PERF: use c-division in nancorr

[MarcoGorelli]

• closes #xxxx (Replace xxxx with the Github issue number)
• Tests added and passed if fixing a bug or adding a new feature
• All code checks passed.
• Added type annotations to new arguments/methods/functions.
• Added an entry in the latest doc/source/whatsnew/vX.X.X.rst file if fixing a bug or adding a new feature.

There's three divisions in this function that can safely be done at the C-level:

    meanx += 1 / nobs * dx
    meany += 1 / nobs * dy

because nobs is at least 1 when we get here (it starts at 0, nobs+=1 is called a few lines
above, and nobs can't decrease), and

    result[xi, yi] = result[yi, xi] = covxy / divisor

because divisor is checked to not be equal to 0 in the preceding line.

This makes a noticeable difference, example:

arr1 = np.random.randn(100, 100)

On main:

%%timeit
nancorr(arr1)

5.39 ms ± 21.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

On this branch:

%%timeit
nancorr(arr1)

3.86 ms ± 58.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

I didn't find much difference in nancorr_spearman, I think because that one calls
the Python function rank_1d within the loop, so that might be a bigger bottleneck there

[MarcoGorelli]

nvm, looks like this changes results (failing tests) - not sure I see why, will look into it, but closing for now

EDIT ok might have fixed this now, just a matter of marking the numerator in the division as float

running the same timing test as above now gives:

main:

5.68 ms ± 123 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

here:

4.97 ms ± 60.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

[jreback]

thanks @MarcoGorelli