Question

给定一个大小为n且整数为m的numpy数组，我想生成数组的所有顺序m长度子序列，最好是二维数组。

实施例：

>>> subsequences(arange(10), 4)

array([[0, 1, 2, 3, 4, 5, 6],
       [1, 2, 3, 4, 5, 6, 7],
       [2, 3, 4, 5, 6, 7, 8],
       [3, 4, 5, 6, 7, 8, 9]])

我能做到这一点的最好方法是

def subsequences(arr, m):
    n = arr.size
    # Create array of indices, essentially solution for "arange" input
    indices = cumsum(vstack((arange(n - m + 1), ones((m-1, n - m + 1), int))), 0)
    return arr[indices]

我缺少一个更好的，最好是内置的功能吗？

Answer 1

scipy.linalg.hankel这样做。

from scipy.linalg import hankel
def subsequences(v, m):
    return hankel(v[:m], v[m-1:])

Answer 2

这是一种非常快速且内存效率高的方法，它只是一个＆＃34;视图＆＃34;进入原始数组：

from numpy.lib.stride_tricks import as_strided

def subsequences(arr, m):
    n = arr.size - m + 1
    s = arr.itemsize
    return as_strided(arr, shape=(m,n), strides=(s,s))

如果需要写入此数组，则应首先创建np.copy，否则您将修改原始数组以及＆＃34;子序列中的相应条目＆＃34;数组也是如此。

此处有更多信息：https://stackoverflow.com/a/4924433/2379410

Answer 3

你走在正确的轨道上。

您可以利用以下广播技巧，从两个1dim indices s创建一个2dim arange数组：

arr = arange(7)[::-1]
arr
=> array([6, 5, 4, 3, 2, 1, 0])
n = arr.size
m = 3

indices = arange(m) + arange(n-m+1).reshape(-1, 1)  # broadcasting rulez
indices
=>
array([[0, 1, 2],
       [1, 2, 3],
       [2, 3, 4],
       [3, 4, 5],
       [4, 5, 6]])

arr[indices]
=>
array([[6, 5, 4],
       [5, 4, 3],
       [4, 3, 2],
       [3, 2, 1],
       [2, 1, 0]])

Answer 4

基于迭代器

from itertools import tee, islice
import collections
import numpy as np

# adapted from https://docs.python.org/2/library/itertools.html
def consumed(iterator, n):
    "Advance the iterator n-steps ahead. If n is none, consume entirely."
    # Use functions that consume iterators at C speed.
    if n is None:
        # feed the entire iterator into a zero-length deque
        collections.deque(iterator, maxlen=0)
    else:
        # advance to the empty slice starting at position n
        next(islice(iterator, n, n), None)
    return iterator


def subsequences(iterable, b):
    return np.array([list(consumed(it, i))[:b] for i, it in enumerate(tee(iterable, len(iterable) - b + 1))]).T

print subsequences(np.arange(10), 4)

基于切片

import numpy as np

def subsequences(iterable, b):
    return np.array([iterable[i:i + b] for i in range(len(iterable) - b + 1)]).T

print subsequences(np.arange(10), 4)

获取numpy数组的所有子序列

4 个答案:

基于迭代器

基于切片