现有重复索引时，numpy会保持分配顺序吗? [英] Will numpy keep order of assignment when existing duplicated indexes?

问题描述

我想按索引数组对数组进行赋值，但是索引重复.

I want to make an assignment to an array by index array, but there are duplicated indexes.

例如:

a = np.arange(5)
index = np.array([1,2,3,1,2,3,1,2,3])
b = np.arange(9)
a[index] = b

两个问题:

1. 对于重复的索引，最新的分配是否总是生效?

1. For duplicated indexes, does the latest assignment always take effect?

在任何情况下都为a[1] == 6，例如对于非常大的数组a?可以a[1] == 0或3吗?

Is a[1] == 6 true for any case, e.g. for very large array a? Is it possible a[1] == 0 or 3?

更具体地说，我使用了用MKL(Anaconda提供)编译的numpy，一些数组操作是并行的.

More specifically, I used numpy compiled with MKL (provided by Anaconda), some array operation are in parallel.

相关文章:处理NumPy分配中的重复索引

如果上面的回答是否，是否有什么可以确保分配始终保持顺序?

If the answer above is no, is there anything can make sure that the assignment always keep in order?

推荐答案

这里是一种保证从相同索引的组中分配到最后一个索引的方法-

Here's one approach to guarantee the assignment into the last indices from the group of identical indices -

# Get sorting indices for index keeping the order with 'mergesort' option
sidx = index.argsort(kind='mergesort')

# Get sorted index array
sindex = index[sidx]

# Get the last indices from each group of identical indices in sorted version
idx = sidx[np.r_[np.flatnonzero(sindex[1:] != sindex[:-1]), index.size-1]]

# Use those last group indices to select indices off index and b to assign
a[index[idx]] = b[idx]

样品运行-

In [141]: a
Out[141]: array([0, 1, 2, 3, 4])

In [142]: index
Out[142]: array([1, 2, 3, 1, 2, 1, 2, 3, 4, 2])

In [143]: b
Out[143]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

In [144]: sidx = index.argsort(kind='mergesort')
     ...: sindex = index[sidx]
     ...: idx = sidx[np.r_[np.flatnonzero(sindex[1:] != sindex[:-1]), index.size-1]]
     ...: a[index[idx]] = b[idx]
     ...: 

In [145]: a
Out[145]: array([0, 5, 9, 7, 8])

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