Ruby从数组中删除非数值并将剩余值转换为浮点数

Question

我有一个用户输入的字符串，我将其变成一个数组，然后遍历该数组并删除非数字值。

看起来我的正则表达式匹配工作了一半时间，而我的to_f从未将数组值设置为浮点型。

假设我输入：“ 1 2 3b c3 4，5t”

puts "Enter Minutes"  
STDOUT.flush  
freq = gets.chomp
freq = freq.split(/\W/) #this creates the array, splitting at non-word chars

p freq #outputs: ["1", "2", "3b", "c3", "4", "", "5t"]

freq.each do |minutes|
        if ( minutes == "" or /\D/.match(minutes) ) then freq.delete(minutes) else minutes.to_f end
end

p freq #outputs: ["1", "2", "c3", "4", "5t"]

我想要的结果是：[1、2、4] #note它们是数字而不是字符

Answer 1

问题是您仅在then条件中而不是在else条件中突变频率。

有许多可为您变异的方法，它们通常以!：结尾

freq = ["1", "2", "3b", "c3", "4", "", "5t"]
=> ["1", "2", "3b", "c3", "4", "", "5t"]

freq.reject! { |minutes| minutes.match(/\D/) || minutes == "" }.map! { |minutes| minutes.to_f }
=> [1.0, 2.0, 4.0]