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I have a 2d array (n x m) from which I would like to produce a 1d array (length n) using a list of row-indices of length n.

For instance:


2d = ([a,b,c],[d,e,f],[g,h,i]) # input array
1d = ([0,2,1]) # row numbers

result = ([a,e,h]) # array of the first row of first column, third row of second column, second row of third column

I have found a way to do this using a list comprehension (iterating over the columns and the indices simultaneously and picking out the value), but there's surely a numpy function that does this?

Robbie Mallett
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3 Answers3

1

Try this:

print([y[x] for x, y in zip(_1d, _2d)])
U13-Forward
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  • As I mentioned in the question "I have found a way to do this using a list comprehension (iterating over the columns and the indices simultaneously and picking out the value), but there's surely a numpy function that does this?" – Robbie Mallett Aug 18 '21 at 16:41
1

Specifying multiple elements by their indices and putting them in a new array is called "advanced indexing".

x = np.array([x for x in 'abcdefghi']).reshape((3,3))

# array([['a', 'b', 'c'],
#        ['d', 'e', 'f'],
#        ['g', 'h', 'i']], dtype='<U1')

d1_indices = np.array([0,1,2])
d2_indices = np.array([0,2,1])

selectx = x[d1_indices, d2_indices]
# array(['a', 'f', 'h'], dtype='<U1')

# selectx[i] = x[d1_indices[i], d2_indices[i]]
BatWannaBe
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0

How about this?

import numpy as np
a = np.arange(9).reshape((3,3))
# [0, 1, 2]
# [3, 4, 5]
# [6, 7, 8]
print(a[[0,2,1], np.arange(3)]) # result : [0 7 5]

Iterating by rows and picking a value based on column index is also possible:

print(a[np.arange(3),[0,2,1]]) # result : [0 5 7]
tpl
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