for 循环，lst[i] = lst[i - 1] - Python

Question

Why doesn't this work?

def func(lst):
    for i in range(len(lst)):
        lst[i] = lst[i - 1]
    return lst


print(func([1, 2, 3, 4]))

Expected output: [4, 1, 2, 3]
Actual output: [4, 4, 4, 4]

why python assign lst[1] = lst[-1] when it should be lst[1] = lst[0]

Answer 1

The following works. lst[i] was reset in the for loop.

def func(lst):
    output = []
    for i in range(len(lst)):
        output.append(lst[i - 1])
    return output


print(func([1, 2, 3, 4]))

Answer 2

You got a logic error here, at first iteration you doing smth like lst[0] = lst[-1] and then pulling this value to all remaining indeces. If you want to shift array left one pos try something like this:

def shifter(lst):
    return lst[-1:] + lst[:-1]

Answer 3

Lists are mutable objects and you are iterating over a single list. On each iteration, the variable lst is pointing to the current state of the list, and inside your loop you are modifying that current state.

Your expected answer would require that Python would copy and remember the initial state of the list at the beginning of the loop, which Python does not do.

def func(lst):
    initial = lst[:]  # copy lst
    for i in range(len(lst)):
        lst[i] = initial[i - 1]
    return lst


print(func([1, 2, 3, 4]))

Answer 4

def func(lst):
for i in range(len(lst)):
    print(i, lst[i-1])
    lst[i] = lst[i - 1]
return lst

If you look at how your code executes, the first iteration, i will be 0, and lst[i-1] will be lst[-1] which points to the last element in the array (4). So after first iteration, your array looks likes this: [4,2,3,4] In the second, i=1, and lst[i-1] will be lst[0] which has a value a 4 now: So now lst[1] will be 4, [4,4,3.4] And so on.... Hope that helps.

Answer 5

You can insert the last item at the first index and delete the last item afterwards:

def func(lst):
    lst[:0] = lst[-1:]
    del lst[-1:]
    return lst

so that func([1, 2, 3, 4]) returns: [4, 1, 2, 3]

Answer 6

Step through your code, one line at a time:

Initially, lst is [1, 2, 3, 4] , so len(lst) is 4 . range(n) returns values form 0 to n - 1 , so your loop will run with the values 0 , 1 , 2 , and 3 .

First loop iteration:

• i is 0
• lst[i] = lst[i - 1] => lst[0] = lst[0 - 1] => lst[0] = lst[-1]
• lst is now [4, 2, 3, 4]

Second loop iteration:

• i is 1
• lst[i] = lst[i - 1] => lst[1] = lst[1 - 1] => lst[1] = lst[0]
• lst is now [4, 4, 3, 4]

Third loop iteration:

• i is 2
• lst[i] = lst[i - 1] => lst[2] = lst[2 - 1] => lst[2] = lst[1]
• lst is now [4, 4, 4, 4]

Fourth loop iteration:

• i is 3
• lst[i] = lst[i - 1] => lst[3] = lst[3 - 1] => lst[3] = lst[2]
• lst is now [4, 4, 4, 4]

Answer 7

Swapping has been proposed. Don't do that:

def func(l):
    for i in range(len(l) // 2):
        l[i], l[-i - 1] = l[-i - 1], l[i]

    return l

# func([1, 2, 3, 4]) -> [4, 3, 2, 1]

To move the last element of a given list/array to the first position, as you've described as your desired output, use blhsing's implementation , if your input is a linked list. Python lists are linked lists, and so blhsing's implementation is an efficient one for those. If your input is a contiguous array, like a Numpy array, use this:

def func(l):

    l_last = l[len(l) - 1]

    for i in range(len(l) - 1, 0, -1):
        l[i] = l[i - 1]

    l[0] = l_last

    return l

Both of our implementations mutate the input list/array.

Why does your function output [4, 4, 4, 4] ?
i_max = len(<list>) - 1
<list>[-i] = <list>[len(<list>) - i] .
for i in range(len(<list>)) => i_n in { 0 , 1 , ... , i_max }
i_n = 0 => <list>[0] = <list>[i_max - 1] .
If <list> = [1, 2, 3, 4] , then [1, 2, 3, 4][0] = [1, 2, 3, 4][3] = [4, 2, 3, 4] .
In the next iterations, you will repeatedly replace the i th with the i-1 th element, so that you get:
[4, 2, 3, 4] => [4, 4, 3, 4] => [4, 4, 4, 4] => [4, 4, 4, 4] .