如何计算列表中某个元素之前的特定元素？

Question

We have a list

list = [1, 1, 1, 0, 1, 0, 0, 1] 

I am trying find a function that would count the number of 0's before each item and then multiply this number by 3.

def formula(n):
    for x in list:
        if x == 1:
            form = n * 3
            return form
#If x is 1, count the number of zeros right before x and multiply this by 3, 

For example for the list above, the first element is a 1 and there are no numbers right before it, the program should compute 0 * 3 = 0, for the second item, which is also a 1, the number right before it is also not a zero, the program should also compute 0 * 3 = 0. The 4th element is 0 so the program should ignore, For the 5th element which is a 1, the number right before it is a 0, the programme to compute 1 * 3 = 3, for the 6th element the number right before it is a 1, the system should compute 0 * 3 = 0. The 7th element is a 0, since x is not equal to 1 the program should not do anything. For the last element which is a 1, the last two numbers before it are zeros, the program should compute 2 * 3 = 6

Answer 1

I believe you are looking for a generator, with a simple counter:

def get_values (list):
    n = 0
    for x in list:
        if x == 1:
            yield n * 3
            n = 0   # reset the counter
        elif x == 0:
            n += 1  # increment the counter

print(list(get_values([1, 1, 1, 0, 1, 0, 0, 1])))
# [0, 0, 0, 3, 6]

Answer 2

Try this,

def formula(l):
    count_zero = 0
    result = []

    for i in l:
        if i == 1:
            result.append(3*count_zero)
            count_zero = 0
        elif i == 0:
            count_zero += 1

    return result

# Test case
l = [1, 1, 1, 0, 1, 0, 0, 1] 
result = formula(l)
print(result)
# [0, 0, 0, 3, 6]

Answer 3

Here is my solution for the problem.

test_list = [1, 1, 1, 0, 1, 0, 0, 1]


def formula(list):
    track = []
    start = 0

    for i,n in enumerate(list):
        count_list_chunk = list[start:i]
        if count_list_chunk.count(0) > 0 and n != 0:
            start = i

        if n != 0:
            track.append( count_list_chunk.count(0)*3 )

    return track

print formula(test_list)
#[ 0, 0, 0, 3, 6]