Count the number of occurrences between markers in a python list

user_528025 0
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I have a boolean (numpy) array. And I want to count how many occurrences of 'True' are between the Falses.

Eg for a sample list:

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

should produce 

ml = [3,3,1]

my initial attempt was to try this snippet:

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

But it keeps appending elements in ml for each F in the b_List.

EDIT

Thank you all for your answers. Sadly I can' accept all the answers as correct. I've accepted Akavall's answer because he referred to my initial attempt (I know what I did wrong now) and also did a comparison between the Mark's and Ashwinis posts.

Please don't take as a define answer the accepted solution, since both the other suggestions introduce alternative methods what work equally well

numpy algorithm python
3 Answers

itertools.groupby provides one easy way to do this:

>>> 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]
user_2709860

Using 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])

Can't say that this is the best NumPy solution, but it is still faster than itertools.groupby:

>>> 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

As @justhalf and @Mark Dickinson pointed out in comments the above code will not work in some cases, so you need to append False on both ends first:

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])
user_8468920

Your original try has some problems:

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

You probably want something like this:

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

Result:

>>> 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]

This solution runs surprisingly fast:

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
user_10780840
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