numpy的：对于一个阵列的每个元素，找到另一个数组索引 [英] Numpy: For every element in one array, find the index in another array

问题描述

我有两个一维数组，X放大器; Y，其中一个比另一个小。我试图找到在x和y的每一个元素的索引。

I have two 1D arrays, x & y, one smaller than the other. I'm trying to find the index of every element of y in x.

我已经发现了两个天真的方式来做到这一点，首先是缓慢的，而第二个内存密集型。

I've found two naive ways to do this, the first is slow, and the second memory-intensive.

indices= []
for iy in y:
    indices += np.where(x==iy)[0][0]

的内存猪

xe = np.outer([1,]*len(x), y)
ye = np.outer(x, [1,]*len(y))
junk, indices = np.where(np.equal(xe, ye))

有一个更快的方法或更少的内存密集型的方法呢？理想情况下，搜索将充分利用这一事实，我们在一个列表搜索不是一回事，但很多事情，因而稍微更适合并行化。
奖励积分，如果你不认为y的每一个元素实际上是x中。

Is there a faster way or less memory intensive approach? Ideally the search would take advantage of the fact that we are searching for not one thing in a list, but many things, and thus is slightly more amenable to parallelization. Bonus points if you don't assume that every element of y is actually in x.

推荐答案

乔金顿说，的 searchsorted（）可以非常快速地搜索元素。为了解决这个问题不是x中的元素，你可以检查与原始y的搜索结果中，并创建一个蒙面的数组：

As Joe Kington said, searchsorted() can search element very quickly. To deal with elements that are not in x, you can check the searched result with original y, and create a masked array:

import numpy as np
x = np.array([3,5,7,1,9,8,6,6])
y = np.array([2,1,5,10,100,6])

index = np.argsort(x)
sorted_x = x[index]
sorted_index = np.searchsorted(sorted_x, y)

yindex = np.take(index, sorted_index, mode="clip")
mask = x[yindex] != y

result = np.ma.array(yindex, mask=mask)
print result

结果是：

[-- 3 1 -- -- 6]

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