问题描述
我试图创建一个函数，将多个变量比较一个整数，并输出一个三个字母的字符串。我想知道是否有一种方法来将它翻译成Python。所以说：
这是由于。 或运算符的优先级比 == 测试的优先级低，因此，
但是，即使这不是的情况，表达式 x或y或z = = 1 实际上被解释为（x或y或z）== 1 ，但这仍然不会做你期望的做。
x或y或z 会计算第一个参数为truey不是 False ，数字0或空（请参阅，以了解Python在布尔上下文中认为是什么的细节。）
所以对于 x = 2; y = 1; z = 0 ， x或y或z 会解析为 2 参数中的第一个真实值。那么 2 == 1 将是 False ，即使 y == 1 将 True 。
I'm trying to make a function that will compare multiple variables to an integer and output a string of three letters. I was wondering if there was a way to translate this into Python. So say:
x = 0 y = 1 z = 3 Mylist = [] if x or y or z == 0 : Mylist.append("c") elif x or y or z == 1 : Mylist.append("d") elif x or y or z == 2 : Mylist.append("e") elif x or y or z == 3 : Mylist.append("f")which would return a list of
["c", "d", "f"]Is something like this possible?
解决方案You misunderstand how boolean expressions work; they don't work like an English sentence and guess that you are talking about the same comparison for all names here. You are looking for:
if x == 1 or y == 1 or z == 1:x and y are otherwise evaluated on their own (False if 0, True otherwise).
You can shorten that to:
if 1 in (x, y, z):or better still:
if 1 in {x, y, z}:using a set to take advantage of the constant-cost membership test (in takes a fixed amount of time whatever the left-hand operand is).
When you use or, python sees each side of the operator as separate expressions. The expression x or y == 1 is treated as first a boolean test for x, then if that is False, the expression y == 1 is tested.
This is due to operator precedence. The or operator has a lower precedence than the == test, so the latter is evaluated first.
However, even if this were not the case, and the expression x or y or z == 1 was actually interpreted as (x or y or z) == 1 instead, this would still not do what you expect it to do.
x or y or z would evaluate to the first argument that is 'truthy', e.g. not False, numeric 0 or empty (see boolean expressions for details on what Python considers false in a boolean context).
So for the values x = 2; y = 1; z = 0, x or y or z would resolve to 2, because that is the first true-like value in the arguments. Then 2 == 1 would be False, even though y == 1 would be True.
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