python中numpy数组的随机大小分块

我想将索引数组分成随机大小的块（从有限的可能大小范围中选取），这些块也在彼此之间混洗。我尝试了以下我发现的方法，但它关注的是同样大小的块

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]

换言之，我如何更改上述函数，使其具有一定数量的块（随机数和相互之间的随机数），并从某个范围随机获取可变大小，例如

[5-10]

，这将起作用：

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))

编辑 我最初的答案并没有给最后一个块的大小设置一个下限，所以有时候它会比指定的小（尽管永远不会大）。据我所知，没有直接的方法来处理这个问题。但是，一般来说，您可以通过拒绝来自该区域的任何样本，从随机分布中删除不需要的区域。换言之，您可以通过丢弃任何建议的块来确保最后一个块足够大：

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]

def getIxs（xsize、nlow、nhigh）：
#在指定范围内至少获取足够的随机块大小，即nlow您可以使用
np.split（数组、索引）

你的问题还不完整。。范围是多少？它是如何计算的？假设可能的尺寸范围是5-10。这是不相关的，但我会将它添加到问题中，在这种情况下，您可能会或可能不会满足n个块数约束，对吧？是的，没错。相关变量是对块大小的限制，应在特定范围内随机选取，例如[5-10]！选项shuffle=False正是我想要的。谢谢！：）@Garini是的，我不确定你是否有意使用shuffle，或者复制的代码中是否携带了cruft。所以我选择了它，只有一个缺陷，我注意到。最后一个分区可以随机突破大小限制。有一种方法可以避免这种情况吗？@Garini通过设计（或多或少）最终的块可以小于下限。没有有效的方法来解决这个问题，但是你可以使用基于“拒绝”的方法来得到你想要的。我已经在上面的一篇编辑文章中发布了一些相关的代码。在我的例子中，我希望在不强加拆分数量的情况下，将拆分的大小保持在一个有限的范围内。另外，我需要在拆分中有有序的数字。在这种情况下，k也应该是一个随机数，但不确定，拆分中的有序数字是什么意思？即使原始数组未排序，也要对拆分进行排序吗？
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]

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])]