遍历布尔值列表以创建新的整数列表

Question

I would like to turn this list of booleans:

bool_list: [False, False, True, False, False, True, False, True, True, False]

Into this list of integers:

int_li = [0, 0, 1, 1, 1, 2, 2, 3, 4, 4]

Basicaly, iterate through the list of booleans and incremente 1 everytime the value is True. Any efficient way to do that ?

Answer 1

If you have Python 3.8 or later, you could make use of the walrus operator to generate your desired result using a list comprehension, making use of the fact that in an integer context, True == 1 and False == 0 :

v = 0
int_li = [v := v+b for b in bool_list]

Output:

[0, 0, 1, 1, 1, 2, 2, 3, 4, 4]

Answer 2

Create a variable (eg count with a value of 0, loop through each item in the list and every time it is True , add +1 to count . Then append count to the new list like so:

bool_list = [False, False, True, False, False, True, False, True, True, False]
int_li = []
count = 0
for item in bool_list:
    if item:
        count += 1
    int_li.append(count)
print(int_li)
# [0, 0, 1, 1, 1, 2, 2, 3, 4, 4]

Answer 3

from itertools import accumulate

int_li = list(map(int, accumulate(bool_list)))

As Daniel Hao pointed out, pure accumulate doesn't work, as the first element will remain bool , resulting in [False, 0, 1, 1, 1, 2, 2, 3, 4, 4] . Mapping to int is the simplest fix I saw, though I don't like it much.

Another way, with an initial 0 that we then disregard:

from itertools import accumulate

_, *int_li = accumulate(bool_list, initial=0)

Answer 4

simply do this with list comprehension in one line:

bool_list = [False, False, True, False, False, True, False, True, True, False]

int_list = [sum(bool_list[:i+1]) for i in range(len(bool_list))]

output:

[0, 0, 1, 1, 1, 2, 2, 3, 4, 4]