通过引用的Ruby调用在每个函数中都不起作用？

Question

excluded = "a", " an", " the", " at", " in", " on ", "since"     
temp = excluded.split(",").each {|val|  val = val.strip}

但是我得到了相同的数组。它没有条纹。我需要用单行

来做这件事

我需要临时输出，如[“a”，“an”，“the”，“at”，“in”，“on”，“since”]

Answer 1

试试这个

 temp =  ["a", " an", " the", " at", " in", " on ", "since"].map(&:strip)

Answer 2

从the docs Array#each可以看到，返回原始接收者（ary.each {|item| block } → ary）。正如其他人已经指出的那样，你想要的是Array#map。

由于在数组上调用split，您当前的代码也应该引发NoMethodError。假设excluded是一个字符串，以下内容将起作用：

excluded.split(",").map(&:strip) #=> ["a", "an", "the", "at", "in", "on", "since"]

您也可以只更改拆分的内容，而不是使用strip：

excluded.split(/,\s*/) #=> ["a", "an", "the", "at", "in", "on", "since"]

Answer 3

你想要这个吗？：

excluded = ["a", " an", " the", " at", " in", " on ", "since"]
stripped_excluded = excluded.collect{ |i| i.strip }

或捷径：

stripped_excluded = excluded.collect(&:strip)

Answer 4

我认为这应该是更简单的方法：

temp = excluded.map(&:strip)
p temp