Numpy数组在屏蔽时丢失尺寸

Question

我想选择数组的某些元素,并根据值执行加权平均计算.但是,使用过滤条件会破坏数组的原始结构.arr它的形状(2, 2, 3, 2)变成了一维的阵列.这对我来说毫无用处,因为并非所有这些元素都需要在以后相互组合(但是它们的子阵列).我怎样才能避免这种扁平化呢？

>>> arr = np.asarray([ [[[1, 11], [2, 22], [3, 33]], [[4, 44], [5, 55], [6, 66]]], [ [[7, 77], [8, 88], [9, 99]], [[0, 32], [1, 33], [2, 34] ]] ])
>>> arr
array([[[[ 1, 11],
         [ 2, 22],
         [ 3, 33]],

        [[ 4, 44],
         [ 5, 55],
         [ 6, 66]]],


       [[[ 7, 77],
         [ 8, 88],
         [ 9, 99]],

        [[ 0, 32],
         [ 1, 33],
         [ 2, 34]]]])
>>> arr.shape
(2, 2, 3, 2)
>>> arr[arr>3]
array([11, 22, 33,  4, 44,  5, 55,  6, 66,  7, 77,  8, 88,  9, 99, 32, 33,
       34])
>>> arr[arr>3].shape
(18,)

Answer 1

查看 numpy.where

http://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html

要保持相同的维度,您需要填充值.在下面的示例中,我使用0,但您也可以使用np.nan

np.where(arr>3, arr, 0)

回报

array([[[[ 0, 11],
         [ 0, 22],
         [ 0, 33]],

        [[ 4, 44],
         [ 5, 55],
         [ 6, 66]]],


       [[[ 7, 77],
         [ 8, 88],
         [ 9, 99]],

        [[ 0, 32],
         [ 0, 33],
         [ 0, 34]]]])

Answer 2

您可以考虑使用a np.ma.masked_array来表示满足条件的元素子集:

import numpy as np

arr = np.asarray([[[[1, 11], [2, 22], [3, 33]],
                   [[4, 44], [5, 55], [6, 66]]],
                  [[[7, 77], [8, 88], [9, 99]],
                   [[0, 32], [1, 33], [2, 34]]]])

masked_arr = np.ma.masked_less(arr, 3)

print(masked_arr)
# [[[[-- 11]
#    [-- 22]
#    [3 33]]

#   [[4 44]
#    [5 55]
#    [6 66]]]


#  [[[7 77]
#    [8 88]
#    [9 99]]

#   [[-- 32]
#    [-- 33]
#    [-- 34]]]]

如您所见,蒙版数组保留其原始尺寸.您可以分别通过.data和.mask属性访问基础数据和掩码.大多数numpy函数都不会考虑屏蔽值,例如:

# mean of whole array
print(arr.mean())
# 26.75

# mean of non-masked elements only
print(masked_arr.mean())
# 33.4736842105

对掩码数组和非掩码数组进行逐元素操作的结果也将保留掩码的值:

masked_arrsum = masked_arr + np.random.randn(*arr.shape)

print(masked_arrsum)
# [[[[-- 11.359989067421582]
#    [-- 23.249092437269162]
#    [3.326111354088174 32.679132708120726]]

#   [[4.289134334263137 43.38559221094378]
#    [6.028063054523145 53.5043991898567]
#    [7.44695154979811 65.56890530368757]]]


#  [[[8.45692625294376 77.36860675985407]
#    [5.915835159196378 87.28574554110307]
#    [8.251106168209688 98.7621940026713]]

#   [[-- 33.24398289945855]
#    [-- 33.411941757624284]
#    [-- 34.964817895873715]]]]

总和仅计算在非屏蔽值上masked_arr- 您可以通过查看以下内容来查看masked_sum.data:

print(masked_sum.data)
# [[[[  1.          11.35998907]
#    [  2.          23.24909244]
#    [  3.32611135  32.67913271]]

#   [[  4.28913433  43.38559221]
#    [  6.02806305  53.50439919]
#    [  7.44695155  65.5689053 ]]]


#  [[[  8.45692625  77.36860676]
#    [  5.91583516  87.28574554]
#    [  8.25110617  98.762194  ]]

#   [[  0.          33.2439829 ]
#    [  1.          33.41194176]
#    [  2.          34.9648179 ]]]]

Answer 3

看看arr>3：

In [71]: arr>3
Out[71]: 
array([[[[False,  True],
         [False,  True],
         [False,  True]],

        [[ True,  True],
         [ True,  True],
         [ True,  True]]],


       [[[ True,  True],
         [ True,  True],
         [ True,  True]],

        [[False,  True],
         [False,  True],
         [False,  True]]]], dtype=bool)

arr[arr>3]选择掩码所在的元素True。您希望该选择具有什么样的结构或形状？平面是唯一有意义的东西，不是吗？ arr本身没有改变。

你可以将不适合掩码的项归零，

In [84]: arr1=arr.copy()
In [85]: arr1[arr<=3]=0
In [86]: arr1
Out[86]: 
array([[[[ 0, 11],
         [ 0, 22],
         [ 0, 33]],

        [[ 4, 44],
         [ 5, 55],
         [ 6, 66]]],


       [[[ 7, 77],
         [ 8, 88],
         [ 9, 99]],

        [[ 0, 32],
         [ 0, 33],
         [ 0, 34]]]])

现在，您可以对各个维度进行权重求和或求平均值。

np.nonzero(或np.where) 也可能有用，为您提供所选术语的索引：

In [88]: np.nonzero(arr>3)
Out[88]: 
(array([0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1]),
 array([0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1]),
 array([0, 1, 2, 0, 0, 1, 1, 2, 2, 0, 0, 1, 1, 2, 2, 0, 1, 2]),
 array([1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1]))