问题
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.
I've found two naive ways to do this, the first is slow, and the second memory-intensive.
The slow way
indices= []
for iy in y:
indices += np.where(x==iy)[0][0]
The memory hog
xe = np.outer([1,]*len(x), y)
ye = np.outer(x, [1,]*len(y))
junk, indices = np.where(np.equal(xe, ye))
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.
回答1:
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
the result is:
[-- 3 1 -- -- 6]
回答2:
I want to suggest one-line solution:
indices = np.where(np.in1d(x, y))[0]
The result is an array with indices for x array which corresponds to elements from y which were found in x.
One can use it without numpy.where if needs.
回答3:
How about this?
It does assume that every element of y is in x, (and will return results even for elements that aren't!) but it is much faster.
import numpy as np
# Generate some example data...
x = np.arange(1000)
np.random.shuffle(x)
y = np.arange(100)
# Actually preform the operation...
xsorted = np.argsort(x)
ypos = np.searchsorted(x[xsorted], y)
indices = xsorted[ypos]
回答4:
I would just do this:
indices = np.where(y[:, None] == x[None, :])[1]
Unlike your memory-hog way, this makes use of broadcast to directly generate 2D boolean array without creating 2D arrays for both x and y.
回答5:
I think this is a clearer version:
np.where(y.reshape(y.size, 1) == x)[1]
than indices = np.where(y[:, None] == x[None, :])[1]. You don't need to broadcast x into 2D.
This type of solution I found to be best because unlike searchsorted() or in1d() based solutions that have seen posted here or elsewhere, the above works with duplicates and it doesn't care if anything is sorted. This was important to me because I wanted x to be in a particular custom order.
回答6:
The numpy_indexed package (disclaimer: I am its author) contains a function that does exactly this:
import numpy_indexed as npi
indices = npi.indices(x, y, missing='mask')
It will currently raise a KeyError if not all elements in y are present in x; but perhaps I should add a kwarg so that one can elect to mark such items with a -1 or something.
It should have the same efficiency as the currently accepted answer, since the implementation is along similar lines. numpy_indexed is however more flexible, and also allows to search for indices of rows of multidimensional arrays, for instance.
EDIT: ive changed the handling of missing values; the 'missing' kwarg can now be set with 'raise', 'ignore' or 'mask'. In the latter case you get a masked array of the same length of y, on which you can call .compressed() to get the valid indices. Note that there is also npi.contains(x, y) if this is all you need to know.
回答7:
Use this line of code :-
indices = np.where(y[:, None] == x[None, :])[1]
回答8:
A more direct solution, that doesn't expect the array to be sorted.
import pandas as pd
A = pd.Series(['amsterdam', 'delhi', 'chromepet', 'tokyo', 'others'])
B = pd.Series(['chromepet', 'tokyo', 'tokyo', 'delhi', 'others'])
# Find index position of B's items in A
B.map(lambda x: np.where(A==x)[0][0]).tolist()
Result is:
[2, 3, 3, 1, 4]
来源：https://stackoverflow.com/questions/8251541/numpy-for-every-element-in-one-array-find-the-index-in-another-array