如何将 x 的值与之前的值进行比较？

Question

how do I compare the values of x so that they are in increasing order

n = 5
i = 1
while i <= n:
    x = int(input())
    i = i + 1

Answer 1

n = 5
i = 1
prev = float('-inf')
while i <= n:
    x = int(input())
    if x < prev:
        print(f'{x} is lesser than {prev}!')
        break
    prev = x
    i += 1

Answer 2

You can only compare values to previous values, so you'll have to keep them around. Since you ask about "increasing order", it appears you want to collect all the inputs:

n = 5
i = 1
x = []
while i <= n:
    x += [int(input())]
    i = i + 1
x = sorted(x)

x = [] sets up x as an empty list. x += [int(input())] does the same as your command, but instead of assigning the result directly to x , it puts it in a small list and adds that to the end of x . The final command just sorts the list in one go.

There's many ways to construct a list though. Instead of x += [int(input())] , you might prefer something like x.append(int(input())) . That's a matter of style, mostly.

Answer 3

While the previous answers are all correct, I prefer:

x = <some random value>
while i <= n:
    prev_x, x = x, int(input())
    ...

This makes it immediately clear that at the same time x is getting a new value, prev_x is getting its previous value.

Others' sensibilities may differ.