Question

我正在编写一个程序，将元素插入到特定索引之前的位置的列表中。例如，

mylist = [1,2,3,4,5]

enter the index = 3
enter the item = 7

输出：

[1,2,3,7,4,5]

由于输入索引是3，python会在位置index-1处附加新项目，即2。

我尝试了类似的东西，但输出似乎有些偏差：

mylist = [1,2,3,4,5]

indexInput = int(input("Enter the index: "))
itemInput = int(input("Enter the item: "))

for i in mylist:
    new_slot = indexInput -1
    new_length = len(mylist) + 1
    if i == new_slot:
         mylist[i] = item
         mylist.append(mylist[i])
print(mylist)

我知道python中的insert（）函数，但我不允许使用它，所以我必须编写很长的代码。非常感谢任何帮助。

Answer 1

你可以使用列表切片：

mylist = [1,2,3,4,5]

indexInput = int(input("Enter the index: "))
itemInput = int(input("Enter the item: "))

if indexInput<0:
  indexInput = len(mylist)+indexInput+1

mylist = mylist[:indexInput] + [itemInput] + mylist[indexInput:]
print(mylist)
# for index 3 and item 7 => [1, 2, 3, 7, 4, 5]
# for index -2 and item 6 => [1, 2, 3, 4, 6, 5]

说明：

list[start:end:step] => list items from index [start] to index [end-1] with a step of 1
if start is not given, 0 is assumed, if end is not given, it goes all the way to the end of the list, if step is not given, it assumes a step of 1
In my code I wrote:
mylist = mylist[:indexInput] + [itemInput] + mylist[indexInput:]
for indexInput=3 and itemInput=7
mylist = mylist[:3] + [7] + mylist[3:]
mylist[:3] => mylist from 0 (start not given), to 3-1, step of 1 (step not given)
list1 + list2 => a list that contains elements from list1 and list2
mylist[:3] + [7] => [...elements of mylist[:3]..., 7]
mylist[3:] => mylist from 3, to len(mylist)-1 (end not given), step of 1 (step not given)

如果你想做很长的路，这是一个解决方案：

mylist = [1,2,3,4,5]

indexInput = int(input("Enter the index: "))
itemInput = int(input("Enter the item: "))
if indexInput<0:
      indexInput = len(mylist)+indexInput+1
leftSide = []
rightSide = []
for i in range(len(mylist)):
  if i<indexInput:
    leftSide.append(mylist[i])
  elif i>indexInput:
    rightSide.append(mylist[i])
  else:
    leftSide.append(itemInput)
    leftSide.append(mylist[i])

mylist = leftSide + rightSide
print(mylist)
# for index 3 and item 7 => [1, 2, 3, 7, 4, 5]
# for index -2 and item 6 => [1, 2, 3, 4, 6, 5]

Answer 2

这是我的代码

mylist = [1, 2, 3, 4, 5]

indexInput = int(input("Enter the index: "))
itemInput = int(input("Enter the item: "))

for idx, val in enumerate(mylist):
    if idx == indexInput:
        mylist = mylist[:idx] + [itemInput] + mylist[idx:]

print(mylist)

希望它可以帮助您，只需使用enumerate函数来获取mylist索引

Answer 3

list = [1,2,3,4,5]
inputIndex = 3
inputItem = 7
temp = 0

for i in range(len(list)):#continuously swap your elements until you reach the end
    if(i == len(list) - 1): # if you're at the end, swap and add the temp to the end
        temp = list[i]
        list[i] = inputItem
        list.append(temp)
    elif(i >= inputIndex):
        temp = list[i]
        list[i] = inputItem
        inputItem = temp

print(list)

这可能是你正在寻找的，如果这是一个你不允许使用insert的作业，那么我假设你可能不允许使用切片

将元素插入列表

3 个答案: