python - >数字和字母的组合

tek*_*agi 0 python arrays string random choice

#!/usr/bin/python
import random
lower_a = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
upper_a = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
num = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']

all = []
all = " ".join("".join(lower_a) + "".join(upper_a) + "".join(num))
all = all.split()
x = 1
c = 1
while x < 10:
        y = []
        for i in range(c):
                a = random.choice(all)
                y.append(a)
        print "".join(y)
        x += 1
        c += 1

我现在有什么输出如下:

5
hE
HAy
1kgy
Pt6JM
2pFuCb
Jv5osaX
5q8PwWAO
SvHWRKfI5

如何让它系统地遍历给定长度的每个字母组合(大写和小写),然后在该长度上加1并重复该过程?

nak*_*tic 5

最好不要重新创建标准库中已有的功能.

看一下标准库模块"itertools".

特别是combination(),permutations()和product()函数.

import itertools

lower_a = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
upper_a = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
num = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']

all = []
all = lower_a + upper_a + num

for r in range(1, 3):
    for s in itertools.product(all, repeat=r):
         print ''.join(s)
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如果您的Python版本较旧,则可能无法访问这些功能.但是,如果您查看Python 2.6的文档,您可以看到如何在Python中实现所有这些函数.例如,itertools.product的实现如下:

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)
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您也可以尝试使用递归解决方案:

lower_a = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
upper_a = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
num = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']

all = []
all = lower_a + upper_a + num

def recursive_product(myList, length, myString = ""):
    if length == 0:
        print myString
        return
    for c in myList:
        recursive_product(myList, length-1, myString + c)

for r in range(1, 3):
    recursive_product(all, r)
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