如果元素不遵循“if”或“语句”，则从列表中删除元素 [英] Deleting elements from a list if they do not follow 'if' 'or' statements

问题描述

我试图摆脱列表中不需要的变量。我需要有两个条件：一个是确保我的数组中的值小于变量A，另一个是确保它们不等于另一个变量B.

I am trying to get rid of unwanted variables in a list. I need to have two condition: one is if making sure the values in my array are smaller than a variable A, and the other is making sure they are not equal to another variable B.

此代码不起作用：

original_Ar = [0,1,2,3,4,5,6,7,8,9,10,11,12]
new_Ar = [s for s in original_Ar if (s != 2) or (s < 10)]

print (new_Ar)

如果我把它分成两个语句（而不是或声明） - 它们确实有效：

while if I split it into two statements (instead of the or statement) - they do work:

original_Ar = [0,1,2,3,4,5,6,7,8,9,10,11,12]
print ([s for s in original_Ar if (s != 2)])
print ([s for s in original_Ar if (s < 10)])

我知道怎么能在一行中做到这一点？

Any idea how can I do that in one line?

推荐答案

你的布尔逻辑混淆了。您希望包含的所有值不等于2 且小于10：

You have your boolean logic mixed up. You want to include all values that are not equal to 2 and are smaller than 10:

new_Ar = [s for s in original_Ar if s != 2 and s < 10]
#           *both* conditions must be true ^^^

否则，你' d包括 s = 2 ，因为它小于10，并且你包括 s = 11 和 s = 12 ，因为两者都不等于两个！

Otherwise, you'd include s = 2, because it is smaller than ten, and you'd include s = 11 and s = 12, because both are not equal to two!

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