在Python中获取列表中每个元组的第一个元素
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/22412258/
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
Get the first element of each tuple in a list in Python
提问by Creak
An SQL query gives me a list of tuples, like this:
SQL 查询为我提供了一个元组列表，如下所示：
[(elt1, elt2), (elt1, elt2), (elt1, elt2), (elt1, elt2), (elt1, elt2), ...]
I'd like to have all the first elements of each tuple. Right now I use this:
我想拥有每个元组的所有第一个元素。现在我用这个：
rows = cur.fetchall()
res_list = []
for row in rows:
res_list += [row[0]]
But I think there might be a better syntax to do it. Do you know a better way?
但我认为可能有更好的语法来做到这一点。你知道更好的方法吗？
采纳答案by Creak
Use a list comprehension:
使用列表理解：
res_list = [x[0] for x in rows]
Below is a demonstration:
下面是一个演示：
>>> rows = [(1, 2), (3, 4), (5, 6)]
>>> [x[0] for x in rows]
[1, 3, 5]
>>>
Alternately, you could use unpacking instead of x[0]:
或者，您可以使用解包而不是x[0]：
res_list = [x for x,_ in rows]
Below is a demonstration:
下面是一个演示：
>>> lst = [(1, 2), (3, 4), (5, 6)]
>>> [x for x,_ in lst]
[1, 3, 5]
>>>
Both methods practically do the same thing, so you can choose whichever you like.
这两种方法实际上做同样的事情，所以你可以选择你喜欢的。
回答by Ruben Bermudez
You can use list comprehension:
您可以使用列表理解：
res_list = [i[0] for i in rows]
This should make the trick
这应该是诀窍
回答by Aegis
res_list = [x[0] for x in rows]
c.f. http://docs.python.org/3/tutorial/datastructures.html#list-comprehensions
参见http://docs.python.org/3/tutorial/datastructures.html#list-comprehensions
For a discussion on why to prefer comprehensions over higher-order functions such as map, go to http://www.artima.com/weblogs/viewpost.jsp?thread=98196.
有关为什么更喜欢推导式而非高阶函数（例如 ）的讨论map，请访问http://www.artima.com/weblogs/viewpost.jsp?thread=98196。
回答by Jimilian
If you don't want to use list comprehension by some reasons, you can use mapand operator.itemgetter:
如果由于某些原因不想使用列表理解，可以使用map和operator.itemgetter：
>>> from operator import itemgetter
>>> rows = [(1, 2), (3, 4), (5, 6)]
>>> map(itemgetter(1), rows)
[2, 4, 6]
>>>
回答by rbonallo
The functional way of achieving this is to unzip the list using:
实现此目的的功能方法是使用以下方法解压缩列表：
sample = [(2, 9), (2, 9), (8, 9), (10, 9), (23, 26), (1, 9), (43, 44)]
first,snd = zip(*sample)
print(first,snd)
(2, 2, 8, 10, 23, 1, 43) (9, 9, 9, 9, 26, 9, 44)
(2, 2, 8, 10, 23, 1, 43) (9, 9, 9, 9, 26, 9, 44)
