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I have the following list:

indices
>>> [21, 43, 58, 64, 88, 104, 113, 115, 120]

I want every occurrence of these values in this list -1 (so 20, 42, 57, etc.) to be zeroed out from a 3D array 'q' I have.

I have tried list comprehensions, for and if loops (see below), but I always get the following error:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

I haven't been able to resolve this.

Any help would be amazing!

>>> for b in q:
...     for u in indices:
...         if b==u:
...             b==0


>>> for u in indices:
...     q = [0 if x==u else x for x in q]
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4
  • Do you want them deleted or just zeroed out? Why isn't you comparison x == u-1? Commented Jul 23, 2017 at 15:24
  • If q is a 3D array, you need one more nested for loop Commented Jul 23, 2017 at 15:25
  • Are you working with numpy? Commented Jul 23, 2017 at 15:25
  • Yes, I am working with numpy. I'd like to zero out every value in indices. trincot, could you explain? Commented Jul 23, 2017 at 15:27

3 Answers 3

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0

I think this is a short and efficient way:

b= b*np.logical_not(np.reshape(np.in1d(b,indices),b.shape))

with np.in1d() we have a boolean array with True where the element in b is in indices. We reshape it to be the as b and then negate, so that we have False (or, if you want, 0) where we want to zero b. Just multiply this matrix element wise with b and you got it

It has the advantage that it works for 1D, 2D, 3D, ... arrays

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1 Comment

Fantastic: inventive, simple and efficient.
0

How about this?

indices = range(1, 10)
>>[1, 2, 3, 4, 5, 6, 7, 8, 9]

q = np.arange(12).reshape(2,2,3)
array([[[ 0,  1,  2],
        [ 3,  4,  5]],

       [[ 6,  7,  8],
        [ 9, 10, 11]]])

def zeroed(row):
    new_indices = map(lambda x: x-1, indices)
    nrow = [0 if elem in new_indices else elem for elem in row]
    return now

np.apply_along_axis(zeroed, 1, q)

array([[[ 0,  0,  0],
        [ 0,  0,  0]],

       [[ 0,  0,  0],
        [ 9, 10, 11]]])
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Comments

0

I tried this and it worked for me:

>>> arr_2D = [3,4,5,6]
>>> arr_3D = [[3,4,5,6],[2,3,4,5],[4,5,6,7,8,8]]
>>> for el in arr_2D:
...    for x in arr_3D:
...       for y in x:
...          if y == el - 1:
...             x.remove(y)
... 
>>> arr_3D
[[6], [], [6, 7, 8, 8]]

Doing it with list comprehensions seams like it might be overkill in this situation.

Or to zero out instead of remove

>>> for el in arr_2D:
...    for x in range(len(arr_3D)):
...       for y in range(len(arr_3D[x])):
...           if arr_3D[x][y] == el - 1:
...               arr_3D[x][y] = 0
... 
>>> arr_3D
[[0, 0, 0, 6], [0, 0, 0, 0], [0, 0, 6, 7, 8, 8]]

Here is the list comprehension:

zero_out = lambda arr_2D, arr_3D: [[0 if x in [el-1 for el in arr_2D] else x for x in y] for y in arr_3D]
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4 Comments

what version of python are you running?
perhaps it is another part of your code that is causing the issue? have you tried running it straight into the interpreter?
Python 3.6.0; yes, everything is processed straight into the interpreter. Problem was solved with J. Torrecilla's answer, but still not sure why this will not work. But as I understood, numpy is causing this ambiguity problem
I was running python 2.7 in this example

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