如何否定析取
我对 Enumerator#reject 在红宝石中。考虑以下代码：
(1..10).select {|i| i % 3 == 0 || i % 5 == 0 } => [3, 5, 6, 9, 10]
以下行不应该是等效的吗？
(1..10).reject {|i| i % 3 != 0 || i % 5 != 0 } => []
如果我只在拒绝方法上使用一个条件，结果将符合预期。但如果我包含 OR 运算符，结果将为空。有人可以帮我澄清一下吗？
(1..10).reject {|i| i % 3 != 0} => [3, 6, 9]
I am a little confused with Enumerator#reject in ruby. Consider the following code:
(1..10).select {|i| i % 3 == 0 || i % 5 == 0 } => [3, 5, 6, 9, 10]
Shouldn't the following line be equivalent?
(1..10).reject {|i| i % 3 != 0 || i % 5 != 0 } => []
If I just use one condition on the reject method, the result is as expected. but If I include the OR operator the result turns out to be empty. Could somebody clarify this for me.
(1..10).reject {|i| i % 3 != 0} => [3, 6, 9]
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您犯了一个基本的逻辑错误：
!(A || B)相当于!A && !B且不等于!A || !B。因此，如果您将第二个示例中的
||更改为&&，那么第二个示例将给出与第一个示例相同的结果：You are making a basic logic mistake:
!(A || B)is equivalent to!A && !Band NOT equivalent to!A || !B.So if you change the
||in your second example to a&&, then your second example would give the same result as the first:您遇到了德摩根定律之一。
p 和 q = Not((不是 p) 或 (不是 q))
p 或者 q = Not((Not p) And (Not q))
很接近，但您忘记更改运算符。
You have run into one of De Morgan's laws.
p And q = Not((Not p) Or (Not q))
p Or q = Not((Not p) And (Not q))
It was close, but you forgot to change the operator.
在第二段代码中，您更改了相等性，因此您需要将
||更改为&&。In the second piece of code, you changed the equality, so you'll need to change the
||to&&.