更改字符串中的每个位置以找到所有可能的组合

Question

我们的目标是找到所有可能的组合，如果我们用某些特定字符更改字符串中的一个位置。

这是我目前的代码：

from itertools import product

string = "abc123"

char = "def2346789"

for i in range(len(string)):
    for char1 in product(char):
        print(string[:i] + char1 + string[i+1:])

char表示将在每个地方使用的字符。

但我对它在第二部分中的工作方式感到困惑，将乐于探索解决方案。

所以结果会是这样的：

dbc123 adc123
ebc123 aec123
fbc123 afc123 
2bc123 a2c123 
3bc123 a3c123
4bc123 a4c123
6bc123 a5c123
7bc123 a6c123
8bc123 a7c123
9bc123 a8c123 etc...

Answer 1

第二部分，我假设你指的是 for 循环。

基本上是：

for every character position in string (called i):
    for every character in char (called char1):
        substitute the character at the current position with char1
        #This is done by printing all characters before the current position 
        #(string[:i]), then printing the new character (char1) then printing all 
        #charactrers after the current position (string[i+1:])

Answer 2

您不需要（不应该）在这里使用 itertools.product。这可能就是让您感到困惑的地方：

string = "abc123"

char = "def2346789"

for i in range(len(string)):
    for c in char:
        print(string[:i] + c + string[i+1:])

Answer 3

使用函数和推导也使代码更具可读性

     string = "abc123"
     char = "def2346789"

     def fncCombinations(a):
           combination_list=[]
           for j in range(len(string)): 
                combination_list.append(string[:j] + a + string[j+1:])
     return combination_list 
     combination_list=[fncCombinations(a) for a in char]
     print(combination_list)

`

Answer 4

你可以使用栈和队列

string="abc123"
char = "def2346789"
queue=[]
stack=[]
combination_list=[]
queue=[x for x in string]
    
for index in range(len(queue)):
    stack.append(queue.pop(0))
    front="".join(stack[:-1])
    back="".join(queue)
    combination_list.append([front+a+back for a in char])

 print(combination_list)

更改字符串中的每个位置以找到所有可能的组合

Change each place in a string to find all possible combinations

python

combinations

product

itertools

python-3.x