单词合并

Question

我是新的用户学习Python。我有查询是否可以完成。如果用户输入一个单词假设“ dance”和字符是“ $”和“@”，则他将获得所有单词和字符组合的可能性['d $@a $@a $@n@$ c@$ e'，， d@$ a@$ n@$ c@$ e，'da $ n $ c $ e'，'d $@an $ c@$ e'，'da@nc $ e']等。组合。有可能实现吗？

import itertools as it

string = 'ABCD'  # input('enter string: ')
fill_char = '$'  # input('enter fill_char: ')

fillers = it.product(['', fill_char], repeat=len(string))

for filler in fillers:
    tmp = it.zip_longest(string, filler)
    print(''.join([''.join(char_and_fill_char) for char_and_fill_char in tmp]))


Output:
ABCD
ABCD$
ABC$D
ABC$D$
AB$CD
AB$CD$
AB$C$D
AB$C$D$
A$BCD
A$BCD$
A$BC$D
A$BC$D$
A$B$CD
A$B$CD$
A$B$C$D
A$B$C$D$

我想要3个字符的组合。如果有人能提供帮助，我会非常奇怪。

Answer 1

这是一个可能的解决方案：
对于更清晰的演示，我只选择了两个字母。基本上，将您的字符串的每个字母拿走，然后将fill_chars的所有可能组合加入其中。然后构建所有列表的产品（每个列表都包含一个字母的所有组合）。
添加了几个评论和打印语句以进行解释。

from itertools import permutations, product
string = 'AB'  # input('enter string: ')
fill_char = '$@'  # input('enter fill_char: ')

s_lst = list(string)
fill_lst = list(fill_char)
powerSet = []

for char in s_lst:
    perm_each_char=[]
    
    for k in range(len(fill_char)+1):
        tmp = [''.join(p) for p in permutations(fill_lst, k)]
        # tmp : permutation result in list, each joined together to string (e.g k=2: ['@
, '$@'])
        tmp2 = [''.join((char,elem)) for elem in tmp]
        # tmp2 : list of current char joined with each item in list of tmp (e.g. k=2: ['A@
, 'A$@'])
        perm_each_char.extend(tmp2)
        print(f"{char=} - {k=} - {perm_each_char=}")
        
    powerSet.append(perm_each_char) # list of lists, each list contains all permutations of one char of the string
    print(f"combined: {powerSet=} \n")
    
all_combinations = list(''.join(x) for x in product(*powerSet))
# now you build the itertools.product with all lists to get every combination
print(f"{all_combinations=}")

Output:

char='A' - k=0 - perm_each_char=['A']
char='A' - k=1 - perm_each_char=['A', 'A
, 'A@']
char='A' - k=2 - perm_each_char=['A', 'A
, 'A@', 'A$@', 'A@
]
combined: powerSet=[['A', 'A
, 'A@', 'A$@', 'A@
]] 

char='B' - k=0 - perm_each_char=['B']
char='B' - k=1 - perm_each_char=['B', 'B
, 'B@']
char='B' - k=2 - perm_each_char=['B', 'B
, 'B@', 'B$@', 'B@
]
combined: powerSet=[['A', 'A
, 'A@', 'A$@', 'A@
], ['B', 'B
, 'B@', 'B$@', 'B@
]] 

all_combinations=['AB', 'AB
, 'AB@', 'AB$@', 'AB@
, 'A$B', 'A$B
, 'A$B@', 'A$B$@', 'A$B@
, 'A@B', 'A@B
, 'A@B@', 'A@B$@', 'A@B@
, 'A$@B', 'A$@B
, 'A$@B@', 'A$@B$@', 'A$@B@
, 'A@$B', 'A@$B
, 'A@$B@', 'A@$B$@', 'A@$B@
]