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
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
其他回答
最明确的
cities.delete_if(&:blank?)
这将删除nil值和空字符串("")值。
例如:
cities = ["Kathmandu", "Pokhara", "", "Dharan", "Butwal", nil]
cities.delete_if(&:blank?)
# => ["Kathmandu", "Pokhara", "Dharan", "Butwal"]
最短路径城市。select(&:present?)
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
1.9.3p194 :001 > ["", "A", "B", "C", ""].reject(&:empty?)
=> ["A", "B", "C"]
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[]?
