价值 || 不是 String 的成员 - scala
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value || is not a member of String - scala
提问by Naren
I am using string filter function like:
我正在使用字符串过滤器功能，如：
val strings = List("hi","I","am","here") //this list is a stream of words from Twitter
val mystrings = strings.filter(word => !word.contains("I" || "sam") // I need filter out certain stop words
to filter certain words. But I am getting compilation error saying that - value ||is not a member of String. Can anybody tell me where I am going wrong?
过滤某些词。但是我收到编译错误，说 - value||不是String. 谁能告诉我我哪里出错了？
回答by Ben Reich
The method containson Stringtakes a Stringas an argument:
containson方法String将 aString作为参数：
"foo".contains("foo") //True
So, the compiler is trying to interpret "I" || "sam"as a String, since that is the argument to contains. In Scala, this statement would be equivalent to "I".||("sam")(calling the ||method on "I"). Since there is no method ||on String, this fails to compile.
因此，编译器试图解释"I" || "sam"为 a String，因为这是 的参数contains。在 Scala 中，此语句等效于"I".||("sam")（调用||on的方法"I"）。由于没有方法||上String，这无法编译。
What you probably meant is !(word.contains("I") || word.contains("sam")). This makes sense, since we are now calling ||on the Booleanthat is returned by word.contains("I"), and ||is a method on Boolean(as documented here). So your whole statement might be:
你可能的意思是!(word.contains("I") || word.contains("sam")). 这是有道理的，因为我们现在呼吁||的Boolean是由返回word.contains("I")，并且||是一个方法Boolean（如记录在这里）。所以你的整个陈述可能是：
strings.filter(word => !(word.contains("I") || word.contains("sam"))
Which is also equivalent to:
这也相当于：
strings.filter(word => !word.contains("I") && !word.contains("sam"))
If you end up with a lot of phrases to filter, you could also flip it around:
如果您最终要过滤很多短语，您也可以翻转它：
val segments = Set("I", "sam")
strings.filter(word => segments.forall(segment => !word.contains(segment)))
//Equivalent
strings.filter(word => !segments.exists(word.contains))
