在 Scala 中将元素添加到 Seq[String]
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Add element to Seq[String] in Scala
提问by user3297367
I'm trying to create a list of words in Scala. I'm new to the language. I have read a lot of posts about how you can't edit immutable objects, but none that has managed to show me how to create the list I need in Scala. I am using var to initialise, but this isn't helping.
我正在尝试在 Scala 中创建一个单词列表。我是语言的新手。我已经阅读了很多关于如何不能编辑不可变对象的帖子，但没有一篇文章能够向我展示如何在 Scala 中创建我需要的列表。我正在使用 var 进行初始化，但这无济于事。
var wordList = Seq.empty[String]
for (x <- docSample.tokens) {
wordList.++(x.word)
}
println(wordList.isEmpty)
I would greatly appreciate some help with this. I understand that objects are immutable in Scala (although vars are not), but what I need is some concise information about why the above always prints "true", and how I can make the list add the words contained in docSample.tokens.word.
我将不胜感激。我知道对象在 Scala 中是不可变的（尽管 vars 不是），但我需要的是一些关于为什么上面总是打印“true”的简明信息，以及如何让列表添加包含在 docSample.tokens.word 中的单词.
回答by Carl
You can use a val and still keep the wordlist immutable like this:
您可以使用 val 并且仍然保持 wordlist 不变，如下所示：
val wordList: Seq[String] =
for {
x <- docSample.tokens
} yield x.word
println(wordList.isEmpty)
Alternatively:
或者：
val wordList: Seq[String] = docSample.tokens.map(x => x.word)
println(wordList.isEmpty)
Or even:
甚至：
val wordList: Seq[String] = docSample.tokens map (_.word)
println(wordList.isEmpty)
回答by Chris Martin
You can append to an immutable Seqand reassign the varto the result by writing
您可以附加到不可变对象Seq并var通过编写将 重新分配给结果
wordList :+= x.word
That expression desugars to wordList = wordList :+ wordin the same way that x += 1desugars to x = x + 1.
该表达式 desugars towordList = wordList :+ word的方式与x += 1desugars to相同x = x + 1。
回答by Bipp
This will work:
这将起作用：
wordList = wordList:+(x.word)
回答by user3297367
After a few hours, I posted a question, and a minute later figured it out.
几个小时后，我发布了一个问题，一分钟后我想通了。
wordList = (x.word)::wordList
wordList = (x.word)::wordList
This code solves it for anyone who comes across the same problem.
此代码为遇到相同问题的任何人解决了它。
