Python返回交换的序列的第一项和最后一项

时间:2018-12-20 05:42:16

标签: python object-slicing

我需要创建一个对序列进行切片的函数,以便交换第一项和最后一项,并且中间部分停留在中间。它需要能够处理字符串/列表/元组。我遇到TypeError错误-无法添加列表+整数。

此:

def exchange_first_last(seq):
    """This returns the first and last item of a sequence exchanged"""
    first = seq[-1]
    mid = seq[1:-1]
    last = seq[0]
    return print(first+mid+last)

产生

(5,[2,3,4],1)

但是我不想要一个元组中的列表,而只是一个流动的序列。

(5,2,3,4,1,)

欢迎任何提示/建议。想法是适当地切片以便处理不同的对象类型。

4 个答案:

答案 0 :(得分:4)

尝试一下:

def exchange_first_last(seq):
    """This returns the first and last item of a sequence exchanged"""
    first = seq[-1:]
    mid = seq[1:-1]
    last = seq[:1]
    return print(first+mid+last)

答案 1 :(得分:1)

稍微修改一下代码,注意方括号:

def exchange_first_last(seq):
    """This returns the first and last item of a sequence exchanged"""
    first = seq[-1]
    mid = seq[1:-1]
    last = seq[0]
    return print([first]+mid+[last])

请注意,它实际上为您提供了一个列表,即[5,2,3,4,1],而不是元组(5,2,3,4,1)

答案 2 :(得分:0)

您可以使用list.extend():

def exchange_first_last(seq):
    """This returns the first and last item of a sequence exchanged"""
    first = seq[-1]
    mid = seq[1:-1]
    last = seq[0]
    merged = []
    merged.extend(first)
    merged.extend(mid)
    merged.extend(last)
    return merged

答案 3 :(得分:0)

您可以交换元素

def swap(l):
     temp = l[0]
     l[0] = l[-1]
     l[-1] = temp
     return l