creating sum of odd indexes python

问题:

I'm trying to create a function equal to the sum of every other digit in a list. For example, if the list is [0,1,2,3,4,5], the function should equal 5+3+1. How could I do this? My knowledge of Python does not extend much farther than while and for loops. Thanks.

回答1:

Here is a simple one-liner:

In [37]: L
Out[37]: [0, 1, 2, 3, 4, 5]

In [38]: sum(L[1::2])
Out[38]: 9

In the above code, L[1::2] says "get ever second element in L, starting at index 1"

Here is a way to do all the heavy lifting yourself:

L = [0, 1, 2, 3, 4, 5]
total = 0
for i in range(len(L)):
    if i%2:  # if this is an odd index
        total += L[i]

Here's another way, using enumerate:

L = [0, 1, 2, 3, 4, 5]
total = 0
for i,num in enumerate(L):
    if i%2:
        total += num

回答2:

>>> arr = [0,1,2,3,4,5]
>>> sum([x for idx, x in enumerate(arr) if idx%2 != 0])
9

This is just a list comprehension that only includes elements in arr that have an odd index.

To illustrate in a traditional for loop:

>>> my_sum = 0
>>> for idx, x in enumerate(arr):
...     if idx % 2 != 0:
...         my_sum += x
...         print("%d was odd, so %d was added. Current sum is %d" % (idx, x, my_sum))
...     else:
...         print("%d was even, so %d was not added. Current sum is %d" % (idx, x, my_sum))
... 
0 was even, so 0 was not added. Current sum is 0
1 was odd, so 1 was added. Current sum is 1
2 was even, so 2 was not added. Current sum is 1
3 was odd, so 3 was added. Current sum is 4
4 was even, so 4 was not added. Current sum is 4
5 was odd, so 5 was added. Current sum is 9