Python 访问列表中元组中的值

Accessing a value in a tuple that is in a list

提问by super9

[(1,2), (2,3), (4,5), (3,4), (6,7), (6,7), (3,8)]

How do I return the 2nd value from each tuple inside this list?

如何从此列表中的每个元组返回第二个值？

Desired output:

期望的输出：

[2, 3, 5, 4, 7, 7, 8]

采纳答案by Ignacio Vazquez-Abrams

With a list comprehension.

具有列表理解能力。

[x[1] for x in L]

回答by gary

Ignacio's answer is what you want. However, as someone also learning Python, let me try to dissect it for you... As mentioned, it is a list comprehension (covered in DiveIntoPython3, for example). Here are a few points:

Ignacio 的答案正是您想要的。然而，作为一个也在学习 Python 的人，让我试着为你剖析它......如前所述，它是一个列表理解（例如在DiveIntoPython3 中介绍）。这里有几点：

[x[1] for x in L]

[x[1] for x in L]

• Notice the []'s around the line of code. These are what define a list. This tells you that this code returns a list, so it's of the listtype. Hence, this technique is called a "list comprehension."
• L is your original list. So you should define L = [(1,2),(2,3),(4,5),(3,4),(6,7),(6,7),(3,8)]prior to executing the above code.
• xis a variable that only exists in the comprehension - try to access xoutside of the comprehension, or type type(x)after executing the above line and it will tell you NameError: name 'x' is not defined, whereas type(L)returns <class 'list'>.
• x[1]points to the seconditem in each of the tuples whereas x[0]would point to each of the first items.
• So this line of code literally reads "return the second item in a tuple for all tuples in list L."

• 请注意[]代码行周围的 。这些是定义列表的内容。这告诉您此代码返回一个列表，因此它属于该list类型。因此，这种技术被称为“列表理解”。
• L 是您的原始列表。所以你应该L = [(1,2),(2,3),(4,5),(3,4),(6,7),(6,7),(3,8)]在执行上面的代码之前定义。
• x是一个仅存在于推导中的变量 - 尝试在推导x之外访问，或type(x)在执行上述行后键入，它会告诉您NameError: name 'x' is not defined，而type(L)返回<class 'list'>.
• x[1]指向每个元组中的第二个项目，而x[0]将指向每个第一个项目。
• 所以这行代码字面意思是“为列表 L 中的所有元组返回元组中的第二项”。

It's tough to tell how much you attempted the problem prior to asking the question, but perhaps you just weren't familiar with comprehensions? I would spend some time reading through Chapter 3 of DiveIntoPython, or any resource on comprehensions. Good luck.

在提出问题之前很难判断您尝试了多少问题，但也许您只是不熟悉理解？我会花一些时间阅读DiveIntoPython 的第 3 章，或任何有关理解的资源。祝你好运。

回答by aaronasterling

A list comprehension is absolutely the way to do this. Another way that shouldbe faster is mapand itemgetter.

列表理解绝对是做到这一点的方法。另一种应该更快的方法是mapand itemgetter。

import operator

new_list = map(operator.itemgetter(1), old_list)

In response to the comment that the OP couldn't find an answer on google, I'll point out a super naive way to do it.

针对 OP 无法在 google 上找到答案的评论，我将指出一种非常幼稚的方法。

new_list = []
for item in old_list:
    new_list.append(item[1])

This uses:

这使用：

1. Declaring a variable to reference an empty list.
2. A for loop.
3. Calling the appendmethod on a list.

1. 声明一个变量来引用一个空列表。
2. 一个 for 循环。
3. 调用append列表上的方法。

If somebody is trying to learn a language and can't put together these basic pieces for themselves, then they need to view it as an exerciseand do it themselves even if it takes twenty hours.

如果有人正在尝试学习一门语言并且无法为自己整理这些基本部分，那么他们需要将其视为练习并自己做，即使需要二十个小时。

One needs to learn how to think about what one wantsand compare that to the available tools. Every element in my second answer should be covered in a basic tutorial. You cannot learn to program without reading one.

人们需要学习如何思考自己想要什么，并将其与可用工具进行比较。我的第二个答案中的每个元素都应该包含在基本教程中。不阅读一本就无法学习编程。

回答by jpp

You can also use sequence unpacking with zip:

您还可以使用序列解包zip：

L = [(1,2),(2,3),(4,5),(3,4),(6,7),(6,7),(3,8)]

_, res = zip(*L)

print(res)

# (2, 3, 5, 4, 7, 7, 8)

This also creates a tuple _from the discarded first elements. Extracting only the second is possible, but more verbose:

这也会_从丢弃的第一个元素创建一个元组。仅提取第二个是可能的，但更详细：

from itertools import islice

res = next(islice(zip(*L), 1, None))

回答by U10-Forward

OR you can use pandas:

或者你可以使用pandas：

>>> import pandas as pd
>>> L = [(1,2),(2,3),(4,5),(3,4),(6,7),(6,7),(3,8)]
>>> df=pd.DataFrame(L)
>>> df[1]
0    2
1    3
2    5
3    4
4    7
5    7
6    8
Name: 1, dtype: int64
>>> df[1].tolist()
[2, 3, 5, 4, 7, 7, 8]
>>> 

Or numpy:

或者numpy：

>>> import numpy as np
>>> L = [(1,2),(2,3),(4,5),(3,4),(6,7),(6,7),(3,8)]
>>> arr=np.array(L)
>>> arr.T[1]
array([2, 3, 5, 4, 7, 7, 8])
>>> arr.T[1].tolist()
[2, 3, 5, 4, 7, 7, 8]
>>>