在Python列表推导中对if/elif语句使用'for'循环 [英] Use a 'for' loop with if/elif statement in Python list comprehension

问题描述

我正在尝试将此for循环转换为列表理解:

I am trying to translate this for loop into a list comprehension:

a = [1,2,3,4,5,6,7,8,9]
result = []
for i in a:
    if i <= 3:
        result.append(1)
    elif i > 4 and i < 7:
        result.append(2)

我已经尝试过了

[1 if i <= 3 else 2 if i > 3 and i < 7 for i in a]

抱怨

File "<ipython-input-155-eebf07a9e0d8>", line 2
    [1 if i <= 3 else 2 if i > 3 and i < 7 for i in a]
                                             ^
SyntaxError: invalid syntax

推荐答案

列表理解:

添加更多条件:D(不，这真的很混乱)

List comprehension:

Add some more conditions :D (no this is really messy)

[
    1 if i <= 3 else 2
    for i in a
    if i != 4 and i < 7
]

我们怎么到这里的?

基本列表组合:[EXPRESSION for TARGET in ITERABLE if CONDITION]

三元表达式:(IF_TRUE if CONDITION else IF_FALSE)

1. 进入for循环.足够简单的for i in a.
2. 添加条件以过滤掉将被忽略的项目.一旦通过CONDITION，列表中的那个位置就必须有一个项目.在这种情况下，我们不希望i是4或大于7.if i != 4 and i < 7.
3. 用物品做您需要的事情.在这种情况下，如果i小于或等于4，我们需要1.否则，我们将取2.1 if i <= 3 else 2.注意:这是一个三元表达式.检查出来！

1. Get the for loop in. Simple enough for i in a.
2. Add conditions that filter out items which will be ignored. Once it gets past CONDITION, there has to be an item in the list at that position. In this case, we don't want i if it's 4 or greater than 7. if i != 4 and i < 7.
3. Do what you need with the item. In this case, we want 1 if i is smaller or equal to 4. Otherwise, we'll take 2. 1 if i <= 3 else 2. Note: this is a ternary expression. Check them out!

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