Delete columns from numpy array depending on a con

问题:

Above all, sorry for my bad English.

I have this array t:

array([[ 0,  1,  2,  0,  4,  5,  6,  7,  8,  9],
       [ 0, 11,  0, 13,  0, 15,  0, 17, 18,  0]])

I would like to delete the columns where the value of second line is null. Here, I would like to delete the columns 0, 2, 4, 6 and 9, to obtain this array:

array([[  1,   0,   5,   7,  8 ],
       [ 11,  13,  15,  17, 18 ]])

I tried with np.sum() but didn't succeed.

回答1:

Similar to Juh_, but more expressive, and avoiding some minor unnecessary performance overhead. A grand total of 12 highly pythonic, explicit and unambigious characters. This is really numpy 101; if you are still trying to wrap your head around this, you would do yourself a favor by reading a numpy primer.

import numpy as np
a = np.array([[ 0,  1,  2,  0,  4,  5,  6,  7,  8,  9],
              [ 0, 11,  0, 13,  0, 15,  0, 17, 18,  0]])
print a[:,a[1]!=0]

回答2:

With numpy.delete:

a = np.array([[0, 1, 2, 0, 4, 5, 6, 7, 8, 9], [0, 11, 0, 13, 0, 15, 0, 17, 18, 0]])

indices = [i for (i,v) in enumerate(a[1]) if v==0]
# [0, 2, 4, 6, 9]

a = np.delete(a, indices, 1)
# array([[ 1,  0,  5,  7,  8], [11, 13, 15, 17, 18]])

回答3:

Simple (fully numpy) solution:

import numpy as np

t = np.array([[ 0, 1, 2, 0, 4, 5, 6, 7, 8, 9], [ 0, 11, 0, 13, 0, 15, 0, 17, 18, 0]])
indices_to_keep = t[1].nonzero()[0]

print t[:,indices_to_keep]
# [[ 1  0  5  7  8]
#  [11 13 15 17 18]]

回答4:

Using np.where:

>>> t.T[np.where(t[1])].T
array([[ 1,  0,  5,  7,  8],
       [11, 13, 15, 17, 18]])

回答5:

I got this working like this:

data = array([[ 0, 1, 2, 0, 4, 5, 6, 7, 8, 9], [ 0, 11, 0, 13, 0, 15, 0, 17, 18, 0]])
res = array([(a, b,) for a, b in zip(data[0], data[1]) if b]).transpose()

got the result

In [23]: res
Out[23]: 
array([[ 1,  0,  5,  7,  8],
       [11, 13, 15, 17, 18]])