<p>The required shape of the tensor <span class="math-container">$T$</span> depends on the shape of other tensors that are involved in the same operations of that same tensor <span class="math-container">$T$</span> and the required/desired shape of the resulting tensor, in the same way that <a href="https://en.wikipedia.org/wiki/Matrix_multiplication" rel="nofollow noreferrer">the number of columns of the matrix <span class="math-container">$M \in \mathbb{R}^{n \times m}$</span> needs to match the number of rows of the matrix <span class="math-container">$M' \in \mathbb{R}^{n' \times m'}$</span> when you perform the matrix multiplication <span class="math-container">$M M'$</span>, i.e. in order for <span class="math-container">$M M'$</span> to be well-defined, <span class="math-container">$m = n'$</span></a>.</p>
<p>So, in your case, although the tensors contain the same elements, it might not be possible to use both in the same operations. Without more context/details, I cannot specifically answer why the first tensor was used in the tutorial you're mentioning.</p>

CC BY-SA 4.0
2021-04-16T13:25:20.160Z
2021-04-16T13:25:20.160Z
nbro
<p>I'm doing a TensorFlow tutorial, where they convert an array of the numbers <code>[1,2,3]</code> to a tensor like this:</p>
<pre><code>const xs = tf.tensor2d([1, 2, 3], [3, 1])
</code></pre>
<p>The shape is <code>[3,1]</code> because there is one row with 3 numbers.
My question is, why would they use a 2D tensor, isn't this just exactly the same as:</p>
<pre><code>const xs = tf.tensor1d([1, 2, 3])
</code></pre>

CC BY-SA 4.0
2021-04-16T09:53:49.690Z
This seems to be just a programming question/issue. Am I right? If yes, then you should ask this question on Stack Overflow, as programming questions are generally off-topic here. Please, if you have some time, take a look at our [on-topic page](https://ai.stackexchange.com/help/on-topic), where we describe what kind of question you can ask here.
nbro
CC BY-SA 4.0
2021-04-16T10:17:26.433Z
I posted it here because it's not really about code but about the difference in 1d or 2d tensors. Unless that is also considered code?
Kokodoko
CC BY-SA 4.0
2021-04-16T10:53:31.787Z
To give a proper answer, I think we will need more context, i.e. please provide the link to the tutorial where the first version of the code is used.
nbro
<machine-learning><tensorflow><math><linear-algebra>
What's the difference between a 1d tensor and a 2d tensor with 1 dimension?
Kokodoko
# What's the difference between a 1d tensor and a 2d tensor with 1 dimension?

(Asked by: username_0 on 2021-04-16)

I'm doing a TensorFlow tutorial, where they convert an array of the numbers [1,2,3] to a tensor like this:
const xs = tf.tensor2d([1, 2, 3], [3, 1])

The shape is [3,1] because there is one row with 3 numbers.
My question is, why would they use a 2D tensor, isn't this just exactly the same as:
const xs = tf.tensor1d([1, 2, 3])

**Tags:** machine-learning, tensorflow, math, linear-algebra

> **Comments:**
> **username_1**: This seems to be just a programming question/issue. Am I right? If yes, then you should ask this question on Stack Overflow, as programming questions are generally off-topic here. Please, if you have some time, take a look at our [on-topic page](https://ai.stackexchange.com/help/on-topic), where we describe what kind of question you can ask here.  
> **username_0**: I posted it here because it's not really about code but about the difference in 1d or 2d tensors. Unless that is also considered code?  
> **username_1**: To give a proper answer, I think we will need more context, i.e. please provide the link to the tutorial where the first version of the code is used.  

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## 1 answers

### Answer by username_1 on 2021-04-16. Score: 1 | Accepted answer

The required shape of the tensor $T$ depends on the shape of other tensors that are involved in the same operations of that same tensor $T$ and the required/desired shape of the resulting tensor, in the same way that [the number of columns of the matrix $M \in \mathbb{R}^{n \times m}$ needs to match the number of rows of the matrix $M' \in \mathbb{R}^{n' \times m'}$ when you perform the matrix multiplication $M M'$, i.e. in order for $M M'$ to be well-defined, $m = n'$](https://en.wikipedia.org/wiki/Matrix_multiplication).

So, in your case, although the tensors contain the same elements, it might not be possible to use both in the same operations. Without more context/details, I cannot specifically answer why the first tensor was used in the tutorial you're mentioning.


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