问题描述
我如何拆分这个字符串.
How do i split this string.
"6885558 8866887777" => ["6", "88", "555", "8", "88", "66", "88", "7777"]
我试过这个，但没有奏效.
I tried this, but it never worked.
ruby-1.8.7-p334 :020 > "111133".split(/(\d)\1+/)
=> ["", "1", "", "3"]
推荐答案
split 将仅使用匹配的任何内容作为分隔符，将其从相关字符串中删除.您正在寻找的是 scan:
split will just use whatever it matches as a delimiter, removing it from the string in question. What you're looking for is scan:
str = "6885558 8866887777"
str.scan(/((\d)\2*)/).map(&:first)
# => ["6", "88", "555", "8", "88", "66", "88", "7777"]
慢慢来，\d 匹配任何数字.它位于第二个捕获组中，因此 \2* 然后匹配相同数字的任何进一步出现.这会产生一个看起来像
Taking it slow, the \d matches any digit. It's in the second capturing group, so \2* then matches any further occurrences of the same digit. This produces an array that looks like
[["6", "6"], ["88", "8"], ["555", "5"], ["8", "8"],
["88", "8"], ["66", "6"], ["88", "8"], ["7777", "7"]]
由于我们只想要每个子数组中的第一项，我们可以使用 map(&:first) 将它们全部收集起来.
Since we only want the first item in each of those sub arrays, we can collect them all with map(&:first).
(注意 str.scan(/(\d)\1*/) 只会从第一个捕获组中生成一个数组，这意味着我们只能从可能重复的数字序列.)
(Note that str.scan(/(\d)\1*/) would simply produce an array out of the first capturing group, which means we'd only get one digit from a sequence of possibly repeated numbers.)
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