Python-如何将IF语句转换为函数以与其他字符串多次调用？

Question

我需要检查另一个变量中是否存在一个变量的排列，如果不存在，则增加一个计数器变量。

我设法创建了一个带for循环的条件语句，以列出所需排列的列表并检查变量是否存在于其中。 我需要将其变成一个函数，以便我可以使代码更整洁，因为我需要该函数来检查多个变量之间的多个变量。

charList = ['A', 'B', 'C'] #actual code has other items in this list

name = "JOHN" #actual variable whose permutations are to be made for checking

checkName = "JOAN" #target variable to check against the permutations of above variable

counter = 0

if checkName is name:
    print('found1')
elif checkName in (name[:i] + c + name[i + 1:] for i in range(len(name)) for c in charList):
    print('found2')
elif checkName in ([name[:i] + c + name[i:] for i in range(len(name)) for c in charList]):
    print('found3')
elif checkName in ([name[0:i] + name[i+1] + name[i] + name[i+2:] for i in range(len(name) - 1)]):
    print('found4')
else:
    counter += 1
    print(counter)

如何使它成为一个函数，以便仅通过使用另一个名称变量就可以直接获得print语句或计数器中的增量的输出？

我只是一个初学者，所以请通过本示例帮助我理解制作函数的概念。 我必须处理两个变量，其中一个循环遍历列表，而作为菜鸟，我不知道该怎么做。

PS我只是在学习，上面的代码只是一个试运行，因此请忽略每个if语句后面的打印功能。

Answer 1

你可以这样尝试

charList = ['A', 'B', 'C'] #actual code has other items in this list

name = "JOHN" #actual variable whose permutations are to be made for checking

checkName = "JOAN" #target variable to check against the permutations of above variable

def checkName_fun(checkName, name):
    counter = 0

    if checkName is name:        
        return ('found1')
    elif checkName in (name[:i] + c + name[i + 1:] for i in range(len(name)) for c in charList):        
        return ('found2')
    elif checkName in ([name[:i] + c + name[i:] for i in range(len(name)) for c in charList]):        
        return ('found3')
    elif checkName in ([name[0:i] + name[i+1] + name[i] + name[i+2:] for i in range(len(name) - 1)]):        
        return ('found4')
    else:
        counter += 1
        return (counter)

checkName_fun(checkName, name)