用指定的字符集替换字符串中的两个字符以创建所有可能的组合

Question

目标是通过更改两个字符来生成具有字符串0a1b2c3d4e5的所有可能组合

到目前为止，我掌握的代码是：

import itertools

pattern_index_zero = "xa1b2c3d4e5"

a = [['0','1', '2', '3', '4', '5', 'a', 'b', 'c', 'd', 'e'] if (c == 'x') else c for c in pattern_index_zero]
b = [''.join(lst) for lst in list(itertools.product(*a))]
for s in b:
    print(s)

输出：

0a1b2c3d4e5
1a1b2c3d4e5
2a1b2c3d4e5
3a1b2c3d4e5
4a1b2c3d4e5
5a1b2c3d4e5
aa1b2c3d4e5
ba1b2c3d4e5
ca1b2c3d4e5
da1b2c3d4e5
ea1b2c3d4e5

它改变了第一个字符，这是可以的，但问题是我想改变第二个，第三个字符，并继续到最后一个

最终目标是生成所有可能的组合来更改字符串中的两个字符，例如：

xx1b2c3d4e5
xaxb2c3d4e5
xa1x2c3d4e5
xa1bxc3d4e5
....
xx1b2c3d4e5
0xxb2c3d4e5
0x1x2c3d4e5
0x1bxc3d4e5

如果x可以是0,1,2,3,4,5,a,b,c,d,e，那么在代码方面，将x的数量增加到3甚至4有多难

EDIT用户Timus有答案。我会尽我所能去探索所有的答案，谢谢大家

Answer 1

您可以从itertools尝试其他工具：

from itertools import combinations

s = "0a1b2c3d4e5"
positions = tuple(range(len(s)))
for i, j in combinations(positions, 2):
    print(s[:i] + s[j] + s[i+1:j] + s[i] + s[j+1:])

编辑：好的，再试一次：）。这就是你想要的吗：

from itertools import combinations, product

s = "0a1b2c3d4e5"
r = "012345abcde"
for i, j in combinations(tuple(range(len(s))), 2):
    for r1, r2 in product(r, r):
        print(s[:i] + r1 + s[i+1:j] + r2 + s[j+1:])

编辑2：如果要排除倍数，可以通过^{统一：

from itertools import combinations, product

s = "0a1b2c3d4e5"
r = "012345abcde"
output = []
for i, j in combinations(tuple(range(len(s))), 2):
    for r1, r2 in product(r, r):
        output.append(s[:i] + r1 + s[i+1:j] + r2 + s[j+1:])
output = set(output)
print(output)

编辑3：我想不出你的问题，这是如何扩展到更高数量的替换。我希望你不介意我添加一个概括。这有点复杂。而且产量增长很快。因此，我没有将其打印到控制台，而是使用了一个文件（带有详细的输出）。也许你感兴趣。没问题，如果不是的话，我玩得很开心。一个问题：你需要这些东西来解决现实世界的问题吗

from itertools import combinations, product

# Base string
base = "0a1b2c3d4e5"
# Length of the base string
len_of_base = len(base)
# String with replacement characters
replacements = "012345abcde"
# Number of replacements
num_of_replacements = 3
output = {}
for istart, *imiddle, iend in combinations(range(len_of_base),
                                           num_of_replacements):
    for rstart, *rrest in product(replacements,
                                  repeat=num_of_replacements):
        new = base[:istart] + rstart
        for i, j, r in zip([istart] + imiddle,
                           imiddle + [iend],
                           rrest):
            new += base[i+1:j] + r
        new += base[iend+1:]
        if new not in output:
            output[new] = {}
            output[new]['positions'] = tuple([istart] + imiddle + [iend])
            output[new]['replacements'] = tuple([rstart] + rrest)

with open('data.txt', 'w') as file:
    for key in output:
        file.write(f"{key} - "
                   f"Positions {output[key]['positions']} "
                   f"replaced with {output[key]['replacements']}\n")

输出：

000b2c3d4e5 - Positions (0, 1, 2) replaced with ('0', '0', '0')
001b2c3d4e5 - Positions (0, 1, 2) replaced with ('0', '0', '1')
002b2c3d4e5 - Positions (0, 1, 2) replaced with ('0', '0', '2')
003b2c3d4e5 - Positions (0, 1, 2) replaced with ('0', '0', '3')
004b2c3d4e5 - Positions (0, 1, 2) replaced with ('0', '0', '4')
005b2c3d4e5 - Positions (0, 1, 2) replaced with ('0', '0', '5')
00ab2c3d4e5 - Positions (0, 1, 2) replaced with ('0', '0', 'a')
00bb2c3d4e5 - Positions (0, 1, 2) replaced with ('0', '0', 'b')
00cb2c3d4e5 - Positions (0, 1, 2) replaced with ('0', '0', 'c')
...