内循环如何在每次迭代中附加？

Question

def per1(seq):
    if not seq:
        return [seq]
    else:
        res=[]
        for i in range(len(seq)):
            rest=seq[:i]+seq[i+1:]
            for x in per1(rest):
                 res.append(seq[i:i+1]+x)

        return res

print(per1('abc'))

The function is to print a permuted list of sequence but I am not sure how the inner for loop is working. I have tried to use trace() to see how the control is moving thru the loop but I couldn't figure how the loop is working the second time. The first time, ' rest ' value is 'bc','c' .

Answer 1

Each time through the for i loop, rest is set to the input string with one character removed. The first time it has the first character removed, the second time it has the second character removed, etc. seq[:i] is every character before i , and seq[i+1:] is every character after i .

Then it calls itself recursively on rest . This returns all the permutations of that shorter string. So when rest is "bc" , this returns ["bc", "cb"] . The for x loop iterates through this list, and concatenates each one to seq[i:i+1] (which is the character that was removed when creating rest ), and appends this to res .

So the first iteration of the main loop sets rest to "bc" , then appends "abc" and "acb" to res .

The second iteration does the same thing, with rest set to "ac" , and seq[i:i+1] being the missing character "b" , so it appends "bac" and "bca" to res .

The third iteration is similar, with rest set to "ab" , so it appends "cab" and "cba" to res .

The special situation is the base case of the recursion. If the input string is empty, we just return a list with that empty string -- an empty string has trivial permutations. When the caller does its for x loop, it will simply concatenate this empty string onto the current letter. This case is needed to prevent infinite recursion. You could also make the case where len(seq) == 1 the base case, since the permutations of a single letter is also trivial.