2

这个问题可能太菜鸟了，但我仍然无法弄清楚如何正确地做到这一点。

我有一个给定的数组[0,0,0,0,0,0,1,1,2,1,0,0,0,0,1,0,1,2,1,0,2,3] （0-5 的任意元素），我想有一个计数器来记录连续出现的零。

1 times 6 zeros in a row
1 times 4 zeros in a row
2 times 1 zero  in a row

=> (2,0,0,1,0,1)

所以字典由n*0作为索引的值和作为值的计数器组成。

最终数组由 500+ 百万个值组成，这些值与上面的一样未排序。

4

3 回答 3

2

这应该可以得到你想要的：

import numpy as np

a = [0,0,0,0,0,0,1,1,2,1,0,0,0,0,1,0,1,2,1,0,2,3]

# Find indexes of all zeroes
index_zeroes = np.where(np.array(a) == 0)[0]

# Find discontinuities in indexes, denoting separated groups of zeroes
# Note: Adding True at the end because otherwise the last zero is ignored
index_zeroes_disc = np.where(np.hstack((np.diff(index_zeroes) != 1, True)))[0]

# Count the number of zeroes in each group
# Note: Adding 0 at the start so first group of zeroes is counted
count_zeroes = np.diff(np.hstack((0, index_zeroes_disc + 1)))

# Count the number of groups with the same number of zeroes
groups_of_n_zeroes = {}
for count in count_zeroes:
    if groups_of_n_zeroes.has_key(count):
        groups_of_n_zeroes[count] += 1
    else:
        groups_of_n_zeroes[count] = 1

groups_of_n_zeroes持有：

{1: 2, 4: 1, 6: 1}
于 2013-05-02T15:34:14.823 回答
1

类似于@fgb，但对出现次数的计数进行了更多的numpythonic处理：

items = np.array([0,0,0,0,0,0,1,1,2,1,0,0,0,0,1,0,1,2,1,0,2,3])
group_end_idx = np.concatenate(([-1],
                                np.nonzero(np.diff(items == 0))[0],
                                [len(items)-1]))
group_len = np.diff(group_end_idx)
zero_lens = group_len[::2] if items[0] == 0 else group_len[1::2]
counts = np.bincount(zero_lens)

>>> counts[1:]
array([2, 0, 0, 1, 0, 1], dtype=int64)
于 2013-05-02T16:20:02.070 回答
0

这似乎非常复杂，但我似乎找不到更好的东西：

>>> l = [0, 0, 0, 0, 0, 0, 1, 1, 2, 1, 0, 0, 0, 0, 1, 0, 1, 2, 1, 0, 2, 3]

>>> import itertools
>>> seq = [len(list(j)) for i, j in itertools.groupby(l) if i == 0]
>>> seq
[6, 4, 1, 1]

>>> import collections
>>> counter = collections.Counter(seq)
>>> [counter.get(i, 0) for i in xrange(1, max(counter) + 1)]
[2, 0, 0, 1, 0, 1]
于 2013-05-02T15:26:23.707 回答