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
纯Ruby:
values = [1,2,3, " ", "", "", nil] - ["", " ", nil]
puts values # [1,2,3]
其他回答
使用拒绝:
>> cities = ["Kathmandu", "Pokhara", "", "Dharan", "Butwal"].reject{ |e| e.empty? }
=> ["Kathmandu", "Pokhara", "Dharan", "Butwal"]
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"]
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]
你可以试试这个
cities.reject!(&: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
