基于另一个数组在Numpy数组中设置值

Question

1 term_map tracks which term is in which position.

In [256]: term_map = np.array([2, 2, 3, 4, 4, 4, 2, 0, 0, 0])

In [257]: term_map
Out[257]: array([2, 2, 3, 4, 4, 4, 2, 0, 0, 0])

2 term_scores tracks the weight of each term at each position.

In [258]: term_scores = np.array([5, 6, 9, 8, 9, 4, 5, 1, 2, 1])

In [259]: term_scores
Out[259]: array([5, 6, 9, 8, 9, 4, 5, 1, 2, 1])

3 Get the unique values and the inverse indices.

In [260]: unqID, idx = np.unique(term_map, return_inverse=True)

In [261]: unqID
Out[261]: array([0, 2, 3, 4])

4 Compute the scores for the unique values.

In [262]: value_sums = np.bincount(idx, term_scores)

In [263]: value_sums
Out[263]: array([  4.,  16.,   9.,  21.])

5 Initialize Array To Update. The indices correspond to the values in the term_map variable.

In [254]: vocab = np.zeros(13)

In [255]: vocab
Out[255]: array([ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.])

6 DESIRED: Insert the values 4 corresponding to the positions listed in 3 into the vocab variable.

In [255]: updated_vocab
Out[255]: array([ 4.,  0.,  16.,  9.,  21.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.])

How do I create 6?

Answer 1

As it turns out, we can avoid the np.unique step to directly get to the desired output by feeding in term_map and term_scores to np.bincount and also mention the length of the output array with its optional argument minlength .

Thus, we could simply do -

final_output = np.bincount(term_map, term_scores, minlength=13)

Sample run -

In [142]: term_map = np.array([2, 2, 3, 4, 4, 4, 2, 0, 0, 0])
     ...: term_scores = np.array([5, 6, 9, 8, 9, 4, 5, 1, 2, 1])
     ...: 

In [143]: np.bincount(term_map, term_scores, minlength=13)
Out[143]: 
array([  4.,   0.,  16.,   9.,  21.,   0.,   0.,   0.,   0.,   0.,   0.,
         0.,   0.])

Answer 2

import numpy as np

term_map = np.array([2, 2, 3, 4, 4, 4, 2, 0, 0, 0])
term_scores = np.array([5, 6, 9, 8, 9, 4, 5, 1, 2, 1])
unqID, idx = np.unique(term_map, return_inverse=True)
value_sums = np.bincount(idx, term_scores)

vocab = np.zeros(13)
vocab[unqID] = value_sums
print(vocab)

OUT: [ 4. 0. 16. 9. 21. 0. 0. 0. 0. 0. 0. 0. 0.]