Python中零之间的列表元素的元素

Question

I have a list:

lst = [1, 2, 3, 5, 0, 0, 9, 45, 3, 0, 1, 7]

And I need the sum of the elements between the 0 s in a new list. I tried

lst1 = []
summ = 0
for i, elem in enumerate(lst):
    if elem != 0:
        summ = summ + elem
    else:
        lst1.append(summ)
        lst1.append(elem)
        summ = 0

but it returns [11, 0, 0, 0, 57, 0] , while I expect [11, 0, 0, 57, 0, 8]

Answer 1

Here's one way to this with itertools.groupby and a list comprehension . The grouping is done by checking if an element is zero, and if not zero, all items in the group are summed:

from itertools import groupby

lst = [1, 2, 3, 5, 0, 0, 9, 45, 3, 0, 1, 7]
f = lambda x: x==0
result = [i for k, g in groupby(lst, f) for i in (g if k else (sum(g),))]
print(result)
# [11, 0, 0, 57, 0, 8]

And of course, if items in your list are only numbers (to avoid generalising and introducing ambuigities), the lambda can be replaced with bool :

result = [i for k, g in groupby(lst, bool) for i in ((sum(g),) if k else g)]

Answer 2

you are appending extra summ when its 0, and missing 1 summ at the end

lst = [1, 2, 3, 5, 0, 0, 9, 45, 3, 0, 1, 7]
lst1 = []
summ = 0
for i, elem in enumerate(lst):
    if elem != 0:
        summ = summ + elem
    else:
        if summ:
            lst1.append(summ)
        lst1.append(elem)
        summ = 0

if summ:
    lst1.append(summ)
# lst1 = [11, 0, 0, 57, 0, 8]

Answer 3

Just add an extra lst1.append(summ) after the loop

lst1 = []
summ = 0
for i, elem in enumerate(lst):
    if elem != 0:
        summ = summ + elem
    else:
        if summ:
            lst1.append(summ)
        lst1.append(elem)
        summ = 0
lst1.append(summ)