Python 将可迭代对象的所有元素添加到列表
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/20592808/
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
Add all elements of an iterable to list
提问by Bartlomiej Lewandowski
Is there a more concise way of doing the following?
是否有更简洁的方法来执行以下操作？
t = (1,2,3)
t2 = (4,5)
l.addAll(t)
l.addAll(t2)
print l # [1,2,3,4,5]
This is what I have tried so far: I would prefer to avoid passing in the list in the parameters.
这是我到目前为止所尝试的：我宁愿避免在参数中传递列表。
def t_add(t,stuff):
for x in t:
stuff.append(x)
采纳答案by Martijn Pieters
Use list.extend(), not list.append()to add all items from an iterable to a list:
使用list.extend(), 不要list.append()将可迭代对象中的所有项目添加到列表中：
l.extend(t)
l.extend(t2)
or
或者
l.extend(t + t2)
or even:
甚至：
l += t + t2
where list.__iadd__(in-place add) is implemented as list.extend()under the hood.
其中list.__iadd__（就地添加）list.extend()在引擎盖下实现。
Demo:
演示：
>>> l = []
>>> t = (1,2,3)
>>> t2 = (4,5)
>>> l += t + t2
>>> l
[1, 2, 3, 4, 5]
If, however, you just wanted to create a list of t + t2, then list(t + t2)would be the shortest path to get there.
但是，如果您只想创建 的列表t + t2，那么list(t + t2)将是到达那里的最短路径。
回答by thefourtheye
stuff.extendis what you want.
stuff.extend是你想要的。
t = [1,2,3]
t2 = [4,5]
t.extend(t2)
# [1, 2, 3, 4, 5]
Or you can do
或者你可以做
t += t2
