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?
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
cities = ["Kathmandu", "Pokhara", "", "Dharan", "Butwal"].delete_if {|c| c.empty? }
其他回答
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]
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[]?
cities = ["Kathmandu", "Pokhara", "", "Dharan", "Butwal"].delete_if {|c| c.empty? }
另一种方法:
> ["a","b","c","","","f","g"].keep_if{|some| some.present?}
=> ["a","b","c","f","g"]
最短路径城市。select(&:present?)