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
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"]
其他回答
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[]?
另一种方法:
> ["a","b","c","","","f","g"].keep_if{|some| some.present?}
=> ["a","b","c","f","g"]
试试这个:
puts ["Kathmandu", "Pokhara", "", "Dharan", "Butwal"] - [""]
cities = ["Kathmandu", "Pokhara", "", "Dharan", "Butwal"].delete_if {|c| c.empty? }
使用严格的join & split更新
cities = ["Kathmandu", "Pokhara", "", "Dharan", "Butwal"]
cities.join(' ').split
结果将是:
["Kathmandu", "Pokhara", "Dharan", "Butwal"]
注意:这对有空间的城市不起作用
