Python 如何获得n个二进制值的所有组合？
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How to get all combination of n binary value?
提问by LWZ
In Python, how can I get all combinations of nbinary values 0and 1?
在 Python 中，如何获得n二进制值0和的所有组合1？
For example, if n = 3, I want to have
例如，如果n = 3，我想要
[ [0,0,0], [0,0,1], [0,1,0], [0,1,1], ... [1,1,1] ] #total 2^3 combinations
How can I do this?
我怎样才能做到这一点？
采纳答案by Volatility
import itertools
lst = list(itertools.product([0, 1], repeat=3))
This will yield a list of tuples (see here)
这将产生一个元组列表（见这里）
You can easily change this to use a variable repeat:
您可以轻松地将其更改为使用变量repeat：
n = 3
lst = list(itertools.product([0, 1], repeat=n))
If you need a list of lists, then you can use the mapfunction (thanks @Aesthete).
如果您需要列表列表，则可以使用该map功能（感谢@Aesthete）。
lst = map(list, itertools.product([0, 1], repeat=n))
Or in Python 3:
或者在 Python 3 中：
lst = list(map(list, itertools.product([0, 1], repeat=n)))
# OR
lst = [list(i) for i in itertools.product([0, 1], repeat=n)]
Note that using mapor a list comprehension means you don't need to convert the product into a list, as it will iterate through the itertools.productobject and produce a list.
请注意，使用map或 列表推导式意味着您不需要将产品转换为列表，因为它会遍历itertools.product对象并生成列表。
回答by MoveFast
Following will give you all such combinations
以下将为您提供所有此类组合
bin = [0,1]
[ (x,y,z) for x in bin for y in bin for z in bin ]
回答by Anil
Without using any in-build functions or smart techniques we can get like this.
无需使用任何内置函数或智能技术，我们就可以得到这样的结果。
def per(n):
for i in range(1<<n):
s=bin(i)[2:]
s='0'*(n-len(s))+s
print (map(int,list(s)))
per(3)
output
输出
[0, 0, 0]
[0, 0, 1]
[0, 1, 0]
[0, 1, 1]
[1, 0, 0]
[1, 0, 1]
[1, 1, 0]
[1, 1, 1]
