Python 对元组列表中的每个值求和
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/14180866/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
sum each value in a list of tuples
提问by Inbar Rose
I have a list of tuples similar to this:
我有一个与此类似的元组列表：
l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]
I want to create a simple one-liner that will give me the following result:
我想创建一个简单的单线，它会给我以下结果：
r = (25, 20) or r = [25, 20] # don't care if tuple or list.
Which would be like doing the following:
这就像执行以下操作：
r = [0, 0]
for t in l:
r[0]+=t[0]
r[1]+=t[1]
I am sure it is something very simple, but I can't think of it.
我相信这是非常简单的事情，但我想不出来。
Note: I looked at similar questions already:
注意：我已经看过类似的问题：
How do I sum the first value in a set of lists within a tuple?
How do I sum the first value in each tuple in a list of tuples in Python?
采纳答案by Ashwini Chaudhary
Use zip()and sum():
使用zip()和sum()：
In [1]: l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]
In [2]: [sum(x) for x in zip(*l)]
Out[2]: [25, 20]
or:
或者：
In [4]: map(sum, zip(*l))
Out[4]: [25, 20]
timeitresults:
timeit结果：
In [16]: l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]*1000
In [17]: %timeit [sum(x) for x in zip(*l)]
1000 loops, best of 3: 1.46 ms per loop
In [18]: %timeit [sum(x) for x in izip(*l)] #prefer itertools.izip
1000 loops, best of 3: 1.28 ms per loop
In [19]: %timeit map(sum, zip(*l))
100 loops, best of 3: 1.48 ms per loop
In [20]: %timeit map(sum, izip(*l)) #prefer itertools.izip
1000 loops, best of 3: 1.29 ms per loop
回答by Alessandro Ruffolo
I want to add something to the given answer:
我想在给定的答案中添加一些内容：
If I have an array of dict e.g.
如果我有一个字典数组，例如
l = [{'quantity': 10, 'price': 5},{'quantity': 6, 'price': 15},{'quantity': 2, 'price': 3},{'quantity': 100, 'price': 2}]
and i want to obtain two (or more) sums of calculated quantity over the values e.g. sum of quantities and of price*quantity
我想获得两个（或更多）计算数量的总和，例如数量和价格*数量的总和
I can do:
我可以：
(total_quantity, total_price) = (
sum(x) for x in zip(*((item['quantity'],
item['price'] * item['quantity'])
for item in l)))
Instead of:
代替：
total_quantity = 0
total_price = 0
for item in l:
total_quantity += item['quantity']
total_price += item['price'] * item['quantity']
Maybe the first solution is less readable, but is more "pythonesque" :)
也许第一个解决方案可读性较差，但更“pythonesque”:)
回答by cipilo
Without using zip
不使用zip
sum(e[0] for e in l), sum(e[1] for e in l)
