如何从数组中删除空白元素?

我有下面的数组

cities = ["Kathmandu", "Pokhara", "", "Dharan", "Butwal"]

我想从数组中删除空白元素，并希望得到以下结果:

cities = ["Kathmandu", "Pokhara", "Dharan", "Butwal"]

有没有像compact这样不需要循环的方法?

当前回答

更新在拒绝和拒绝!

注意:我遇到了这个问题，并在irb控制台上用ruby-3.0.1检查了这些方法。我也检查了ruby文档，但没有提到这一点。我不确定从哪个ruby版本变化就在那里。非常感谢来自社区的任何帮助。

在ruby-3.0.1中，我们可以使用reject或reject!

cities = ["Kathmandu", "Pokhara", "", "Dharan", "Butwal"]
cities.reject{ |e| e.empty? }
=> ["Kathmandu", "Pokhara", "Dharan", "Butwal"]

或简写

cities.reject(&:empty?)
=> ["Kathmandu", "Pokhara", "Dharan", "Butwal"]

无论我们是否有一个空值，两者都会返回[]?

其他回答

试试这个:

puts ["Kathmandu", "Pokhara", "", "Dharan", "Butwal"] - [""]

如果你的数组中有混合类型，下面是一个解决方案:

[nil,"some string here","",4,3,2]

解决方案:

[nil,"some string here","",4,3,2].compact.reject{|r| r.empty? if r.class == String}

输出:

=> ["some string here", 4, 3, 2]

纯Ruby:

values = [1,2,3, " ", "", "", nil] - ["", " ", nil]
puts values # [1,2,3]

删除空值:

 ['a', nil, 'b'].compact  ## o/p =>  ["a", "b"]

删除空字符串:

   ['a', 'b', ''].select{ |a| !a.empty? } ## o/p => ["a", "b"]

删除空字符串和空字符串:

['a', nil, 'b', ''].select{ |a| a.present? }  ## o/p => ["a", "b"]

这里还有一种方法可以实现这一点

我们可以使用presence和select

cities = ["Kathmandu", "Pokhara", "", "Dharan", nil, "Butwal"]

cities.select(&:presence)

["Kathmandu", "Pokhara", "Dharan", "Butwal"]

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