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saulopo opened this issue May 6, 2025 · 4 comments
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DOC: indices on numpy.fft.ifft #28914

saulopo opened this issue May 6, 2025 · 4 comments
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@saulopo
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saulopo commented May 6, 2025

Issue with current documentation:

https://numpy.org/doc/stable/reference/generated/numpy.fft.ifft.html#numpy.fft.ifft

In the description "The input should be ordered in the same way as is returned by fft, i.e.,

  • a[0] should contain the zero frequency term,
  • a[1:n//2] should contain the positive-frequency terms,
  • a[n//2 + 1:] should contain the negative-frequency terms, in increasing order starting from the most negative frequency",

shouldn't "a[n//2 + 1:]" be "a[n//2:]" ? The former item (a[1:n//2]) ranges from 1 to n//2-1.

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@eendebakpt
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I think you are right. Would you like to open a PR to fix this?

@saulopo
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saulopo commented May 8, 2025

Hmm maybe not a good idea as I am not skilled with it.. but I can see the tutorial when I have a chance

@vtavana
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vtavana commented May 9, 2025 •
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@saulopo The next line in the doc says that "For an even number of input points, A[n//2] represents the sum of ..." so the missing element you mentioned is explained separately
It is explained here in more detail: https://numpy.org/doc/stable/reference/routines.fft.html#:~:text=The%20values%20in,largest%20negative%20frequency.

However, I believe the correct way to write it is

* ``a[0]`` ...,
* ``a[1:(n+1)//2]`` ..., (This one should be modified to be correct with both `n` even or odd 
* ``a[n//2 + 1:]`` ...

For an even number of input points, ``a[n//2]`` ...

For example

import numpy as np
n=5   # odd
a = np.arange(n)
a[1:(n+1)//2]
# array([1, 2])
a[n//2 + 1:]
# array([3, 4])

n=6  # even
a = np.arange(n)
a[1:(n+1)//2]
# array([1, 2])
a[n//2 + 1:]
# array([4, 5])
a[n//2]
# np.int64(3)

@saulopo
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saulopo commented May 10, 2025

Thanks @vtavana , I see what you mean.

@saulopo saulopo closed this as completed May 10, 2025
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