假设我创建了一个由n个键组成的字典.每个键都映射到长度一致的整数列表.我现在想做的是一个新列表，它表示字典列表中每个点的整数之和.为了说明:
Let's assume I have a created a dict that is made up of n keys. Each key is mapped to a list of integers of a consistent length. What I want to make now is a new list that represents the sum of the integers at each point in lists of the dict. To illustrate:
my_dict = {'a': [1, 2, 3, 4], 'b': [2, 3, 4, 5], 'c': [3, 4, 5, 6]} total_sum_list = [] for key in my_dict.keys(): total_sum_list += ###some way of adding the numbers together预期输出:
total_sum_list = [6,9,12,15]如上所述，我不确定如何设置此for循环，以便可以创建类似total_sum_list的列表.我曾经尝试过对列表进行理解，但是到目前为止，我的努力还没有成功.有什么建议吗?
As demonstrated above, I am not sure how to set up this for loop so that I can create a list like total_sum_list. I have tried putting together a list comprehension, but my efforts have not been successful thus far. Any suggestions?
推荐答案您需要的是对列表进行转置，以便可以对列求和.因此，在字典值(键可以忽略)上使用zip，在列表理解中使用sum:
What you need is to transpose the lists so you can sum the columns. So use zip on the dictionary values (keys can be ignored) and sum in list comprehension:
一行:
total_sum_list = [sum(x) for x in zip(*my_dict.values())]结果:
[6, 9, 12, 15]工作原理:
zip交织值.我正在使用参数解包来传递dict值是zip的参数(如zip(a,b,c)).因此，当您这样做时:
zip interleaves the values. I'm using argument unpacking to pass the dict values are arguments to zip (like zip(a,b,c)). So when you do:
for x in zip(*my_dict.values()): print(x)您得到(如tuple):
(1, 3, 2) (2, 4, 3) (3, 5, 4) (4, 6, 5)数据随时可以求和(即使顺序不同，但我们不在乎，因为加法是可交换的:))
data are ready to be summed (even in different order, but we don't care since addition is commutative :))