Use python built-in library to solve permutation and combination problems

Cartesian product: itertools.product(*iterables[, repeat])

import itertools
for i in itertools.product('BCDEF', repeat = 2):
    print(''.join(i),end=",")
print('\n')

# 输出 BB BC BD BE BF CB CC CD CE CF DB DC DD DE DF EB EC ED EE EF FB FC FD FE FF

　　The Cartesian product of two tuples:

import itertools
a = (1, 2)
b = ('A', 'B', 'C')
c = itertools.product(a,b)
for i in c:
    print(i,end=",")

# 输出(1, 'A') (1, 'B') (1, 'C') (2, 'A') (2, 'B') (2, 'C')

 

 

排列：itertools.permutations(iterable[, r])

import itertools
for i in itertools.permutations('BCD', 2):
    print(''.join(i),end=",")
# 输出 BC BD CB CD DB DC
print('\n')

 

组合：itertools.combinations(iterable, r)

import itertools
for i in itertools.combinations('BCDEF', 2):
    print(''.join(i),end=" ")
# 输出 BC BD BE BF CD CE CF DE DF EF
print('\n')

 

Combination (including self-repetition): itertools.combinations_with_replacement(iterable, r)

import itertools
for i in itertools.combinations_with_replacement('ABC', 3):
    print (''.join(i),end=' ')

# 输出 AAA AAB AAC ABB ABC ACC BBB BBC BCC CCC
print('\n')

For example

'BCDEF五个字母组合问题'
import itertools

print("1个组合：")
for i, val in enumerate(list(itertools.combinations('BCDEF', 1))):
    print("序号：%s   值：%s" % (i + 1, ''.join(val)))
print("2个组合：")
for i, val in enumerate(list(itertools.combinations('BCDEF', 2))):
    print("序号：%s   值：%s" % (i + 1, ''.join(val)))
print("3个组合：")
for i, val in enumerate(list(itertools.combinations('BCDEF', 3))):
    print("序号：%s   值：%s" % (i + 1, ''.join(val)))
print("4个组合：")
for i, val in enumerate(list(itertools.combinations('BCDEF', 4))):
    print("序号：%s   值：%s" % (i + 1, ''.join(val)))
print("5个组合：")
for i, val in enumerate(list(itertools.combinations('BCDEF', 5))):
    print("序号：%s   值：%s" % (i + 1, ''.join(val)))