Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
775 views
in Technique[技术] by (71.8m points)

python - Numpy-vectorized function to repeat blocks of consecutive elements

Numpy has а repeat function, that repeats each element of the array a given (per element) number of times.

I want to implement a function that does similar thing but repeats not individual elements, but variably sized blocks of consecutive elements. Essentially I want the following function:

import numpy as np

def repeat_blocks(a, sizes, repeats):
    b = []    
    start = 0
    for i, size in enumerate(sizes):
        end = start + size
        b.extend([a[start:end]] * repeats[i])
        start = end
    return np.concatenate(b)

For example, given

a = np.arange(20)
sizes = np.array([3, 5, 2, 6, 4])
repeats = np.array([2, 3, 2, 1, 3])

then

repeat_blocks(a, sizes, repeats)

returns

array([ 0,  1,  2, 
        0,  1,  2,

        3,  4,  5,  6,  7, 
        3,  4,  5,  6,  7, 
        3,  4,  5,  6,  7, 

        8,  9, 
        8,  9,

        10, 11, 12, 13, 14, 15,

        16, 17, 18, 19,
        16, 17, 18, 19,
        16, 17, 18, 19 ])

I want to push these loops into numpy in the name of performance. Is this possible? If so, how?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Here's one vectorized approach using cumsum -

# Get repeats for each group using group lengths/sizes
r1 = np.repeat(np.arange(len(sizes)), repeats)

# Get total size of output array, as needed to initialize output indexing array
N = (sizes*repeats).sum() # or np.dot(sizes, repeats)

# Initialize indexing array with ones as we need to setup incremental indexing
# within each group when cumulatively summed at the final stage. 
# Two steps here:
# 1. Within each group, we have multiple sequences, so setup the offsetting
# at each sequence lengths by the seq. lengths preceeeding those.
id_ar = np.ones(N, dtype=int)
id_ar[0] = 0
insert_index = sizes[r1[:-1]].cumsum()
insert_val = (1-sizes)[r1[:-1]]

# 2. For each group, make sure the indexing starts from the next group's
# first element. So, simply assign 1s there.
insert_val[r1[1:] != r1[:-1]] = 1

# Assign index-offseting values
id_ar[insert_index] = insert_val

# Finally index into input array for the group repeated o/p
out = a[id_ar.cumsum()]

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...