python排列組合

1. 如何用python列出N個數字的所有排列組合

>> from itertools import combinations, permutations
>> permutations([1, 2, 3], 2)
<itertools.permutations at 0x7febfd880fc0>
# 可迭代對象
>> list(permutations([1, 2, 3], 2)) #排列
[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]
>> list(combinations([1, 2, 3], 2)) #組合
[(1, 2), (1, 3), (2, 3)]

2. 怎麼樣讓python計算任意六位數的排列組合

from itertools import proct
for i in proct(range(2),repeat=6):
print i

3. 如何在python中列出由一定個數0和1組成的list的排列組合

importitertools
src=[0,0,1]
printset(itertools.permutations(src,len(src)))

這個是正解！

4. Python實現的排列組合計算操作示例

Python實現的排列組合計算操作示例
本文實例講述了Python實現的排列組合計算操作。分享給大家供大家參考，具體如下：
1. 調用 scipy 計算排列組合的具體數值
>> from scipy.special import comb, perm
>> perm(3, 2)
6.0
>> comb(3, 2)
3.0
2. 調用 itertools 獲取排列組合的全部情況數
>> from itertools import combinations, permutations
>> permutations([1, 2, 3], 2)
<itertools.permutations at 0x7febfd880fc0>
# 可迭代對象
>> list(permutations([1, 2, 3], 2))
[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]
>> list(combinations([1, 2, 3], 2))
[(1, 2), (1, 3), (2, 3)]

5. python 實現6個數的排列組合,每個數可選值為0和1

fromitertoolsimportproct
foriinproct(range(2),repeat=6):
printi

6. python新手問題：排列組合與輸出最大數

python2.7語法

第一個問題:

7. Python取隨機數排列組合

沒懂你的意思，既然是排列組合，就是針對3個固定元素。
如果3個元素本身就是隨機的，就不用排列了，下面的getplans函數也就沒有意義，直接循環2-5行代碼就好。

8. 通過python實現0,1,2,3四個數排列組合可以組成多少個數

##縮進格式看圖

import itertools

c=0

for e in itertools.permutations('0123',4):

print(''.join(e).lstrip('0'),end=',')

c+=1

print(' 共%d個'%(c))

9. 在python中如何實現列表中元素的所有排列組合如輸入為['1','2','3']和2輸出為['

#!/usr/bin/python
#Two method for generate a list whose item is all possible permutation and combination come from every item of many list.

A = ['1', '2']
B = ['a', 'b', 'c']
C = ['A', 'B', 'C', 'D']

retList = []
for a in A:
for b in B:
for c in C:
retList.append((a,b,c))
print retList

print '*' * 40

def myfunc(*lists):
#list all possible composition from many list, each item is a tuple.
#Here lists is [list1, list2, list3], return a list of [(item1,item2,item3),...]

#len of result list and result list.
total = rece(lambda x, y: x * y, map(len, lists))
retList = []

#every item of result list.
for i in range(0, total):
step = total
tempItem = []
for l in lists:
step /= len(l)
tempItem.append(l[i/step % len(l)])
retList.append(tuple(tempItem))

return retList

print myfunc(A,B,C)