标签:itertools too rod rto tools ble bbb aac join
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
两个元组进行笛卡尔积:
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‘)
import itertools
for i in itertools.permutations(‘BCD‘, 2):
print(‘‘.join(i),end=",")
# 输出 BC BD CB CD DB DC
print(‘\n‘)
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‘)
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‘)
‘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)))
标签:itertools too rod rto tools ble bbb aac join
原文地址:https://www.cnblogs.com/xiao-apple36/p/10861830.html