如何在Python中的元组列表中对每个元组中的第一个值求和？

I have a list of tuples (always pairs) like this:

我有一个像这样的元组列表(总是对):

[(0, 1), (2, 3), (5, 7), (2, 1)]

I'd like to find the sum of the first items in each pair, i.e.:

我想找到每对中第一项的总和,即:

0 + 2 + 5 + 2

How can I do this in Python? At the moment I'm iterating through the list:

我怎么能用Python做到这一点?目前我正在遍历列表:

sum = 0
for pair in list_of_pairs:
   sum += pair[0]

I have a feeling there must be a more Pythonic way.

我觉得必须有更多的Pythonic方式。

7 个解决方案

#1

A version compatible with Python 2.3 is

与Python 2.3兼容的版本是

sum([pair[0] for pair in list_of_pairs])

or in recent versions of Python, see this answer or this one.

或者在Python的最新版本中,请参阅此答案或此答案。

#2

sum(i for i, j in list_of_pairs)

will do too.

也会这样做。

#3

I recommend:

sum(i for i, _ in list_of_pairs)

Note:

Using the variable _(or __ to avoid confliction with the alias of gettext) instead of j has at least two benefits:

使用变量_(或__以避免与gettext的别名冲突)而不是j具有至少两个好处:

_(which stands for placeholder) has better readability

_(代表占位符)具有更好的可读性

pylint won't complain: "Unused variable 'j'"

pylint不会抱怨:“未使用的变量'j'”

#4

If you have a very large list or a generator that produces a large number of pairs you might want to use a generator based approach. For fun I use itemgetter() and imap(), too. A simple generator based approach might be enough, though.

如果您有一个非常大的列表或生成大量对的生成器,您可能希望使用基于生成器的方法。为了好玩,我也使用itemgetter()和imap()。但是,基于简单生成器的方法可能就足够了。

import operator
import itertools
idx0 = operator.itemgetter(0)
list_of_pairs = [(0, 1), (2, 3), (5, 7), (2, 1)]
sum(itertools.imap(idx0, list_of_pairs)

Edit: itertools.imap() is available in Python 2.3. So you can use a generator based approach there, too.

编辑:itertools.imap()在Python 2.3中可用。所以你也可以在那里使用基于生成器的方法。

#5

Obscure (but fun) answer:

模糊(但有趣)答案:

>>> sum(zip(*list_of_pairs)[0])
9

Or when zip's are iterables only this should work:

或者当zip是迭代时,这应该是有用的:

>>> sum(zip(*list_of_pairs).__next__())
9

#6

Below is sample code, you can also specify the list range.

下面是示例代码,您还可以指定列表范围。

def test_lst_sum():
    lst = [1, 3, 5]
    print sum(lst)  # 9
    print sum(lst[1:])  # 8

    print sum(lst[5:])  # 0  out of range so return 0
    print sum(lst[5:-1])  # 0

    print sum(lst[1: -1])  # 3

    lst_tp = [('33', 1), ('88', 2), ('22', 3), ('44', 4)]
    print sum(x[1] for x in lst_tp[1:])  # 9

#7

If you don't mind converting it to a numpy array, you can use np.sum over axis=0 as given here

如果你不介意将它转换为numpy数组,你可以使用np.sum over axis = 0,如下所示

#1