题
我想删除 numpy.array 中选定的列。这就是我所做的:
n [397]: a = array([[ NaN, 2., 3., NaN],
.....: [ 1., 2., 3., 9]])
In [398]: print a
[[ NaN 2. 3. NaN]
[ 1. 2. 3. 9.]]
In [399]: z = any(isnan(a), axis=0)
In [400]: print z
[ True False False True]
In [401]: delete(a, z, axis = 1)
Out[401]:
array([[ 3., NaN],
[ 3., 9.]])
在此示例中,我的目标是删除所有包含 NaN 的列。我希望最后一个命令会导致:
array([[2., 3.],
[2., 3.]])
我怎样才能做到这一点?
解决方案
鉴于其名称,我认为标准方法应该是 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
根据 numpy 的文档页面, ,参数为 numpy.delete
如下:
numpy.delete(arr, obj, axis=None)
arr
指的是输入数组,obj
指的是哪些子数组(例如列/行号或数组的切片)和axis
指的是任一列(axis = 1
) 或逐行 (axis = 0
) 删除操作。
其他提示
来自 numpy文档的示例:
>>> 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]])
另一种方法是使用蒙面数组:
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.]]
np.ma.masked_invalid方法返回一个掩码数组,其中nans和infs被掩盖:
print(np.ma.masked_invalid(a))
[[-- 2.0 3.0 --]
[1.0 2.0 3.0 9.0]]
np.ma.compress_cols方法返回一个二维数组,其中任何列都包含一个 屏蔽值被抑制:
a=np.ma.compress_cols(np.ma.masked_invalid(a))
print(a)
# [[ 2. 3.]
# [ 2. 3.]]
请参阅 操纵蒙面阵列
这将创建另一个没有这些列的数组:
b = a.compress(logical_not(z), axis=1)
来自 Numpy文档
np.delete(arr,obj,axis = None) 沿着删除的轴返回一个包含子数组的新数组。
>>> 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])
在您的情况下,您可以使用以下内容提取所需数据:
a[:, -z]
QUOT <!>; <!> -z QUOT;是布尔数组<!>“z <!>”的逻辑否定。这与:
相同a[:, logical_not(z)]
>>> A = array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
>>> A = A.transpose()
>>> A = A[1:].transpose()
删除包含NaN的Matrix列。 这是一个冗长的答案,但希望很容易理解。
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.]]
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