python中numpy数组的随机大小块

Question

我希望将索引数组划分为随机大小的块(从有限的可能大小范围内获取)，这些块也在彼此之间进行混合。我尝试了以下我找到的here，但它专注于同样大小的块。

a = np.arange(1, 100)

def chunk(xs, n): # to chunk the array xs in n parts
    ys = list(xs)
    random.shuffle(ys)
    size = len(ys) // n
    leftovers= ys[size*n:]
    for c, xtra in enumerate(leftovers):
        yield ys[c*size:(c+1)*size] + [ xtra ]
    for c in xrange(c+1,n):
        yield ys[c*size:(c+1)*size]

换句话说，如何将上述函数更改为具有一定数量的块(随机数和彼此之间的混叠)，并从一个范围内随机选取一个可变大小，例如[5-10]

Answer 1

这会有用的：

from itertools import chain
import numpy as np

a = np.arange(1, 100)
def chunk(xs, nlow, nhigh, shuffle=True):
    xs = np.asarray(xs)
    if shuffle:
        # shuffle, if you want
        xs = xs.copy()
        np.random.shuffle(xs)

    # get at least enough random chunk sizes in the specified range, ie nlow <= n <= nhigh
    ns = np.random.randint(nlow, nhigh+1, size=xs.size//nlow)
    # add up the chunk sizes to get the indices at which we'll slice up the input array
    ixs = np.add.accumulate(ns)
    # truncate ixs so that its contents are all valid indices with respect to xs
    ixs = ixs[:np.searchsorted(ixs, xs.size)]

    # yield slices from the input array
    for start,end in zip(chain([None], ixs), chain(ixs, [None])):
        yield xs[start:end]

list(chunk(a, 5, 10))

输出：

[array([67, 79, 17, 62, 12, 37, 70, 24]),
 array([98, 48, 88, 59, 47]),
 array([52, 60, 89, 23, 43, 44]),
 array([ 7, 27, 33, 74, 49,  2]),
 array([ 6, 51, 40, 13, 56, 45]),
 array([31,  3, 55, 10, 11, 46,  9, 42, 34]),
 array([53, 22, 95, 41, 19, 32,  4, 69, 86]),
 array([93, 68, 57, 65, 92, 76, 28, 63, 64, 58]),
 array([91, 66, 18, 99, 21]),
 array([36, 83, 15, 78,  1, 81, 97, 84]),
 array([61, 71, 25, 94, 87, 20, 85, 38]),
 array([ 8, 96, 75, 30, 77, 14, 72, 29]),
 array([35, 90, 82, 73, 39,  5, 26, 50, 16]),
 array([80, 54])]

编辑

我最初的答案并没有对最后一个块的大小设置一个下限，所以有时它会比指定的要小(尽管永远不会变大)。据我所知，没有直接的方法来处理这件事。但是，通常您可以从随机分布中删除一个不需要的区域，只需拒绝来自该区域的任何样本。换句话说，您可以确保最后一个块足够大，只需抛出它不存在的任何建议块：

def getIxs(xsize, nlow, nhigh):
    # get at least enough random chunk sizes in the specified range, ie nlow <= n <= nhigh
    ns = np.random.randint(nlow, nhigh+1, size=xsize//nlow)

    # add up the chunk sizes to get the indices at which we'll slice up the input array
    ixs = np.add.accumulate(ns)

    # truncate ixs so that its contents are all valid indices with respect to xs
    ixs = ixs[:np.searchsorted(ixs, xsize)]

    return ixs

def chunk(xs, nlow, nhigh):
    xs = np.asarray(xs)

    ixs = getIxs(xs.size, nlow, nhigh)

    # rerun getIxs until the size of the final chunk is large enough
    while (xs.size - ixs[-1]) < nlow:
        ixs = getIxs(xs.size, nlow, nhigh)

    # yield slices from the input array
    for start,end in zip(chain([None], ixs), chain(ixs, [None])):
        yield xs[start:end]

这种方法应该保持每个块大小的总体随机性。

Answer 2

您可以使用np.split(array,indices)

import random
a = np.arange(100)
np.random.shuffle(a)
ind = sorted(random.sample(range(len(a)),k=np.random.randint(low=1,high=10)))
np.split(a,ind)



  [array([41, 19, 85, 51, 34]),
 array([71, 27]),
 array([36, 16, 18, 74, 43, 96, 45, 97, 54, 75, 89, 48, 33, 32, 63, 98,  5,
        80, 30, 17, 86, 14, 67]),
 array([ 9, 70, 84, 99, 39]),
 array([59, 20, 78, 61, 49, 37, 93]),
 array([ 1, 79, 81, 69, 40, 42, 29,  8,  3, 68, 87, 66,  4, 21, 91, 92, 31]),
 array([83, 15, 56,  2, 64, 95, 12,  0, 90, 77, 57, 60, 38, 76, 94, 22, 24,
         6, 46, 65, 50, 62, 28, 44, 73, 13, 26, 72,  7, 53, 82, 47, 58, 35,
        52, 25, 88, 11, 10, 55, 23])]