这终于对我有用。

Question

如何在给定列表中交换数字？

例如：

list = [5,6,7,10,11,12]

我想将12与5交换。

是否有内置的Python函数可以让我这样做？

Answer 1

>>> lis = [5,6,7,10,11,12]
>>> lis[0], lis[-1] = lis[-1], lis[0]
>>> lis
[12, 6, 7, 10, 11, 5]

上述表达式的

Order of evaluation：

expr3, expr4 = expr1, expr2

RHS上的第一项收集在元组中，然后收集tuple is unpacked并分配给LHS上的项目。

>>> lis = [5,6,7,10,11,12]
>>> tup = lis[-1], lis[0]
>>> tup
(12, 5)
>>> lis[0], lis[-1] = tup
>>> lis
[12, 6, 7, 10, 11, 5]

Answer 2

您可以使用此代码进行交换，

list[0],list[-1] = list[-1],list[0]

Answer 3

您可以使用＆＃34; *＆＃34;操作

my_list = [1,2,3,4,5,6,7,8,9]
a, *middle, b = my_list
my_new_list = [b, *middle, a]
my_list
[1, 2, 3, 4, 5, 6, 7, 8, 9]
my_new_list
[9, 2, 3, 4, 5, 6, 7, 8, 1]

阅读here了解详情。

Answer 4

使用您要更改的号码的索引。

In [38]: t = [5,6,7,10,11,12]

In [40]: index5 = t.index(5) # grab the index of the first element that equals 5

In [41]: index12 = t.index(12) # grab the index of the first element that equals 12

In [42]: t[index5], t[index12] = 12, 5 # swap the values

In [44]: t
Out[44]: [12, 6, 7, 10, 11, 5]

然后你可以进行快速交换功能

def swapNumbersInList( listOfNums, numA, numB ):
    indexA = listOfNums.index(numA)
    indexB = listOfNums.index(numB)

    listOfNums[indexA], listOfNums[indexB] = numB, numA

# calling the function
swapNumbersInList([5,6,7,10,11,12], 5, 12)

Answer 5

另一种方式（不那么可爱）：

mylist   = [5, 6, 7, 10, 11, 12]
first_el = mylist.pop(0)   # first_el = 5, mylist = [6, 7, 10, 11, 12]
last_el  = mylist.pop(-1)  # last_el = 12, mylist = [6, 7, 10, 11]
mylist.insert(0, last_el)  # mylist = [12, 6, 7, 10, 11]
mylist.append(first_el)    # mylist = [12, 6, 7, 10, 11, 5]

Answer 6

array = [5,2,3,6,1,12]
temp = ''
lastvalue = 5

temp = array[0]
array[0] = array[lastvalue]
array[lastvalue] = temp

print(array)

希望这有帮助:)

Answer 7

这终于对我有用。

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

交换列表中的第一个和最后一个项目

7 个答案:

这终于对我有用。