(n)$, then $I_\lambda$ is the trivial ideal $S$ itself.
Therefore, in the rest of this paper, we assume that $\lambda_2 >0$.

For partitions $\lambda=(\lambda_1,\dots,\lambda_m)$ and $\mu =(\mu_1,\dots,\mu_l)$ of $n$, we write $\mu \trianglelefteq \lambda$ if $\lambda$ is larger than or equal to $\mu$ with respect to the dominance order,
that is,
$$ \mu_1+ \cdots + \mu_k  \leq \lambda_1+ \cdots+\lambda_k
\ \ \ \mbox{ for all }k.
$$ By $\mu \vartriangleleft \lambda$, we mean that $\mu \trianglelefteq \lambda$ and
$\mu \ne \lambda$. Now we can introduce an unpublished result of Haiman and Woo.

\begin{theorem}[Haiman-Woo, c.f.~\cite{OMY}]\label{HW}
With the above situation,
$$\{ f_T \mid T \in \tab \mu, \mu \vdash n, \mu \trianglelefteq \lambda \}$$
forms a universal Gr\"obner basis of $I_\lambda$.
\end{theorem}

Let $\operatorname{in}_<(I_\lambda)$ be the initial ideal of $I_\lambda$ with respect to a monomial order $<$ on $S$.
 Then there exists a permutation $\sigma \in {\mathfrak S}_n$ such that $\operatorname{in}_< (I_\lambda)=\operatorname{in}_{\prec}(I_\lambda)$, where $\prec$ is the lexicographic order with $x_{\sigma(1)} \prec x_{\sigma(2)} \prec \cdots \prec x_{\sigma(n)}$, see the proof of \cite[Theorem~1.1]{OMY}.
 Unless otherwise specified, we fix $\sigma$ and use the  lexicographic order $\prec$.
 Therefore we simply denote $\operatorname{in}_\prec(I_\lambda)$ by $\ini{I_\lambda}$.

If $i$ and $j$ are in the same column of $T$,
we have $f_{\tau T} =-f_T$ for the transposition $\tau= (i \, j)$.
In this sense, to consider $f_T$, we may assume that $T$ is \textbf{column standard} (with respect to $\sigma$), that is, all columns are increasing from top to bottom with respect to the order $\sigma(1) \prec \sigma(2) \prec \cdots \prec \sigma(n)$.  If a column standard tableau $T$ is also row standard (i.e., all rows are increasing from left to right with respect to $\prec$), we say $T$ is \textbf{standard}. Let $\stab \lambda$ be the set of all standard tableaux of shape $\lambda$.  It is a classical result that $\{ f_T \mid T \in \stab \lambda \}$ is a basis of the vector space spanned by $\{ f_T \mid T \in \tab \lambda \}$.
Hence we have $I_\lambda= \langle \, f_T \mid T \in \stab \lambda \, \rangle.$
We also remark that, if $T$ is column standard and if the number $i$ is in the $d_i$-th row of $T$
for $i=1,2,\dots,n$, then we have $\ini {f_T}=\prod_{i=1}^n x_i^{d_i-1}$.
In particular, for column standard tableaux $T$ and $T'$,
$\ini {f_T} = \ini {f_{T'}} $ if and only if
$j$-th rows of $T$ and $T'$ have the same numbers for any $j$ (that is, $T$ and $T'$ are {\bf row equivalent}).

The next result follows from Theorem~\ref{HW} and \cite[Lemma~4.3.1]{L}. Note that the following set is still far from a minimal Gr\"obner basis in general.

\begin{corollary}[{\cite[Remark~3.8]{OMY}}]\label{pre minimal GB}
With the above situation,
$$\left\{ f_T \mid T \in \stab \mu,\, \mu \vdash n, \, \mu \trianglelefteq\lambda, \, \mu_1 =\lambda_1 \right\}$$
forms a Gr\"obner basis of $I_\lambda$ with respect to $\prec$.
\end{corollary}

\begin{lemma}\label{gap for mu}
Let $\lambda=(\lambda_1, \ldots, \lambda_m)$ be a partition of $n$ with $\lambda_m>0$, and set $k:=\min \{ \, \lambda_{i-1}-\lambda_i \mid i=2,3, \ldots, m \, \}$. If $\ini{f_T}$ for $T \in \stab \mu$ with $\mu \trianglelefteq \lambda$ is a minimal generator of $\ini{I_\lambda}$ and $\sigma(n)$ is in the $j$-th row of $T$, then either $j=1$ or $j\ge 2$ and $\mu_{j-1}-\mu_j \ge k$.
\end{lemma}

\begin{proof}
Assume that $j\ge 2$ and $(0\le ) \,\mu_{j-1}-\mu_j < k$.
By Lemma~\ref{pre minimal GB}, we have $\mu \vartriangleleft \lambda$ and $\mu_1 =\lambda_1$.
Assume that $j=2$.
Since $\mu_2 \le \lambda_2$, it follows that $k > \mu_1-\mu_2 =\lambda_1-\mu_2 \ge \lambda_1-\lambda_2\ge k$,
a contradiction. Thus we have $j \ge 3$.
Since $\mu_1 = \lambda_1 > \lambda_2 \ge \mu_2$ and $j-1 \ge 2$,
there exists $1 < l < j$ such that
$$
\mu_{l-1}>\mu_l =\mu_{l+1}=\cdots =\mu_{j-1}.
$$
Note that $\sum_{i=1}^{l-1} \mu_i \le \sum_{i=1}^{l-1} \lambda_i$.
Next we will show that
\begin{equation}\label{L J-1}
    \sum_{i=1}^s \mu_i < \sum_{i=1}^s \lambda_i \mbox{ for all } l \le s \le j-1.
\end{equation}

\noindent
First we prove it  for $l \le s \le j-2$ (when $l \le j-2$).
If $\sum_{i=1}^l \mu_i=\sum_{i=1}^l \lambda_i$,
then we have $\mu_l \ge \lambda_l$. Since $\lambda_l -  \lambda_{l+1} \ge k >0$ and $\mu_l=\mu_{l+1}$, we have $\mu_{l+1} > \lambda_{l+1}$.
Hence $\sum_{i=1}^{l+1} \mu_i > \sum_{i=1}^{l+1} \lambda_i$,  but it contradicts $\mu \vartriangleleft \lambda$.
Thus we have $\sum_{i=1}^l \mu_i < \sum_{i=1}^l \lambda_i$.
Similarly, for $l \le s \le j-2$, we have $\sum_{i=1}^s \mu_i < \sum_{i=1}^s \lambda_i$.
It remains to show that $\sum_{i=1}^{j-1} \mu_i < \sum_{i=1}^{j-1} \lambda_i$.
Assume the contrary, that is,  $\sum_{i=1}^{j-1} \mu_i = \sum_{i=1}^{j-1} \lambda_i$.
Since $\sum_{i=1}^{j-2} \mu_i \le \sum_{i=1}^{j-2} \lambda_i$ (unless $l=j-1$, the inequality is strict), we have $\mu_{j-1} \ge \lambda_{j-1}$. Moreover, since
$$\mu_{j-1}-\mu_j < k \le \lambda_{j-1}-\lambda_j\le\mu_{j-1}-\lambda_j,$$
we have $\mu_j > \lambda_j$ and hence $\sum_{i=1}^j \mu_i > \sum_{i=1}^j \lambda_i$. This is a contradiction.
Summing up, we have $\sum_{i=1}^s \mu_i < \sum_{i=1}^s \lambda_i$ for all $l \le s \le j-1$.

Since $\sigma(n)$ is in the $j$-th row of $T \in \stab \mu$, we have $\mu_j >\mu_{j+1}$.
We define the partition $\nu \vdash n$ by $\nu_l=\mu_l+1$, $\nu_j =\mu_j-1$, and $\nu_i=\mu_i$ for $i \ne l,j$ (since $\mu_{l-1}>\mu_l$ and $\mu_j >\mu_{j+1}$, $\nu$ is actually a partition).
By (\ref{L J-1}) above, we have $\mu \vartriangleleft \nu \trianglelefteq \lambda$. For the tableau $T$, lifting $\sigma(n)$ to the $l$-th row, we get a new tableau $T' \in \stab \nu$. Then $\ini{f_{T'}} \in \ini{I_\lambda}$ strictly divides $\ini{f_T}$, but it contradicts the assumption that $\ini{f_T}$ is a minimal generator of $\ini{I_\lambda}$.
\end{proof}

\begin{corollary}\label{cor of gap for mu}
With the same notation as Lemma~{\rm \ref{gap for mu}}, assume that $\sigma(n)$ is in (the right most box of) the $j$-th row
of $T \in \stab{\mu}$ with $j \ge 2$
and $\ini{f_T}$ is a minimal generator of $\ini{I_\lambda}$.
If the number just above $\sigma(n)$ (i.e., the one in the $(j-1, \mu_j)$ position) is $\sigma(i)$, then we have $i < n-k$.
\end{corollary}

\begin{proof}
Since $\mu_{j-1}-\mu_j \ge k$ by the lemma,
there are at least $k$ boxes in
the right of the box filled by $\sigma(i)$.
These boxes are filled by $\sigma(l)$ for $i < l <n$.
Since $\# \{ \sigma(l) \mid i < l <n\} =n-i-1 \ge k$,
we have $i \le n-k-1$.
\end{proof}

For a monomial ${\mathsf m} :=\prod_{i=1}^n x_i^{a_i} \in S$, set $\deg_i {\mathsf m}=a_i$.
For a partition $\lambda \vdash n$ and $1 \le i \le n$, set
$$d_\lambda(i):=\sum_{T \in \stab \lambda} \deg_i (\ini {f_T}).$$

\begin{lemma}\label{d_lam}
With the above notation,
we have $$d_\lambda(\sigma(i))\le d_\lambda(\sigma(i+1))$$ for all $1 \le i <n$. Moreover, the inequality is strict if and only if there exists $T \in \stab \lambda$ such that $\sigma(i)$ and $\sigma(i+1)$ are in the same column of $T$.
\end{lemma}

\begin{proof}
Take $T \in \stab \lambda$.
If $\sigma(i)$ and $\sigma(i+1)$ are in the same row of $T$, we have $\deg_{\sigma(i)}( \ini{f_T})= \deg_{\sigma(i+1)} (\ini{f_T})$.
Next, assume that  $\sigma(i)$ and $\sigma(i+1)$ are in different rows and different columns of $T$. For the transposition $\tau =(\sigma(i) \ \, \sigma(i+1))$, we have $\tau T \in \stab \lambda$, and
$$\deg_{\sigma(i)} (\ini{f_T})+ \deg_{\sigma(i)} (\ini{f_{\tau T}})= \deg_{\sigma(i+1)} (\ini{f_T})+ \deg_{\sigma(i+1)} (\ini{f_{\tau T}}).$$
Finally, if $\sigma(i)$ and $\sigma(i+1)$ are in the same column of $T$, we have $\deg_{\sigma(i)} (\ini{f_T}) < \deg_{\sigma(i+1)} (\ini{f_T})$.
\end{proof}

For a partition $\lambda=(\lambda_1, \ldots,\lambda_m) \vdash n$ with $\lambda_1 \ge 2$, we define ${\widehat{\lambda}}=({\widehat{\lambda}}_1, \ldots,{\widehat{\lambda}}_l) \vdash (n-1)$ inductively as follows. Set ${\widehat{\lambda}}_1 =\lambda_1-1$, and
$$
{\widehat{\lambda}}_i = \min \left \{ {\widehat{\lambda}}_{i-1}, \ \sum_{j=1}^i \lambda_j -  \sum_{j=1}^{i-1} {\widehat{\lambda}}_j-1 \right \}
$$
for $i \ge 2$. If $\lambda_2 < \lambda_1$, then ${\widehat{\lambda}}_1=\lambda_1-1$ and ${\widehat{\lambda}}_i= \lambda_i$ for $i \ge 2$. If $\lambda_1=\lambda_2=\cdots =\lambda_i$, then ${\widehat{\lambda}}_1={\widehat{\lambda}}_2=\cdots ={\widehat{\lambda}}_i=\lambda_1-1$.

\begin{lemma}\label{elimination}
For a partition $\lambda\vdash n$,
we have
$$I_{\widehat{\lambda}} = I_\lambda \cap K[x_{\sigma(1)}, \ldots, x_{\sigma(n-1)}].$$
\end{lemma}

\begin{proof}
Set $S':= K[x_{\sigma(1)}, \ldots, x_{\sigma(n-1)}]$ and
$J:=I_\lambda \cap S'$.
Since our order $\prec$ is a lexicographic order, it is an  elimination order, that is, $\ini{f_T} \in S'$ implies $f_T \in S'$. By \cite[Proposition~15.29]{E},
\begin{equation}\label{GB elimination}
    \{ \, f_T \mid T \in \stab \mu,\, \mu \vdash n, \, \mu \trianglelefteq\lambda, \,
    \ini{f_T} \in S' \, \}
\end{equation}
is a Gr\"obner basis of $J$.
Here, as the monomial order on $S'$, we use the restriction of $\prec$ to $S'$. Take $f_T$ for $T \in \stab \mu$ with $\mu \vdash n$ and  $\mu \trianglelefteq\lambda$.
Clearly, $\ini{f_T} \in S'$ if and only if $\sigma(n)$ is in the first row of $T$. If this is the case, we have $\mu_1 > \mu_2$. Moreover, removing $\sigma(n)$ from $T$, we get $T' \in \stab{\mu'}$ satisfying $\ini{f_T}=\ini{f_{T'}}$, where $\mu' \vdash (n-1)$ is given by $\mu'_1 =\mu_1 -1$ and $\mu'_i =\mu_i$ for $i \ge 2$. The partition ${\widehat{\lambda}}$ is the maximum element among these $\mu'$ in the dominance order.
Hence
$$ \{ \, f_{T'} \mid T' \in \stab \nu,\, \nu \vdash (n-1), \, \nu \trianglelefteq{\widehat{\lambda}}\,
\},$$
which is a Gr\"obner basis of $I_{\widehat{\lambda}}$,
coincides with the set \eqref{GB elimination}.
Thus we have $J=I_{\widehat{\lambda}}$.
\end{proof}

\section{The proofs of the main results}
The following proposition is just a special case of Theorem~\ref{n!/(m+1)!} below.
However, for better exposition, we prove it independently.
Later, in Theorem~\ref{n!/(m+1)!} below, we will see that the converse of the proposition also holds.

\begin{proposition}\label{n!}
If a partition $\lambda=(\lambda_1, \ldots, \lambda_m)$ of $n$ satisfies  $\lambda_{i-1}=\lambda_i>0$ for some $i$, considering all monomial orders of $S$
{\rm (}equivalently, considering all $\sigma \in {\mathfrak S}_n${\rm )},
the Specht ideal $I_\lambda$ admits $n!$ distinct initial ideals.
\end{proposition}

\begin{proof}
We prove the statement by induction on $n$.
It suffices to show that different $\sigma, \sigma' \in {\mathfrak S}_n$ give different initial ideals of $I_\lambda$, in other words, we can recover $\sigma$ from $\ini {I_\lambda}$.
Let $i= \max \{ j \mid \lambda_{j-1}=\lambda_j >0\}$.
Then there exists $T \in \stab \lambda$ such that $\sigma(n)$ (resp. $\sigma(n-1)$) is in the right most box of the $i$-th (resp. $(i-1)$-st) row, that is, in the $(i, \lambda_i)$ (resp. $(i-1, \lambda_i)$) position.
Since $\sigma(n)$ and $\sigma(n-1)$ are in the same column of $T$, we have $d_\lambda(\sigma(n))> d_\lambda(\sigma(n-1))$, and hence $d_\lambda(\sigma(n))> d_\lambda(\sigma(i))$ for all $1 \le i <n$ by Lemma~\ref{d_lam}.
Thus we can detect $\sigma(n)$.

As we have seen in the proof of   Lemma~\ref{elimination}, we have
$$\ini{I_{\widehat{\lambda}}} = \ini{I_\lambda} \cap K[x_{\sigma(1)}, \ldots, x_{\sigma(n-1)}],$$
where we use the lexicographic order with $x_{\sigma(1)} \prec x_{\sigma(2)} \prec \cdots \prec x_{\sigma(n-1)}$  as the monomial order in $K[x_{\sigma(1)}, \ldots, x_{\sigma(n-1)}]$.
If $\lambda_1>\lambda_2$, then we have $i\ge 3$ and ${\widehat{\lambda}}_{i-1}=\lambda_{i-1}=\lambda_i={\widehat{\lambda}}_i$.
If $\lambda_1=\lambda_2$, then ${\widehat{\lambda}}_1={\widehat{\lambda}}_2$.
Hence ${\widehat{\lambda}}$ always satisfies the assumption of the proposition.
By induction hypothesis, we can detect $\sigma(1),\ldots, \sigma(n-1)$ from $\ini{I_{\widehat{\lambda}}}$ (hence, from $\ini{I_\lambda}$).
\end{proof}

\begin{example}
If $\lambda=(2,2)$, $I_\lambda$ admits $4!$ distinct initial ideals, and we can recover the permutation $\sigma$ from $\ini {I_\lambda}$ by Proposition~\ref{n!}. However, it is easy to see that
$$d_\lambda(\sigma(2))= d_\lambda(\sigma(3))= 1,$$
in other words, $d_\lambda(-)$ does not distinguish $\sigma(2)$ from $\sigma(3)$.
Hence we have to consider $\mu$ with $\mu \vartriangleleft \lambda$.
\end{example}

We are in a position to give the number of distinct initial ideals of Specht ideals.

\begin{theorem}\label{n!/(m+1)!}
Let $\lambda=(\lambda_1, \ldots, \lambda_m)$ be a partition of $n$ with $\lambda_m>0$, and set
$k:=\min \{ \, \lambda_{i-1}-\lambda_i \mid i=2,3, \ldots, m \, \}$.
Then the Specht ideal $I_\lambda$ admits exactly $n!/(k+1)!$ distinct initial ideals under all monomial orders of $S$ {\rm (}equivalently, all $\sigma \in {\mathfrak S}_n${\rm )}.
\end{theorem}

\begin{proof}
By Proposition~\ref{n!},
we may assume $k>0$, and hence $\lambda_1 > \cdots > \lambda_m$.
Take $p$ with $k=\lambda_{p-1} -\lambda_p$.
To the Young diagram of $\lambda$,
we put $\sigma(n-k)$ in the right most box of the $p$-th row (i.e., in the $(p, \lambda_p)$ position),
and $\sigma(n-k-1)$ just above it (i.e., in the $(p-1, \lambda_p)$ position).
In the right of the box filled by $\sigma(n-k-1)$,
there are  $k$ boxes, and we fill them by  $\sigma(n-k+1), \sigma(n-k+2), \ldots, \sigma(n)$
in the suitable order.
Filling the remaining boxes in a suitable way, we get
$T \in \stab{\lambda}$ such that  $\sigma(n-k-1)$ and $\sigma(n-k)$ are in the same column of $T$.
Thus $d_\lambda(\sigma(n-k-1)) < d_\lambda(\sigma(n-k))$, and hence we can detect the sets
$A_1 := \{ \sigma(1), \sigma(2), \ldots, \sigma(n-k-1) \}$
and
$A_2:=\{ \sigma(n-k), \sigma(n-k+1), \ldots, \sigma(n) \}$
from $\ini{I_\lambda}$.
Since $\sharp A_2 = k+1$,  it is enough to show that
\begin{itemize}
    \item[(a)]
we cannot distinguish $\sigma(n-k), \sigma(n-k+1), \ldots, \sigma(n)$ by $\ini{I_\lambda}$,
and
    \item[(b)]
we can detect $\sigma(1), \sigma(2), \ldots, \sigma(n-k-1)$ from $\ini{I_\lambda}$.
\end{itemize}

\smallskip

\noindent
(a)
It suffices to show the following statement.

\begin{itemize}
\item [($*$)] If some two elements of $A_2$ are in the same column of  some  $T \in \stab{\mu}$ with $\mu \trianglelefteq \lambda$, then $\ini{f_T}$ is not a minimal generator of $\ini{I_\lambda}$.
\end{itemize}
The proof of $(*)$ is by induction on $n$.
Assume that $\ini{f_T}$ with $T \in \stab{\mu}$ is a minimal generator of $\ini{I_\lambda}$ and some two elements $\sigma(\alpha), \sigma(\beta)
\in A_2$
are in the same column of $T$.
By Corollary~\ref{cor of gap for mu},
we have $n\neq \alpha, \beta$.
Assume that $\sigma(n)$ is in the $j$-th row of $T$.
Removing $\sigma(n)$ from $T$, we have a standard tableau $T'$ of shape $\mu' \vdash (n-1)$ with $\mu'_j=\mu_j -1$ and $\mu'_i=\mu_i$ for $i \ne j$.
Since $\lambda_j >\lambda_{j+1}$,
we can define the partition $\lambda' \vdash (n-1)$ by $\lambda'_j=\lambda_j-1$ and $\lambda'_i=\lambda_i$ for $i \ne j$.
Clearly, $\mu' \trianglelefteq \lambda'$.

Since $k':=\min \{ \, \lambda'_{i-1}-\lambda'_i \mid i=2,3, \ldots, m \, \} \ge k-1$ and
$$
\sigma(\alpha), \sigma(\beta)
\in
A_2 \setminus \{ \sigma(n)\}=\{\sigma(n-k), \ldots, \sigma(n-1)\}
\subset \{\sigma((n-1)-k'), \ldots, \sigma(n-1)\}
,$$ we can apply the induction hypothesis, that is, the condition $(*)$ for $T'$ and $\ini{I_{\lambda'}}$.
Hence $\ini{f_{T'}}$ is not a minimal generator of $\ini{I_{\lambda'}}$.
In other words, there exists $\nu' \vdash (n-1)$ with $\nu' \trianglelefteq \lambda'$, and $T_1' \in \stab{\nu'}$ such that $\ini{f_{T'_1}}$ strictly divides $\ini{f_{T_1}}$.
Since $\sum_{i=1}^j v'_i \le \sum_{i=1}^j \lambda'_i < \sum_{i=1}^j \lambda_i$,
there exists $l \le j$ such that the sequence $\nu=(\nu_1, \nu_2, \ldots)$ given by $\nu_l=\nu'_l+1$ and $\nu_i=\nu'_i$ for $i \ne l$ is a partition of $n$ satisfying $\nu \trianglelefteq \lambda$. Adding $\sigma(n)$ to the $l$-th row of $T_1'$, we get a standard tableau $T_1 \in \stab{\nu}$.
Clearly, $\ini{f_{T_1}} \in \ini{I_\lambda}$ strictly divides $\ini{f_T}$, but it contradicts the assumption that $\ini{f_T}$ is a minimal generator of $\ini{I_\lambda}$.

\smallskip

\noindent
(b)
Set
$
l:= \lambda_1 - \lambda_2$ ($ \ge k$).

\smallskip

\noindent
\textbf{Case 1} ($l >k$)\textbf{.}
First, we will show that
\begin{equation}\label{n-l to n-k}
d_\lambda(\sigma(n-l)) < d_\lambda(\sigma(n-l+1)) < \cdots < d_\lambda(\sigma(n-k-1)).
\end{equation}
It suffices to show that, for $j$ with $n-l < j < n-k$, there exists $T \in \stab{\lambda}$ such that  $\sigma(j-1)$ and $\sigma(j)$ are in the same column of $T$.
Take $p$ with $k=\lambda_{p-1} -\lambda_p$.
Then $p\ge 3$.
To the Young diagram of $\lambda$, we put $\sigma(j)$ in the right most box of the $p$-th row (i.e., in the $(p, \lambda_p)$ position), and $\sigma(j-1)$ just above it (i.e., in the $(p-1, \lambda_p)$ position). In the right of the box filled by $\sigma(j-1)$, there are  $k\, (=\lambda_{p-1} -\lambda_p)$ boxes, and we fill them by  $\sigma(j+1), \sigma(j+2), \ldots, \sigma(j+k)$ in the  suitable order. Next, we fill the boxes in the first row from the right most one by $\sigma(j+k+1), \sigma(j+k+2), \ldots, \sigma(n)$ in the suitable order. Since
$$\# \{\sigma(j+k+1), \sigma(j+k+2), \ldots, \sigma(n) \} =n-j-k < l,$$
these numbers are contained in the ``peninsula" part of the first row.
Filling the remaining boxes in a suitable way, we get a desired standard tableau.

By \eqref{n-l to n-k}, we can detect the set $\{ \sigma(1), \sigma(2), \ldots, \sigma(n-l) \}$ from $\ini{I_\lambda}$.
Hence we can take the ideal
$I_\lambda \cap K[x_{\sigma(1)}, x_{\sigma(2)}, \ldots, x_{\sigma(n-l)}]$, which equals $I_\mu$ by the repeated use of Lemma~\ref{elimination}.
Here $\mu$ is the partition of $(n-l)$ given by $\mu_1=\lambda_1-l=\lambda_2$ and $\mu_i=\lambda_i$ for $i \ge 2$.
Since $\mu_1=\mu_2$, we can detect $\sigma(1), \ldots, \sigma(n-l)$ from $\ini{I_\mu}$ (hence, from $\ini{I_\lambda}$) by Proposition~\ref{n!}.
Combining with $\eqref{n-l to n-k}$, we can detect $\sigma(1), \sigma(2), \ldots, \sigma(n-k-1)$ from $\ini{I_\lambda}$.

\smallskip

\noindent
\textbf{Case 2} ($l  = k$)\textbf{.}
Recall that we can detect the set
$A_1 = \{ \sigma(1), \sigma(2), \ldots, \sigma(n-k-1) \}$.
As we have seen in (a),
we cannot detect $\sigma(n-k)$ from $A_2$,
in other words, the variables $x_{\sigma(n-k)}, \dots, x_{\sigma(n)}$ appear
in the initial ideal $\ini{I_\lambda}$ in the same way.
Take any $r \in A_2$, and consider
$$I_\mu =I_\lambda \cap K[x_{\sigma(1)}, x_{\sigma(2)}, \ldots, x_{\sigma(n-k-1)}, x_r],$$
where $\mu \vdash (n-l)$ is the partition given in Case 1.
By Proposition~\ref{n!}, we can detect $\sigma(1), \sigma(2), \ldots, \sigma(n-k-1)$ from $\ini{I_\lambda}$.
\end{proof}

Let $\{0\} \ne I \subset K[x_1,\ldots,x_n]$ be a homogeneous ideal.
Given a vector ${\mathbf w} \in {\mathbb R}^n$,
let $\operatorname{in}_{\mathbf w} (I)$ denote the initial form ideal of $I$ with respect to ${\mathbf w}$.
Note that $\operatorname{in}_{\mathbf w} (I)$ is not necessarily a monomial ideal.
For example, if ${\mathbf w} = {\bf 0}$, then  $\operatorname{in}_{\mathbf w} (I)=I$.
Since $I$ is homogeneous, for any vector ${\mathbf w}_1 \in {\mathbb R}^n$,
there exists a nonnegative vector ${\mathbf w}_2 \in {\mathbb R}^n$ such that
$\operatorname{in}_{{\mathbf w}_1} (I) = \operatorname{in}_{{\mathbf w}_2} (I)$.
Given a vector ${\mathbf w} \in {\mathbb R}^n$, let
$$
C[{\mathbf w}]:= \{ {\mathbf w}' \in {\mathbb R}^n \mid  \operatorname{in}_{\mathbf w} (I) = \operatorname{in}_{{\mathbf w}'} (I) \}.
$$
In general, $C[{\mathbf w}]$ is a relatively open convex polyhedral cone (\cite[Proposition~2.3]{Stu}).
A \textbf{fan} is a polyhedral complex consisting of the cones from the origin.
Let
$$
{\rm GF}(I) :=
\left\{\left.
\overline{C[{\mathbf w}]} \ \right| \  {\mathbf w} \in {\mathbb R}^n
\right\},
$$
where $\overline{C[{\mathbf w}]}$ is the closure of $C[{\mathbf w}]$.
Then ${\rm GF}(I)$ is a fan, and called the \textbf{Gr\"obner fan} of $I$.
Note that ${\rm GF}(I)$ is \textbf{complete}, i.e.,
$$\bigcup_{C \in {\rm GF}(I)} C  = {\mathbb R}^n.$$
The \textbf{normal fan} of a polytope $P \subset {\mathbb R}^n$ is a fan
that is dual to $P$.
A convex polytope $P \subset {\mathbb R}^n$ is called a \textbf{state polytope} of $I$
if ${\rm GF} (I)$ is the normal fan of $P$.
There is a one to one correspondence between the initial ideals of $I$ and the vertices of the state polytope of $I$.

Given a vector $(u_1,\dots,u_n) \in {\mathbb R}^n$
let $P_n (u_1,\dots,u_n)$ be the convex hull of the set
$$
\{
(u_{\sigma(1)}, u_{\sigma(2)}, \dots, u_{\sigma(n)})
\in {\mathbb R}^n \mid \sigma \in {\mathfrak S}_n
\}
.
$$
In particular, $\Pi_n :=P_n(1,2,\dots,n)$ is called  the \textbf{permutohedron}
of order $n$.
It is known that
the normal fan of the permutohedron $\Pi_n$ is
the \textbf{braid fan} ${\rm Br}_n$ that is the complete fan in ${\mathbb R}^n$ given by the
hyperplanes $x_i - x_j = 0$ for all $i\ne j $.
Each maximal cone of ${\rm Br}_n$ is of the form
$$
\{ {\mathbf w} \in {\mathbb R}^n \mid w_{\sigma(1)} \le w_{\sigma(2)}  \le \cdots \le
w_{\sigma(n)}
\}
$$
for some $\sigma \in {\mathfrak S}_n $.
See \cite[Section 3.2]{PRW} for details.
For $0 \le k < n$, let
$$\Pi_{n,k} := P_n(1,2,\ldots,n-k-1, n-k,\dots,n-k).$$
Then each maximal cone of the normal fan of $\Pi_{n,k}$ is of the form
$$
C_{\sigma,k} :=
\{ {\mathbf w} \in {\mathbb R}^n \mid w_{\sigma(1)} \le w_{\sigma(2)} \le \cdots \le w_{\sigma(n-k)},\dots,w_{\sigma(n)}
\}
$$
for some $\sigma \in {\mathfrak S}_n$.
A \textbf{generalized permutohedron} \cite{Pos}
is a polytope
obtained by moving the vertices of
a permutohedron while keeping the same edge directions.

\begin{proposition}[{\cite{PRW}}]
\label{gp}
A polytope $P\subset {\mathbb R}^n$ is
a generalized permutohedron if and only if the normal fan of $P$ is
refined by the braid fan ${\rm Br}_n$.
\end{proposition}

Since ${\rm Br}_n$ refines the normal fan of $\Pi_{n,k}$ for all $0 \le k <n$,
by Proposition~\ref{gp},
each $\Pi_{n,k}$ is a generalized permutohedron.

\begin{theorem}\label{genpermuto}
Let $\lambda=(\lambda_1, \ldots, \lambda_m)$ be a partition of $n$ with $\lambda_m>0$, and set
$k=\min \{ \, \lambda_{i-1}-\lambda_i \mid i=2,3, \ldots, m \, \}$.
Then the generalized permutohedron $\Pi_{n,k}$ is a state polytope of $I_\lambda$.
In particular, ${\rm GF}(I_\lambda)$ is the normal fan of $\Pi_{n,k}$, and
refined by the braid fan ${\rm Br}_n$.
\end{theorem}

\begin{proof}
From Proof of Theorem~\ref{n!/(m+1)!},
$I_\lambda$ admits exactly $n!/(k+1)!$ distinct initial ideals, and
monomial orders for each initial ideal correspond to the cone
$$
\{ {\mathbf w} \in {\mathbb R}^n \mid w_{\sigma(1)} < w_{\sigma(2)}  < \cdots < w_{\sigma(n-k)} , \dots ,w_{\sigma(n)}
\}
$$
for some $\sigma \in {\mathfrak S}_n$.
Since its closure is $C_{\sigma,k}$, it follows that
 ${\rm GF}(I_\lambda)$ is the normal fan of $\Pi_{n,k}$
as desired.
\end{proof}

\begin{remark}
In the situation of Theorem~\ref{genpermuto}, the largest  possible value of $k$ is $n-2$, which occurs when $\lambda=(n-1,1)$ (we are assuming that $\lambda \ne (n)$).
In this case, the state polytope $\Pi_{n,2}$ is an $(n-1)$-simplex. We also remark that the dimension of $\Pi_{n,k}$ is always $n-1$ for all $0 \le k \le n-2$.
\end{remark}

\begin{corollary}\label{permuto}
Let $\lambda=(\lambda_1, \ldots, \lambda_m)$ be a partition of $n$ with $\lambda_m>0$.
If $\lambda_{i-1}=\lambda_i>0$ for some $i$, then
the permutohedron $\Pi_n$ of order $n$ is a state polytope of $I_\lambda$,
and ${\rm GF}(I_\lambda)$ is the braid fan ${\rm Br}_n$.
\end{corollary}

\noindent
\textbf{Acknowledgments}:
The authors are grateful to professors Yasuhide Numata, Satoshi Murai, Ryota Okazaki, Kosuke Shibata, Akihito Wachi and Junzo Watanabe
for the useful discussions
at the MFO-RIMS Tandem Workshop ``Symmetries on polynomial ideals and varieties".
In particular, Lemma~\ref{elimination} was given in the discussion.
The first author is partially supported by KAKENHI 18H01134.
The second author is partially supported by KAKENHI 19K03456.

redpajama_set_name: RedPajamaArXiv


text:
Q: editing using regular expressions So I have a string and I also have a regular Expression for that string. I simply want to add 8 to one of the numbers in that string. I Would like to know how I can change the value of a group and then return the match object to a string.
here is a example of what I want to do.
m = re.match(r"(\w+) (\w+)", "Isaac Newton, physicist")
m.group(1) = 'john'
//then some how return the value back to a string

so the string will be "john Newton, physicist"

A: You need to use re.sub, rather than re.match, and use the captured groups from the match in your repl. For example:
>>> import re
>>> re.sub(r"(\w+) (\w+)", r"john \2", "Isaac Newton, physicist")
'john Newton, physicist'

Here r"john \2" means "replace the match with the word 'john' and the second captured group from the pattern".

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