Python 更改字符串中的每个位置以查找所有可能的组合
我们的目标是找到所有可能的组合,如果我们用一些特定的字符改变字符串中的一个位置 以下是我目前掌握的代码:Python 更改字符串中的每个位置以查找所有可能的组合,python,python-3.x,combinations,product,itertools,Python,Python 3.x,Combinations,Product,Itertools,我们的目标是找到所有可能的组合,如果我们用一些特定的字符改变字符串中的一个位置 以下是我目前掌握的代码: 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
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
表示将在每个位置使用的字符
但我不知道它将如何工作在第二部分,将享受探索解决方案
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...
对于第二部分,我假设您指的是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:])
您不(不应该)需要在此处使用itertools.product
。这可能就是让你困惑的地方:
string = "abc123"
char = "def2346789"
for i in range(len(string)):
for c in char:
print(string[:i] + c + string[i+1:])
使用函数和理解也可以使代码更具可读性
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)
`您可以使用堆栈和队列
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)
您能否澄清哪一部分是“第二部分”@Moosefeather对于loopsCan,您提供了一些输出示例?