I have a NumPy array [1,2,3,4,5,6,7,8,9,10,11,12,13,14] and want to have an array structured like [[1,2,3,4], [2,3,4,5], [3,4,5,6], ..., [11,12,13,14]].
Sure this is possible by looping over the large array and adding arrays of length four to the new array, but I'm curious if there is some secret 'magic' Python method doing just this :)
解决方案
The fastest way seems to be to preallocate the array, given as option 7 right at the bottom of this answer.
>>> import numpy as np
>>> A=np.array([1,2,3,4,5,6,7,8,9,10,11,12,13,14])
>>> A
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14])
>>> np.array(zip(A,A[1:],A[2:],A[3:]))
array([[ 1, 2, 3, 4],
[ 2, 3, 4, 5],
[ 3, 4, 5, 6],
[ 4, 5, 6, 7],
[ 5, 6, 7, 8],
[ 6, 7, 8, 9],
[ 7, 8, 9, 10],
[ 8, 9, 10, 11],
[ 9, 10, 11, 12],
[10, 11, 12, 13],
[11, 12, 13, 14]])
>>>
You can easily adapt this to do it for variable chunk size.
>>> n=5
>>> np.array(zip(*(A[i:] for i in range(n))))
array([[ 1, 2, 3, 4, 5],
[ 2, 3, 4, 5, 6],
[ 3, 4, 5, 6, 7],
[ 4, 5, 6, 7, 8],
[ 5, 6, 7, 8, 9],
[ 6, 7, 8, 9, 10],
[ 7, 8, 9, 10, 11],
[ 8, 9, 10, 11, 12],
[ 9, 10, 11, 12, 13],
[10, 11, 12, 13, 14]])
You may wish to compare performance between this and using itertools.islice.
>>> from itertools import islice
>>> n=4
>>> np.array(zip(*[islice(A,i,None) for i in range(n)]))
array([[ 1, 2, 3, 4],
[ 2, 3, 4, 5],
[ 3, 4, 5, 6],
[ 4, 5, 6, 7],
[ 5, 6, 7, 8],
[ 6, 7, 8, 9],
[ 7, 8, 9, 10],
[ 8, 9, 10, 11],
[ 9, 10, 11, 12],
[10, 11, 12, 13],
[11, 12, 13, 14]])
My timing results:
1. timeit np.array(zip(A,A[1:],A[2:],A[3:]))
10000 loops, best of 3: 92.9 us per loop
2. timeit np.array(zip(*(A[i:] for i in range(4))))
10000 loops, best of 3: 101 us per loop
3. timeit np.array(zip(*[islice(A,i,None) for i in range(4)]))
10000 loops, best of 3: 101 us per loop
4. timeit numpy.array([ A[i:i+4] for i in range(len(A)-3) ])
10000 loops, best of 3: 37.8 us per loop
5. timeit numpy.array(list(chunks(A, 4)))
10000 loops, best of 3: 43.2 us per loop
6. timeit numpy.array(byN(A, 4))
10000 loops, best of 3: 100 us per loop
# Does preallocation of the array help? (11 is from len(A)+1-4)
7. timeit B=np.zeros(shape=(11, 4),dtype=np.int32)
loops, best of 3: 2.19 us per loop
timeit for i in range(4):B[:,i]=A[i:11+i]
10000 loops, best of 3: 20.9 us per loop
total 23.1us per loop
As len(A) increases (20000) 4 and 5 converge to be equivalent speed (44 ms). 1,2,3 and 6 all remain about 3 times slower (135 ms). 7 is much faster (1.36 ms).
版权声明：本文内容由互联网用户自发贡献，该文观点仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容，请联系我们，一经查实，本站将立刻删除。
如需转载请保留出处：https://51itzy.com/kjqy/66715.html