Question

我有一个布尔（numpy）数组。而且我想知道Falses之间出现了多少次'True'。

例如，对于样本列表：

b_List = [T,T,T,F,F,F,F,T,T,T,F,F,T,F]

应该产生

ml = [3,3,1]

我最初的尝试是尝试这个代码段：

i = 0
ml = []
for el in b_List:
  if (b_List):
    i += 1
  ml.append(i)
  i = 0

但它会在b_List中为每个F添加以ml为单位的元素。

修改

谢谢大家的回答。可悲的是，我可以'接受所有答案都是正确的。我接受了Akavall的答案，因为他提到了我最初的尝试（我知道我现在做错了什么），并且还对Mark和Ashwinis的帖子进行了比较。

请不要将接受的解决方案作为定义答案，因为其他建议都引入了同样适用的替代方法

Answer 1

itertools.groupby提供了一种简单的方法：

>>> import itertools
>>> T, F = True, False
>>> b_List = [T,T,T,F,F,F,F,T,T,T,F,F,T,F]
>>> [len(list(group)) for value, group in itertools.groupby(b_List) if value]
[3, 3, 1]

Answer 2

使用NumPy：

>>> import numpy as np
>>> a = np.array([ True,  True,  True, False, False, False, False,  True,  True, True, False, False,  True, False], dtype=bool)
>>> np.diff(np.insert(np.where(np.diff(a)==1)[0]+1, 0, 0))[::2]
array([3, 3, 1])

>>> a = np.array([True, False, False, True, True, False, False, True, False])
>>> np.diff(np.insert(np.where(np.diff(a)==1)[0]+1, 0, 0))[::2]
array([1, 2, 1])

不能说这是最好的NumPy解决方案，但它仍然比itertools.groupby更快：

>>> lis = [ True,  True,  True, False, False, False, False,  True,  True, True, False, False,  True, False]*1000
>>> a = np.array(lis)
>>> %timeit [len(list(group)) for value, group in groupby(lis) if value]
100 loops, best of 3: 9.58 ms per loop
>>> %timeit np.diff(np.insert(np.where(np.diff(a)==1)[0]+1, 0, 0))[::2]
1000 loops, best of 3: 1.4 ms per loop

>>> lis = [ True,  True,  True, False, False, False, False,  True,  True, True, False, False,  True, False]*10000
>>> a = np.array(lis)
>>> %timeit [len(list(group)) for value, group in groupby(lis) if value]
1 loops, best of 3: 95.5 ms per loop
>>> %timeit np.diff(np.insert(np.where(np.diff(a)==1)[0]+1, 0, 0))[::2]
100 loops, best of 3: 14.9 ms per loop

正如@justhalf和@Mark Dickinson在评论中指出的那样，上述代码在某些情况下不起作用，因此您需要先在两端添加False：

In [28]: a                                                                                        
Out[28]: 
array([ True,  True,  True, False, False, False, False,  True,  True,
        True, False, False,  True, False], dtype=bool)

In [29]: np.diff(np.where(np.diff(np.hstack([False, a, False])))[0])[::2]
Out[29]: array([3, 3, 1])

Answer 3

你原来的尝试有一些问题：

i = 0
ml = []
for el in b_List:
    if (b_List): # b_list is a list and will evaluate to True
                 # unless you have an empty list, you want if (el)
        i += 1
    ml.append(i) # even if the above line was correct you still get here
                 # on every iteration, and you don't want that
    i = 0

你可能想要这样的东西：

def count_Trues(b_list):
    i = 0
    ml = []
    prev = False
    for el in b_list:
        if el:
            i += 1
            prev = el
        else:
            if prev is not el:
                ml.append(i)
                i = 0
            prev = el
    if el:
        ml.append(i)
    return m

结果：

>>> T, F = True, False
>>> b_List = [T,T,T,F,F,F,F,T,T,T,F,F,T,F] 
>>> count_Trues(b_List)
[3, 3, 1]
>>> b_List.extend([T,T])
>>> count_Trues(b_List)
[3, 3, 1, 2]
>>> b_List.extend([F])
>>> count_Trues(b_List)
[3, 3, 1, 2]

此解决方案运行得非常快：

In [5]: T, F = True, False

In [6]: b_List = [T,T,T,F,F,F,F,T,T,T,F,F,T,F] 

In [7]: new_b_List = b_List * 100

In [8]: import numpy as np

# Ashwini Chaudhary's Solution
In [9]: %timeit np.diff(np.insert(np.where(np.diff(new_b_List)==1)[0]+1, 0, 0))[::2]
1000 loops, best of 3: 299 us per loop

In [11]: %timeit count_Trues(new_b_List)
1000 loops, best of 3: 130 us per loop

In [12]: new_b_List = b_List * 1000

# Ashwini Chaudhary's Solution 
In [13]: %timeit np.diff(np.insert(np.where(np.diff(new_b_List)==1)[0]+1, 0, 0))[::2]
100 loops, best of 3: 2.25 ms per loop

In [14]: %timeit count_Trues(new_b_List)
100 loops, best of 3: 1.33 ms per loop

计算python列表中标记之间的出现次数

3 个答案: