この配列がある場合:
["1","2","3,a","4,b"]
どうすればこの配列を取得できますか?
["1","2","3","4"].
すべての整数があると仮定すると、(そうでない場合は、アイデアが得られると思います:))
["1","2","3,a","4,b"].collect {|i| i.to_i}
次のような配列を使用します。
ary = ["1","2","3,a","4,b"]
私は使用します:
ary.map{ |s| s[/\A(\d+)/, 1] }
結果は次のとおりです。
[ [0] "1", [1]「2」、 [2]「3」、 [3]「4」 ]
文字列の先頭にある数値を検索し、それらを新しい配列で返すだけです。
試す:
["1","2","3,a","4,b"].map(&:to_i)
=> [1, 2, 3, 4]
文字列の配列を取得するには
["1","2","3,a","4,b"].map(&:to_i).map(&:to_s)
=> ["1", "2", "3", "4"]
これも機能するはずです...
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"]
この削除オプションを試してください:
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
更新されたコード(各ループの使用 - Tin Man のおかげ)
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