以另一个顺序排列一个numpy数组 [英] Order one numpy array by another

问题描述

我有一个确定元素顺序的数组:

I have an array that determines an ordering of elements:

order = [3, 1, 4, 2]

然后我想对另一个更大的数组(仅包含那些元素)进行排序:

And then I want to sort another, larger array (containing only those elements):

a = np.array([4, 2, 1, 1, 4, 3, 1, 3])    

，使得order中最先出现的元素在结果中排​​名第一，等等.
在直接的Python中，我可以通过一个关键函数来做到这一点:

such that the element(s) that come first in order come first in the results, etc.
In straight Python, I would do this with a key function:

sorted(a, key=order.index)
[3, 3, 1, 1, 1, 4, 4, 2]

我怎样才能用numpy做到这一点? numpy数组是否有类似的键函数"概念?

How can I do this (efficiently) with numpy? Is there a similar notion of "key function" for numpy arrays?

推荐答案

特定情况:Ints

对于ints，我们可以使用 bincount -

Specific case : Ints

For ints, we could use bincount -

np.repeat(order,np.bincount(a)[order])

样品运行-

In [146]: sorted(a, key=order.index)
Out[146]: [3, 3, 1, 1, 1, 4, 4, 2]

In [147]: np.repeat(order,np.bincount(a)[order])
Out[147]: array([3, 3, 1, 1, 1, 4, 4, 2])

一般情况

方法1

使用bincount-

# https://stackoverflow.com/a/41242285/ @Andras Deak
def argsort_unique(idx):
    n = idx.size
    sidx = np.empty(n,dtype=int)
    sidx[idx] = np.arange(n)
    return sidx

sidx = np.argsort(order)
c = np.bincount(np.searchsorted(order,a,sorter=sidx))
out = np.repeat(order, c[argsort_unique(sidx)])

方法#2-A

对于order中的所有元素都在a-

With np.unique and searchsorted for the case when all elements from order are in a -

unq, count = np.unique(a, return_counts=True)
out = np.repeat(order, count[np.searchsorted(unq, order)])

方法#2-B

要涵盖所有情况，我们需要再执行一步-

To cover for all cases, we need one extra step -

unq, count = np.unique(a, return_counts=1)
sidx = np.searchsorted(unq, order)
out = np.repeat(order, np.where(unq[sidx] == order,count[sidx],0))

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