Python NumPy 数组中满足值和索引条件的元素的索引
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Indexes of elements in NumPy array that satisfy conditions on the value and the index
提问by user3329302
I have a NumPy array, A. I want to know the indexes of the elements in A equal to a value and which indexes satisfy some condition:
我有一个 NumPy 数组，A. 我想知道 A 中元素的索引等于一个值以及哪些索引满足某些条件：
import numpy as np
A = np.array([1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4])
value = 2
ind = np.array([0, 1, 5, 10]) # Index belongs to ind
Here is what I did:
这是我所做的：
B = np.where(A==value)[0] # Gives the indexes in A for which value = 2
print(B)
[1 5 9]
mask = np.in1d(B, ind) # Gives the index values that belong to the ind array
print(mask)
array([ True, True, False], dtype=bool)
print B[mask] # Is the solution
[1 5]
The solution works, but I find it complicated. Also, in1ddoes a sort which is slow. Is there a better way of achieving this?
该解决方案有效，但我觉得它很复杂。此外，in1d做一个很慢的排序。有没有更好的方法来实现这一目标？
采纳答案by Robb
If you flip the operation order around you can do it in one line:
如果您翻转操作顺序，您可以在一行中完成：
B = ind[A[ind]==value]
print B
[1 5]
Breaking that down:
打破它：
#subselect first
print A[ind]
[1 2 2 3]
#create a mask for the indices
print A[ind]==value
[False True True False]
print ind
[ 0 1 5 10]
print ind[A[ind]==value]
[1 5]
回答by Paul
The second two steps can be replaced by intersect1D. Probably also does a sort. Don't know how you'd avoid that unless you can guarantee your ind array is ordered.
后两步可以用intersect1D代替。大概也做了一个排序。不知道如何避免这种情况，除非您可以保证您的 ind 数组是有序的。
回答by John Zwinck
B = np.where(A==value)[0] #gives the indexes in A for which value = 2
print np.intersect1d(B, ind)
[1 5]
回答by Praveen
How about deferring the np.whereuntil the end, like so:
推迟np.where到最后怎么样，像这样：
res = (A == value)
mask = np.zeros(A.size)
mask[ind] = 1
print np.where(res * z)[0]
That shouldn't require any sorting.
那应该不需要任何排序。
回答by wwii
This is a bit different - I didn't do any timing tests.
这有点不同 - 我没有做任何计时测试。
>>>
>>> A = np.array([1,2,3,4,1,2,3,4,1,2,3,4])
>>> ind = np.array([0,1,5,10])
>>> b = np.ix_(A==2)
>>> np.intersect1d(ind, *b)
array([1, 5])
>>>
Although after looking at @Robb's solution, that's probably the way to do it.
尽管在查看了@Robb 的解决方案之后，这可能就是这样做的方法。
