ruby - 拆分数组并获取所需的元素

Question

如果我有这个数组：

["1","2","3,a","4,b"]

我怎样才能从中得到这个数组？

["1","2","3","4"].

Answer 1

假设你有所有整数，（如果没有，我猜你明白了:)）

["1","2","3,a","4,b"].collect {|i| i.to_i}

Answer 2

使用如下数组：

ary = ["1","2","3,a","4,b"]

我会使用：

ary.map{ |s| s[/\A(\d+)/, 1] }

结果是：

[
    [0] "1",
    [1] “2”，
    [2] “3”，
    [3] “4”
]

它只是在字符串的开头找到数字并将它们返回到一个新数组中。

Answer 3

尝试：

["1","2","3,a","4,b"].map(&:to_i)
=> [1, 2, 3, 4]

获取字符串数组

["1","2","3,a","4,b"].map(&:to_i).map(&:to_s)
=> ["1", "2", "3", "4"]

Answer 4

这也应该有效......

array=["1","2","3,a","4,b"]
  for i in 0..array.length()-1
    if array[i].include?(",")
      ab=array[i].split(",")
      array[i]=ab[0]
    end

  end
print array #["1","2","3","4"]

Answer 5

试试这个删除选项：

irb(main):014:0> a = ["101","2zd","3,a","10,ab"]
=> ["101", "2zd", "3,a", "10,ab"]
irb(main):015:0> count = a.count
irb(main):016:0> for i in 0..count 
irb(main):017:1> puts a[i].delete("^0-9")
irb(main):018:1> end
101
2
3
10

更新的代码（每个循环的使用 - 感谢 Tin Man）

irb(main):005:0>  a = ["101","2zd","3,a","10,ab"]
=> ["101", "2zd", "3,a", "10,ab"]
irb(main):006:0> b = []
=> []
irb(main):007:0> a.each do |t|
irb(main):008:1* b << t.delete("^0-9")
irb(main):009:1> end
=> ["101", "2zd", "3,a", "10,ab"]
irb(main):010:0> puts b
101
2
3
10
=> nil