Get all subsequences of a numpy array

Question

Given a numpy sized array nand an integer m, I want to generate all consecutive subsequences of the length of the marray, preferably as a two-dimensional array.

Example:

>>> 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]])

the best way i can do this is

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]

Is there a better, preferably built-in function that I miss?

Answer 1

Here's a very fast and memory efficient method that is just a “representation” in the original array:

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))

np.copy, , "".

: fooobar.com/questions/64243/...

Answer 2

scipy.linalg.hankel .

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

Answer 3

You were on the right track.

You can use the following broadcast trick to create a 2dim array indicesof two 1dim aranges:

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

Iterator based

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)

Slice Based

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)