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
你可以试试这个
cities.reject!(&:empty?)
其他回答
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
在我的项目中,我使用delete:
cities.delete("")
有很多方法可以做到这一点,其中之一是拒绝
noEmptyCities = cities.reject { |c| c.empty? }
你也可以用reject!,这将改变城市的位置。如果拒绝某项,它将返回cities作为返回值,如果没有拒绝,则返回nil。如果你不小心,这可能是一个陷阱(感谢ninja08在评论中指出这一点)。
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"]
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"]