I have tried to get first letters of words of a string using ruby. the following is what i have written.
我尝试使用ruby获取字符串的首字母。以下是我写的。
puts "world is a better place".split.collect{|w| w[0].capitalize}.join()
Is there a much more concise way of producing the same result?
有没有更简洁的方法来产生同样的结果?
1 个解决方案
#1
6
Using regular expression:
使用正则表达式:
"world is a better place".scan(/\b[a-z]/i).join
# => "wiabp"
"world is a better place".scan(/\b[a-z]/i).join.upcase
# => "WIABP"
\b matches word boundary. (between word character and non-word character). [a-z] match any alphabet.
\ b匹配单词边界。(介于字与字之间)[a - z]匹配任何字母表。
\b[a-z] matches the first alphabet letter of word.
\b[a-z]匹配单词的第一个字母。
NOTE Above code will not work if there's a word(?) that starts with non-alphabet character. Also does not work if there's a word that contains a punctuation in it. (For example: World is 1 better-place.)
注意，如果有一个以非字母表字符开头的单词(?)，上面的代码将不起作用。如果有一个包含标点符号的词，也不起作用。(例如:世界是一个更好的地方。)
UPDATE
更新
Using String#gsub with capturing group you will get the same result:
使用带捕获组的字符串#gsub，您将得到相同的结果:
"world is a better place".gsub(/\s*(\S)\S*/, '\1').upcase
# => "WIABP"
#1
6
Using regular expression:
使用正则表达式:
"world is a better place".scan(/\b[a-z]/i).join
# => "wiabp"
"world is a better place".scan(/\b[a-z]/i).join.upcase
# => "WIABP"
\b matches word boundary. (between word character and non-word character). [a-z] match any alphabet.
\ b匹配单词边界。(介于字与字之间)[a - z]匹配任何字母表。
\b[a-z] matches the first alphabet letter of word.
\b[a-z]匹配单词的第一个字母。
NOTE Above code will not work if there's a word(?) that starts with non-alphabet character. Also does not work if there's a word that contains a punctuation in it. (For example: World is 1 better-place.)
注意，如果有一个以非字母表字符开头的单词(?)，上面的代码将不起作用。如果有一个包含标点符号的词，也不起作用。(例如:世界是一个更好的地方。)
UPDATE
更新
Using String#gsub with capturing group you will get the same result:
使用带捕获组的字符串#gsub，您将得到相同的结果:
"world is a better place".gsub(/\s*(\S)\S*/, '\1').upcase
# => "WIABP"