为什么我的变量“让”打印b而不是？

Question

对于enumerate（）函数中的给定字符串，当i = 0时，它应该打印第一个元素。但我不明白为什么这次打印第二个元素。在我看来，在i = 0的第一次迭代中，它应该为给定的字符串（'abc'）打印'a'。但它正在打印'b'。帮帮我理解另外，在第一次迭代中，当[i = 0]时，s [：i]的值是'b'？

def permute(s):
    out = []

    # Base Case
    if len(s) == 1:
        out = [s]

    else:
        # For every letter in string
        for i, let in enumerate(s):

        # Step 2 and 3 

            for perm in permute(s[:i] + s[i+1:]):
                print ("current i is:", i)
                print ("current let is:", let)
                print ("current perm is:", perm)
                print ("current s[:i] is:", s[:1])

                # Add it to output
                out += [let + perm]
                print("current out is", out)
                print("\n")

    return out

permute('abc')

Answer 1

一些伪代码来说明出了什么问题。

for perm in permute( s[:0] + s[1:] = 'bc'):
    for i, let in enumerate('bc'):
        for perm in permute ('bc'[:i] + 'bc'[i+1:]):
            print stuff

            #print(i=0)
            #print(let = b)
            #print(perm = c)
            #printinseconditeration(i=1, let = c, perm=b)

换句话说，由于递归，您的排列不会从您认为它们开始的位置开始。

Answer 2

当你在你的enumerate循环中，你调用permute时，它实际上重新运行你刚才所做的事情并且状态不再是你认为的那样......：

def permute(s):
    out = []

    # Base Case
    if len(s) == 1:
        out = [s]

    else:
        # For every letter in string
        for i, let in enumerate(s):
            print 'If this is a first call to permute, things are like you expect...'
            print 'Otherwise, something looks weird ;-)'
            print ("current i is:", i)
            print ("current let is:", let)
            print ("current s[:i] is:", s[:1])
        # Step 2 and 3
            for perm in permute(s[:i] + s[i+1:]):
                print 'This only comes after another call to permute'

                print ("current i is:", i)
                print ("current let is:", let)
                print ("current s[:i] is:", s[:1])

                # Add it to output
                out += [let + perm]

    return out

permute('abc')
# If this is a first call to permute, things are like you expect...
# Otherwise, something looks weird ;-)
# ('current i is:', 0)
# ('current let is:', 'a')
# ('current s[:i] is:', 'a')
# If this is a first call to permute, things are like you expect...
# Otherwise, something looks weird ;-)
# ('current i is:', 0)
# ('current let is:', 'b')
# ('current s[:i] is:', 'b')
# This only comes after another call to permute
# ('current i is:', 0)
# ('current let is:', 'b')
# ('current s[:i] is:', 'b')
# etc...

不太确定你想要实现什么，但删除对permute的递归调用对我来说似乎是一个好主意：

def permute(s):
    # Base Case
    if len(s) == 1:
        out = [s]
    else:
        out = []
        for i, let in enumerate(s):
            print ("current i is:", i)
            print ("current let is:", let)
            print ("current s[:i] is:", s[:1])
            out += [let + s[:i] + s[i+1:]]
            print("current out is", out)
            print("\n")
    return out


foo = permute('abc')
print foo
# ['abc', 'bac', 'cab']