如何删除 numpy.array 中的列

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This creates another array without those columns:

  b = a.compress(logical_not(z), axis=1)

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Another way is to use masked arrays:

import numpy as np
a = np.array([[ np.nan,   2.,   3., np.nan], [  1.,   2.,   3., 9]])
print(a)
# [[ NaN   2.   3.  NaN]
#  [  1.   2.   3.   9.]]

The np.ma.masked_invalid method returns a masked array with nans and infs masked out:

print(np.ma.masked_invalid(a))
[[-- 2.0 3.0 --]
[1.0 2.0 3.0 9.0]]

The np.ma.compress_cols method returns a 2-D array with any column containing a masked value suppressed:

a=np.ma.compress_cols(np.ma.masked_invalid(a))
print(a)
# [[ 2.  3.]
#  [ 2.  3.]]

manipulating-a-maskedarray

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最佳答案

Given its name, I think the standard way should be delete:

import numpy as np


A = np.delete(A, 1, 0)  # delete second row of A
B = np.delete(B, 2, 0)  # delete third row of B
C = np.delete(C, 1, 1)  # delete second column of C

According to numpy's documentation page, the parameters for numpy.delete are as follow:

numpy.delete(arr, obj, axis=None)

arr refers to the input array,
obj refers to which sub-arrays (e.g. column/row no. or slice of the array) and
axis refers to either column wise (axis = 1) or row-wise (axis = 0) delete operation.

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Example from the numpy documentation:

>>> a = numpy.array([[ 0,  1,  2,  3],
[ 4,  5,  6,  7],
[ 8,  9, 10, 11],
[12, 13, 14, 15]])


>>> numpy.delete(a, numpy.s_[1:3], axis=0)                       # remove rows 1 and 2


array([[ 0,  1,  2,  3],
[12, 13, 14, 15]])


>>> numpy.delete(a, numpy.s_[1:3], axis=1)                       # remove columns 1 and 2


array([[ 0,  3],
[ 4,  7],
[ 8, 11],
[12, 15]])

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In your situation, you can extract the desired data with:

a[:, -z]

"-z" is the logical negation of the boolean array "z". This is the same as:

a[:, logical_not(z)]

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From Numpy Documentation

np.delete(arr, obj, axis=None) Return a new array with sub-arrays along an axis deleted.

>>> arr
array([[ 1,  2,  3,  4],
[ 5,  6,  7,  8],
[ 9, 10, 11, 12]])
>>> np.delete(arr, 1, 0)
array([[ 1,  2,  3,  4],
[ 9, 10, 11, 12]])


>>> np.delete(arr, np.s_[::2], 1)
array([[ 2,  4],
[ 6,  8],
[10, 12]])
>>> np.delete(arr, [1,3,5], None)
array([ 1,  3,  5,  7,  8,  9, 10, 11, 12])

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>>> A = array([[ 1,  2,  3,  4],
[ 5,  6,  7,  8],
[ 9, 10, 11, 12]])


>>> A = A.transpose()


>>> A = A[1:].transpose()

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Removing Matrix columns that contain NaN. This is a lengthy answer, but hopefully easy to follow.

def column_to_vector(matrix, i):
return [row[i] for row in matrix]
import numpy
def remove_NaN_columns(matrix):
import scipy
import math
from numpy import column_stack, vstack


columns = A.shape[1]
#print("columns", columns)
result = []
skip_column = True
for column in range(0, columns):
vector = column_to_vector(A, column)
skip_column = False
for value in vector:
# print(column, vector, value, math.isnan(value) )
if math.isnan(value):
skip_column = True
if skip_column == False:
result.append(vector)
return column_stack(result)


### test it
A = vstack(([ float('NaN'), 2., 3., float('NaN')], [ 1., 2., 3., 9]))
print("A shape", A.shape, "\n", A)
B = remove_NaN_columns(A)
print("B shape", B.shape, "\n", B)


A shape (2, 4)
[[ nan   2.   3.  nan]
[  1.   2.   3.   9.]]
B shape (2, 2)
[[ 2.  3.]
[ 2.  3.]]