Estimate normal distribution from dnorm in R The function dnorm(x) in R gives you the value of the probability density function in the points x of a certain normal distribution (mean = 0 and SD = 1 by default), returning a vector of the same length than x.
However, I want to do the opposite: given a vector that approximates a probability density function (like the result of dnorm), I want to get the mean and standard deviation of the normal distribution represented by that given probability density.
Code example of what I would like to do:
pdf = dnorm(seq(-3,3,0.1), mean = 0, sd=1)
## get_normal would be supposes to return an list/vector containing the mean and SD, which in this particular case should be close to 0 and 1 respectively.
var_parameters = get_normal(pdf)


For a normal density function $f,$ if you have a grid of points X and corresponding density values $y = f(x),$
then you can use numerical integration to find $\mu$ and $\sigma.$ [See Note (2) at the end.]
If you have many realizations $X_i$ from the distribution, you can estimate the population mean $\mu$ by the sample mean $\bar X$ and the population SD $\sigma$ by the sample SD $S.$
Another possibility is to use a kernel density estimator (KDE) of $f$ based on a sufficiently large
sample. In R the procedure density gives points $(x, y)$ that can be used to plot a density estimator.
set.seed(718)
x = rnorm(100, 50, 7)
mean(x);  sd(x)
[1] 50.62287
[1] 6.443036

hist(x, prob=T, col="skyblue2")
 rug(x);  lines(density(x), col="red")


In R, the KDE consists of 512 points with values summarized as below:
density(x)

Call:
        density.default(x = x)

Data: x (100 obs.);     Bandwidth 'bw' = 2.309

       x               y            
 Min.   :31.36   Min.   :1.974e-05  
 1st Qu.:41.69   1st Qu.:3.239e-03  
 Median :52.03   Median :2.371e-02  
 Mean   :52.03   Mean   :2.417e-02  
 3rd Qu.:62.36   3rd Qu.:4.378e-02  
 Max.   :72.70   Max.   :5.566e-02  

You can estimate $\mu$ and $\sigma$ corresponding to the KDE as follows:
xx = density(x)$x
yy = density(x)$y              # (xx, yy) is KDE plot point
sum(xx*yy)/sum(yy)
[1] 50.62329                   # aprx pop mean = 50
sum((xx-50.62)^2 * yy)/sum(yy)
[1] 46.42294                   # aprx pop variance = 49
sqrt(sum((xx-50.62)^2 * yy)/sum(yy))
[1] 6.813438                   # aprx pop SD = 7

Because $\bar X$ and $S$ are sufficient statistics for $\mu$ and $\sigma,$
it is hard to imagine that $\hat \mu$ and $\hat \sigma$ re-claimed from
a KDE (based on data) would be systematically better than the sample mean $\bar X = 50.62$ and SD $S = 6.44.$
I mention the KDE method because it seems possibly related to your question.
Notes: (1) Of course there are also methods for estimating $\bar X$ and $S$ from a histogram, but they can be very inaccurate for small samples.
(2) Here is a numerical evaluation of $\mu \approx \int_0^{100} x\varphi(x,50,7)\, dx,$
using the sum of areas of 1000 rectangles.
m = 1000
w = (100-0)/m
x = seq(0+w/2, 100-w/2, len=m) 
f = x*dnorm(x, 50, 7)
sum(w*f)
[1] 50   # mu
f2 = (x-50)^2*dnorm(x,50,7)
sum(w*f2)
[1] 49   # sigma^2

