I have a bunch of lists I want to append to a single list that is sort of the "main" list in a program I'm trying to write. Is there a way to do this in one line of code rather than like 10? I'm a beginner so I have no idea...
我有一堆列表要添加到单个列表中，这个列表是我正在尝试编写的程序中的“主要”列表。有没有办法在一行代码而不是10代码中执行此操作？我是初学者，所以我不知道......
For a better picture of my question, what if I had these lists:
为了更好地了解我的问题，如果我有这些列表怎么办：
x = [1, 2, 3]
y = [4, 5, 6]
z = [7, 8, 9]
And want to append y and z to x. Instead of doing:
并希望将y和z附加到x。而不是做：
x.append(y)
x.append(z)
Is there a way to do this in one line of code? I already tried:
有没有办法在一行代码中执行此操作？我已经尝试过：
x.append(y, z)
And it wont work.
它不会工作。
6 个解决方案
#1
28
x.extend(y+z)
should do what you want
应该做你想做的事
or
要么
x += y+z
or even
甚至
x = x+y+z
#2
14
Extending my comment
扩展我的评论
In [1]: x = [1, 2, 3]
In [2]: y = [4, 5, 6]
In [3]: z = [7, 8, 9]
In [4]: from itertools import chain
In [5]: print list(chain(x,y,z))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
#3
5
You can use sum function with start value (empty list) indicated:
您可以使用sum函数和指示的起始值（空列表）：
a = sum([x, y, z], [])
This is especially more suitable if you want to append an arbitrary number of lists.
如果要附加任意数量的列表，这尤其适用。
#4
0
equivalent to above answer, but sufficiently different to be worth a mention:
相当于上面的答案，但足够不同值得一提：
my_map = {
'foo': ['a', 1, 2],
'bar': ['b', '2', 'c'],
'baz': ['d', 'e', 'f'],
}
list(itertools.chain(*my_map.values()))
['d', 'e', 'f', 'a', 1, 2, 'b', '2', 'c']
in the above expression, * is important for groking result as args to chain, this is same as prior chain(x,y,z). Also, note the result is hash-ordered.
在上面的表达式中，*对于将结果作为args转换为链是重要的，这与先前的链（x，y，z）相同。另外，请注意结果是哈希排序的。
#5
0
If you prefer a slightly more functional approach, you could try:
如果您更喜欢功能稍强的方法，可以尝试：
import functools as f
x = [1, 2, 3]
y = [4, 5, 6]
z = [7, 8, 9]
x = f.reduce(lambda x, y: x+y, [y, z], x)
This will enable you to concatenate any number of lists onto list x.
这将使您能够将任意数量的列表连接到列表x上。
If you would just like to concatenate any number of lists together (i.e. not onto some base list), you can simplify to:
如果您只想将任意数量的列表连接在一起（即不在某些基本列表中），您可以简化为：
import functools as f
from operator import add
big_list = f.reduce(add, list_of_lists)
Take note that our BFDL has his reservations with regard to lambdas, reduce, and friends: https://www.artima.com/weblogs/viewpost.jsp?thread=98196
请注意，我们的BFDL对lambdas，reduce和朋友有所保留：https：//www.artima.com/weblogs/viewpost.jsp？thread = 98196
To complete this answer, you can read more about reduce in the documentation: https://docs.python.org/3/library/functools.html#functools.reduce
要完成此答案，您可以在文档中阅读有关reduce的更多信息：https：//docs.python.org/3/library/functools.html#functools.reduce
I quote: "Apply function of two arguments cumulatively to the items of sequence, from left to right, so as to reduce the sequence to a single value."
我引用：“将两个参数的函数累加到序列项中，从左到右，以便将序列减少为单个值。”
P.S. Using sum() as described in https://*.com/a/41752487/532513 is super compact, it does seem to work with lists, and is really fast (see https://*.com/a/33277438/532513 ) but help(sum) in Python 3.6 has the following to say:
附：使用https://*.com/a/41752487/532513中描述的sum（）非常紧凑，它似乎与列表一起使用，并且非常快（请参阅https://*.com/a/33277438/ 532513）但Python 3.6中的帮助（总和）有以下说法：
This function is intended specifically for use with numeric values and may reject non-numeric types.
此函数专门用于数值，可能会拒绝非数字类型。
Although this is slightly worrying, I will probably keep it as my first option for concatenating lists.
虽然这有点令人担忧，但我可能会将它作为连接列表的第一选择。
#6
0
In one line , it can be done in following ways
在一行中，可以通过以下方式完成
x.extend(y+z)
or
要么
x=x+y+z
#1
28
x.extend(y+z)
should do what you want
应该做你想做的事
or
要么
x += y+z
or even
甚至
x = x+y+z
#2
14
Extending my comment
扩展我的评论
In [1]: x = [1, 2, 3]
In [2]: y = [4, 5, 6]
In [3]: z = [7, 8, 9]
In [4]: from itertools import chain
In [5]: print list(chain(x,y,z))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
#3
5
You can use sum function with start value (empty list) indicated:
您可以使用sum函数和指示的起始值（空列表）：
a = sum([x, y, z], [])
This is especially more suitable if you want to append an arbitrary number of lists.
如果要附加任意数量的列表，这尤其适用。
#4
0
equivalent to above answer, but sufficiently different to be worth a mention:
相当于上面的答案，但足够不同值得一提：
my_map = {
'foo': ['a', 1, 2],
'bar': ['b', '2', 'c'],
'baz': ['d', 'e', 'f'],
}
list(itertools.chain(*my_map.values()))
['d', 'e', 'f', 'a', 1, 2, 'b', '2', 'c']
in the above expression, * is important for groking result as args to chain, this is same as prior chain(x,y,z). Also, note the result is hash-ordered.
在上面的表达式中，*对于将结果作为args转换为链是重要的，这与先前的链（x，y，z）相同。另外，请注意结果是哈希排序的。
#5
0
If you prefer a slightly more functional approach, you could try:
如果您更喜欢功能稍强的方法，可以尝试：
import functools as f
x = [1, 2, 3]
y = [4, 5, 6]
z = [7, 8, 9]
x = f.reduce(lambda x, y: x+y, [y, z], x)
This will enable you to concatenate any number of lists onto list x.
这将使您能够将任意数量的列表连接到列表x上。
If you would just like to concatenate any number of lists together (i.e. not onto some base list), you can simplify to:
如果您只想将任意数量的列表连接在一起（即不在某些基本列表中），您可以简化为：
import functools as f
from operator import add
big_list = f.reduce(add, list_of_lists)
Take note that our BFDL has his reservations with regard to lambdas, reduce, and friends: https://www.artima.com/weblogs/viewpost.jsp?thread=98196
请注意，我们的BFDL对lambdas，reduce和朋友有所保留：https：//www.artima.com/weblogs/viewpost.jsp？thread = 98196
To complete this answer, you can read more about reduce in the documentation: https://docs.python.org/3/library/functools.html#functools.reduce
要完成此答案，您可以在文档中阅读有关reduce的更多信息：https：//docs.python.org/3/library/functools.html#functools.reduce
I quote: "Apply function of two arguments cumulatively to the items of sequence, from left to right, so as to reduce the sequence to a single value."
我引用：“将两个参数的函数累加到序列项中，从左到右，以便将序列减少为单个值。”
P.S. Using sum() as described in https://*.com/a/41752487/532513 is super compact, it does seem to work with lists, and is really fast (see https://*.com/a/33277438/532513 ) but help(sum) in Python 3.6 has the following to say:
附：使用https://*.com/a/41752487/532513中描述的sum（）非常紧凑，它似乎与列表一起使用，并且非常快（请参阅https://*.com/a/33277438/ 532513）但Python 3.6中的帮助（总和）有以下说法：
This function is intended specifically for use with numeric values and may reject non-numeric types.
此函数专门用于数值，可能会拒绝非数字类型。
Although this is slightly worrying, I will probably keep it as my first option for concatenating lists.
虽然这有点令人担忧，但我可能会将它作为连接列表的第一选择。
#6
0
In one line , it can be done in following ways
在一行中，可以通过以下方式完成
x.extend(y+z)
or
要么
x=x+y+z