FFT of a 2d array with NAN
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I'm trying to apply FFT on a 2D array with NAN. I don't want to replace the NAN values with anything. Is there a way to apply FFT and ignore the NAN values?
arr=np.array([[1, 2, 3], [4, NAN, 6], [1, 2, NAN]])
f=np.fft.fft2(arr)
2d fft nan
edited Nov 24 '18 at 22:44
SleuthEye
10.8k22046
asked Nov 23 '18 at 13:22
NedoNedo
1
add a comment |
I'm trying to apply FFT on a 2D array with NAN. I don't want to replace the NAN values with anything. Is there a way to apply FFT and ignore the NAN values?
arr=np.array([[1, 2, 3], [4, NAN, 6], [1, 2, NAN]])
f=np.fft.fft2(arr)
2d fft nan
edited Nov 24 '18 at 22:44
SleuthEye
10.8k22046
asked Nov 23 '18 at 13:22
NedoNedo
1
How would you expect an FFT that "ignores NAN" to behave? As if those were zeros? Also, any reason why you don't want to replace the NAN values? Do you simply want to preserve your source, or do you have memory constraints that would make a temporary copy where NAN are replaced impractical?
– SleuthEye
Nov 24 '18 at 2:55
add a comment |
I'm trying to apply FFT on a 2D array with NAN. I don't want to replace the NAN values with anything. Is there a way to apply FFT and ignore the NAN values?
arr=np.array([[1, 2, 3], [4, NAN, 6], [1, 2, NAN]])
f=np.fft.fft2(arr)
2d fft nan
edited Nov 24 '18 at 22:44
SleuthEye
10.8k22046
asked Nov 23 '18 at 13:22
NedoNedo
1
I'm trying to apply FFT on a 2D array with NAN. I don't want to replace the NAN values with anything. Is there a way to apply FFT and ignore the NAN values?
arr=np.array([[1, 2, 3], [4, NAN, 6], [1, 2, NAN]])
f=np.fft.fft2(arr)
2d fft nan
2d fft nan
edited Nov 24 '18 at 22:44
SleuthEye
10.8k22046
asked Nov 23 '18 at 13:22
NedoNedo
1
edited Nov 24 '18 at 22:44
SleuthEye
10.8k22046
asked Nov 23 '18 at 13:22
NedoNedo
1
edited Nov 24 '18 at 22:44
SleuthEye
10.8k22046
edited Nov 24 '18 at 22:44
SleuthEye
10.8k22046
edited Nov 24 '18 at 22:44
SleuthEye
10.8k22046
10.8k22046
asked Nov 23 '18 at 13:22
NedoNedo
1
asked Nov 23 '18 at 13:22
NedoNedo
1
asked Nov 23 '18 at 13:22
NedoNedo
1
1
How would you expect an FFT that "ignores NAN" to behave? As if those were zeros? Also, any reason why you don't want to replace the NAN values? Do you simply want to preserve your source, or do you have memory constraints that would make a temporary copy where NAN are replaced impractical?
– SleuthEye
Nov 24 '18 at 2:55
add a comment |
How would you expect an FFT that "ignores NAN" to behave? As if those were zeros? Also, any reason why you don't want to replace the NAN values? Do you simply want to preserve your source, or do you have memory constraints that would make a temporary copy where NAN are replaced impractical?
– SleuthEye
Nov 24 '18 at 2:55
How would you expect an FFT that "ignores NAN" to behave? As if those were zeros? Also, any reason why you don't want to replace the NAN values? Do you simply want to preserve your source, or do you have memory constraints that would make a temporary copy where NAN are replaced impractical?
– SleuthEye
Nov 24 '18 at 2:55
How would you expect an FFT that "ignores NAN" to behave? As if those were zeros? Also, any reason why you don't want to replace the NAN values? Do you simply want to preserve your source, or do you have memory constraints that would make a temporary copy where NAN are replaced impractical?
– SleuthEye
Nov 24 '18 at 2:55
add a comment |
1 Answer
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oldest
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No. An FFT will propagate any unknown input to every output. This because in a transform such as a DFT, every output value directly depends on every input value, and thus can not be computed without known values for every input.
Or, an alternate answer is Yes: If any input to an FFT is an NaN, just output the FFT result as an appropriately sized array completely filled with NaNs.
answered Nov 25 '18 at 19:42
hotpaw2hotpaw2
61.9k1072132
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
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votes
No. An FFT will propagate any unknown input to every output. This because in a transform such as a DFT, every output value directly depends on every input value, and thus can not be computed without known values for every input.
Or, an alternate answer is Yes: If any input to an FFT is an NaN, just output the FFT result as an appropriately sized array completely filled with NaNs.
answered Nov 25 '18 at 19:42
hotpaw2hotpaw2
61.9k1072132
add a comment |
No. An FFT will propagate any unknown input to every output. This because in a transform such as a DFT, every output value directly depends on every input value, and thus can not be computed without known values for every input.
Or, an alternate answer is Yes: If any input to an FFT is an NaN, just output the FFT result as an appropriately sized array completely filled with NaNs.
answered Nov 25 '18 at 19:42
hotpaw2hotpaw2
61.9k1072132
add a comment |
No. An FFT will propagate any unknown input to every output. This because in a transform such as a DFT, every output value directly depends on every input value, and thus can not be computed without known values for every input.
Or, an alternate answer is Yes: If any input to an FFT is an NaN, just output the FFT result as an appropriately sized array completely filled with NaNs.
answered Nov 25 '18 at 19:42
hotpaw2hotpaw2
61.9k1072132
No. An FFT will propagate any unknown input to every output. This because in a transform such as a DFT, every output value directly depends on every input value, and thus can not be computed without known values for every input.
Or, an alternate answer is Yes: If any input to an FFT is an NaN, just output the FFT result as an appropriately sized array completely filled with NaNs.
answered Nov 25 '18 at 19:42
hotpaw2hotpaw2
61.9k1072132
edited Nov 25 '18 at 19:48
answered Nov 25 '18 at 19:42
hotpaw2hotpaw2
61.9k1072132
answered Nov 25 '18 at 19:42
hotpaw2hotpaw2
61.9k1072132
answered Nov 25 '18 at 19:42
hotpaw2hotpaw2
61.9k1072132
61.9k1072132
add a comment |
add a comment |
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How would you expect an FFT that "ignores NAN" to behave? As if those were zeros? Also, any reason why you don't want to replace the NAN values? Do you simply want to preserve your source, or do you have memory constraints that would make a temporary copy where NAN are replaced impractical?
– SleuthEye
Nov 24 '18 at 2:55