Cleanest way to replace np.array value with np.nan by user defined index
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Track title: CC B Schuberts Piano Sonata No 16 D
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Chapters
00:00 Question
01:31 Accepted answer (Score 4)
02:20 Answer 2 (Score 2)
03:02 Thank you
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Full question
https://stackoverflow.com/questions/3626...
Answer 1 links:
[np.logical_and]: http://docs.scipy.org/doc/numpy-1.10.0/r...
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https://meta.stackexchange.com/help/lice...
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Tags
#python #arrays
#avk47
ACCEPTED ANSWER
Score 4
You are nearly there.
You can pass arrays of indices to arrays. You probably know this with 1D-arrays.
With a 2D-array you need to pass the array a tuple of lists (one tuple for each axis; one element in the lists (which have to be of equal length) for each array-element you want to chose). You have a list of tuples. So you have just to "transpose" it.
t1 = zip(*t)
gives you the right shape of your index array; which you can now use as index for any assignment, for example: value[t1] = np.NaN
(There are lots of nice explanation of this trick (with zip and *) in python tutorials, if you don't know it yet.)
ANSWER 2
Score 2
You can use np.logical_and
arr = np.zeros((20,20))
You can select by location, this is just an example location.
arr[4:8,4:8] = 1
You can create a mask the same shape as arr
mask = np.ones((20,20)).astype(bool)
Then you can use the np.logical_and.
mask = np.logical_and(mask, arr == 1)
And finally, you can replace the 1s with the np.nan
arr[mask] = np.nan