How do I filter a specific number in python?答案

【问题标题】：How do I filter a specific number in python?How do I filter a specific number in python?
【发布时间】：2022-12-01 20:57:10
【问题描述】：

current code:

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

for i in list:
    Dev.step(2)
    if i == 2 or 1 or 0:
        Dev.turnLeft()
        Dev.step(Dev.x-Item[i].x)
        Dev.step(Dev.x-15)
        Dev.turnRight()
    else:
        Dev.turnRight()
        Dev.step(Item[i].x-Dev.x)
        Dev.step(15-Dev.x)
        Dev.turnLeft()

How do I create an if statement for the Dev / Character do something for a specific list element or filter the list elements. Example I want, if the number of 'i' is equal to 2 or 1 or 0 the Dev will turnLeft. So the output of the list is seperated with the other numbers.

Example: [2,1,0] and [4,5]

Create an if statement for a specific list elements / numbers.

【问题讨论】：

use of inoperator like if i in [2,1,0]: ?
Does this answer your question? Why does "a == x or y or z" always evaluate to True? How can I compare "a" to all of those?

标签： python

【解决方案1】：

The condition below should represent this:

if i in {0, 1, 2}:
    #do logic

【讨论】：

【解决方案2】：

You have to rewrite i for every condition you are evaluating:

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

for i in list:
    Dev.step(2)
    if i == 2 or i == 1 or i == 0:
        Dev.turnLeft()
        Dev.step(Dev.x-Item[i].x)
        Dev.step(Dev.x-15)
        Dev.turnRight()
    else:
        Dev.turnRight()
        Dev.step(Item[i].x-Dev.x)
        Dev.step(15-Dev.x)
        Dev.turnLeft()

【讨论】：

【解决方案3】：

Your if statement won't work because of i == 2 or 1 or 0. See, when you use or, it checks if each statement is true. So you need to use i == for each number. (If this is a bit confusing, feel free to read more about this here)
You should replace it to:

if i == 2 or i == 1 or i == 0:

Full code:

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

for i in list:
    Dev.step(2)
    if i == 2 or i == 1 or i == 0:
        Dev.turnLeft()
        Dev.step(Dev.x-Item[i].x)
        Dev.step(Dev.x-15)
        Dev.turnRight()
    else:
        Dev.turnRight()
        Dev.step(Item[i].x-Dev.x)
        Dev.step(15-Dev.x)
        Dev.turnLeft()

【讨论】：