Python 如何将两个列表合并为一个列表？
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/3471999/
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
How do I merge two lists into a single list?
提问by sureshvv
I have
我有
a = [1, 2]
b = ['a', 'b']
I want
我想要
c = [1, 'a', 2, 'b']
采纳答案by John La Rooy
[j for i in zip(a,b) for j in i]
回答by Mark Byers
If the order of the elements much match the order in your example then you can use a combination of zipand chain:
如果元素的顺序与您的示例中的顺序非常匹配，那么您可以使用zip和chain的组合：
from itertools import chain
c = list(chain(*zip(a,b)))
If you don't care about the order of the elements in your result then there's a simpler way:
如果您不关心结果中元素的顺序，那么有一种更简单的方法：
c = a + b
回答by Nick T
If you care about order:
如果您关心订单：
#import operator
import itertools
a = [1,2]
b = ['a','b']
#c = list(reduce(operator.add,zip(a,b))) # slow.
c = list(itertools.chain.from_iterable(zip(a,b))) # better.
print cgives [1, 'a', 2, 'b']
print c给 [1, 'a', 2, 'b']
回答by John Machin
Parsing
解析
[j for i in zip(a,b) for j in i]
in your head is easy enough if you recall that the forand ifclauses are done in order, followed a final append of the result:
如果你记得forandif子句是按顺序完成的，然后是结果的最后附加，那么在你的头脑中就很容易了：
temp = []
for i in zip(a, b):
for j in i:
temp.append(j)
and would be easier had it have been written with more meaningful variable names:
如果用更有意义的变量名编写它会更容易：
[item for pair in zip(a, b) for item in pair]
回答by SeanM
An alternate method using index slicing which turns out to be faster and scales better than zip:
一种使用索引切片的替代方法，结果证明它比 zip 更快且扩展性更好：
def slicezip(a, b):
result = [0]*(len(a)+len(b))
result[::2] = a
result[1::2] = b
return result
You'll notice that this only works if len(a) == len(b)but putting conditions to emulate zip will not scale with a or b.
您会注意到，这仅适用len(a) == len(b)于模拟 zip 的条件不会随 a 或 b 缩放的情况。
For comparison:
比较：
a = range(100)
b = range(100)
%timeit [j for i in zip(a,b) for j in i]
100000 loops, best of 3: 15.4 μs per loop
%timeit list(chain(*zip(a,b)))
100000 loops, best of 3: 11.9 μs per loop
%timeit slicezip(a,b)
100000 loops, best of 3: 2.76 μs per loop
回答by Paul Fabing
def main():
drinks = ["Johnnie Walker", "Jose Cuervo", "Jim Beam", "Hyman Daniels,"]
booze = [1, 2, 3, 4, 5]
num_drinks = []
x = 0
for i in booze:
if x < len(drinks):
num_drinks.append(drinks[x])
num_drinks.append(booze[x])
x += 1
else:
print(num_drinks)
return
main()
主要的（）
回答by Arvind S.P
c = []
c.extend(a)
c.extend(b)
回答by Samer Aamar
Here is a standard / self-explaining solution, i hope someone will find it useful:
这是一个标准/不言自明的解决方案，我希望有人会发现它有用：
a = ['a', 'b', 'c']
b = ['1', '2', '3']
c = []
for x, y in zip(a, b):
c.append(x)
c.append(y)
print (c)
output:
输出：
['a', '1', 'b', '2', 'c', '3']
Of course, you can change it and do manipulations on the values if needed
当然，如果需要，您可以更改它并对值进行操作
