numpy.take和numpy.choose有什么区别？

Question

似乎numpy.take(array, indices)和numpy.choose(indices, array)返回相同的内容：由array索引的indices的子集。

这两者之间是否存在微妙的差异，还是我错过了一些更重要的东西？是否有理由偏爱另一个？

Answer 1

numpy.take(array, indices)和numpy.choose(indices, array)在1-D阵列上表现相似，但这只是巧合。正如jonrsharpe所指出的，它们在高维数组上表现不同。

numpy.take

numpy.take(array, indices)从展平版array中挑选出元素。 （结果元素当然不一定来自同一行。）

例如，

numpy.take([[1, 2], [3, 4]], [0, 3])

返回

array([1, 4])

numpy.choose

numpy.choose(indices, set_of_arrays)从数组indices[0]中取出元素0，从数组indices[1]中取出元素1，从数组indices[2]中取出元素2，依此类推。 （这里，array实际上是一组数组。）

例如

numpy.choose([0, 1, 0, 0], [[1, 2, 3, 4], [4, 5, 6, 7]])

返回

array([1, 5, 3, 4])

因为元素0来自数组0，元素1来自数组1，元素2来自数组0，元素3来自数组0。

更多信息

这些说明已经过简化 - 完整说明可在此处找到：numpy.take，numpy.choose。例如，当numpy.take和numpy.choose为1-D时，indices和array的行为相似，因为numpy.choose首先广播array。

Answer 2

它们当然不是等价的，正如你可以通过给两个方法提供相同的参数（切换）看到的那样：

>>> a = np.array([[1, 2, 3, 4], 
                  [5, 6, 7, 8], 
                  [9, 10, 11, 12], 
                  [13, 14, 15, 16]])
>>> np.choose([0, 2, 1, 3], a)
array([ 1, 10,  7, 16]) # one from each row
>>> np.take(a, [0, 2, 1, 3])
array([1, 3, 2, 4]) # all from same row

我建议您阅读take和choose上的文档。

Answer 3

文档中对 np.choose() 行为的解释。

• numpy.choose(a, choices, out=None, mode='raise')[source]

np.choose(a,c) == np.array([c[a[I]][I] for I in ndi.ndindex(a.shape)])

索引数组和选项

import numpy as np

print("--------------------------------------------------------------------------------")
print("Index array 'a' and array of choice(s).")
print("--------------------------------------------------------------------------------")
a = np.array([4, 3, 2, 1])
print("'a' is {} shape is {}\n".format(
    a, a.shape
))
    
c = np.arange(60).reshape((5, 3, 4))
_choice = c[0]
print("An example choice is \n{} \nshape is {}\n".format(
    _choice,
    _choice.shape
))

--------------------------------------------------------------------------------
Index array 'a' and array of choice(s).
--------------------------------------------------------------------------------
'a' is [4 3 2 1] shape is (4,)

An example choice is 
[[ 0  1  2  3]
 [ 4  5  6  7]
 [ 8  9 10 11]] 
shape is (3, 4)

广播

<块引用>

给定一个由整数组成的“索引”数组 (a) 和一个由 n 个数组组成的序列 (choices), a 和每个选择数组首先广播，必要时， 到共同形状的数组；调用这些 Ba 和 Bchoices[i], i = 0,...,n-1 我们有，必然，Ba.shape == Bchoices[i].shape 为 每个我。

Ba.shape == Bchoices[i].shape。

print("--------------------------------------------------------------------------------")
print("np.choose() broadcasts 'a' and 'choice' to match (here only on a)")
print("--------------------------------------------------------------------------------")
a = np.vstack([
    np.array(a) for i in range(_choice.shape[0])
])
print("Broadcast shape of 'a' is {}\n 'a' is \n{}\n.".format(a.shape, a))

--------------------------------------------------------------------------------
np.choose() broadcasts 'a' and 'choice' to match (here only on a)
--------------------------------------------------------------------------------
Broadcast shape of 'a' is (3, 4)
 'a' is 
[[4 3 2 1]
 [4 3 2 1]
 [4 3 2 1]]

从选项中选择值

c[a[I]][I] for I in ndi.ndindex(a.shape)

<块引用>

假设 i（在那个范围内）是 (j0, j1, …, jm) 处的值 Ba 中的位置 - 然后是新数组中相同位置的值 是 Bchoices[i] 中相同位置的值；

print("--------------------------------------------------------------------------------")
print("Simulate np.choose() behavior of choosing elements c[a[I]][I]]]:")
print("np.array([ c[a[I]][I] for I in np.lib.index_tricks.ndindex(a.shape) ])")
print("--------------------------------------------------------------------------------")
result = np.array([]).astype(int)
for I in np.lib.index_tricks.ndindex(a.shape):
    print("Selecting the element {} at c[{},{}]".format(
        c[a[I]][I].astype(int), a[I], I
    ))
    result = np.concatenate([result, [c[a[I]][I].astype(int)]])
    print("chosen items: {}".format(result))
    
print("\nResult is \n{}\n".format(
    result.reshape(a.shape)
))

--------------------------------------------------------------------------------
Simulate np.choose() behavior of choosing elements c[a[I]][I]]]:
np.array([ c[a[I]][I] for I in np.lib.index_tricks.ndindex(a.shape) ])
--------------------------------------------------------------------------------
Selecting the element 48 at c[4,(0, 0)]
chosen items: [48]
Selecting the element 37 at c[3,(0, 1)]
chosen items: [48 37]
Selecting the element 26 at c[2,(0, 2)]
chosen items: [48 37 26]
Selecting the element 15 at c[1,(0, 3)]
chosen items: [48 37 26 15]
Selecting the element 52 at c[4,(1, 0)]
chosen items: [48 37 26 15 52]
Selecting the element 41 at c[3,(1, 1)]
chosen items: [48 37 26 15 52 41]
Selecting the element 30 at c[2,(1, 2)]
chosen items: [48 37 26 15 52 41 30]
Selecting the element 19 at c[1,(1, 3)]
chosen items: [48 37 26 15 52 41 30 19]
Selecting the element 56 at c[4,(2, 0)]
chosen items: [48 37 26 15 52 41 30 19 56]
Selecting the element 45 at c[3,(2, 1)]
chosen items: [48 37 26 15 52 41 30 19 56 45]
Selecting the element 34 at c[2,(2, 2)]
chosen items: [48 37 26 15 52 41 30 19 56 45 34]
Selecting the element 23 at c[1,(2, 3)]
chosen items: [48 37 26 15 52 41 30 19 56 45 34 23]

Result is 
[[48 37 26 15]
 [52 41 30 19]
 [56 45 34 23]]

np.choose(a,c)

print("--------------------------------------------------------------------------------")
print("Run np.choose(a, c):")
print("--------------------------------------------------------------------------------")
print(np.choose(a, c))

--------------------------------------------------------------------------------
Run np.choose(a, c):
--------------------------------------------------------------------------------
[[48 37 26 15]
 [52 41 30 19]
 [56 45 34 23]]