计算在 False 布尔值之间组合在一起的 True 布尔值的数量并在列表中返回

Question

I have a list of booleans:

[False, True, True, True, False, False, True, True, True, True, True, False, True, True, True, True, True, True, True, False]

I am looking for a way to count the number of True values grouped together, separated by False values. For the above example, I would be looking for an output of [3, 5, 7] .

I have tried using groupby, and various counting methods but all I have been able to get so far is the total number of True values or the number of groups of True there are. My actual data will be a much longer list of booleans, is there a way to efficiently get this type of output?

Thank you!

Answer 1

from itertools import groupby

bools = [False, True, True, True, False, False, True, True, True, True, True, False, True, True, True, True, True, True, True, False]

print([len(list(group)) for key, group in groupby(bools) if key])

Output:

[3, 5, 7]
>>> 

Answer 2

list = [False, True, True, True, False, False, True, True, True, True, True, 
False, True, True, True, True, True, True,
        True, False]
groupList = []
flag = False
for i in list:
    if i:
        if flag is False:
            groupList.append(1)
            flag = True
        else:
            groupList[-1]+=1
    else:
        flag=False

print(groupList)

Answer 3

You can use groupby ; you just need to process the result.

>>> x = [False, True, True, True, False, False, True, True, True, True, True, False, True, True, True, True, True, True, True, False]
>>> [sum(1 for _ in sg) for v, sg in groupby(x) if v]
[3, 5, 7]

---

The sequence produced by groupby consists of tuples of the key value and the subgroup corresponding to the key value. In this case, each list value is the key, so you get tuples like (False, _grouper(...)) . _grouper is an implementation detail of groupby ; all you need to worry about is that it is an iterable of like values in the original list.

Once you have the groupby instance, you can unpack each tuple, test if the first element is True , then compute the length of the corresponding _grouper . You can't use len , because an iterable isn't necessarily finite or pre-computed, so you use sum to add up a sequence of 1s for each element in the iterable. (Alternatively, you could use len(list(sg)) .)

Answer 4

is this what you mean

booleans=[False, True, True, True, False, False, True, False, True, True, True, False, 
True, True, True, True, True, True, True, False]
counter=0
for i in range(1, len(booleans)-1):
    if(booleans[i-1]==booleans[i+1] and booleans[i+1]==False):
        counter=counter+1
print(counter)