Ruby 查找字符串左右两侧的单词

我想把单词放在字符串的左边和右边。例如，如果整个字符串为：

"Programming is my passion"

我想提取

“is”

：

“Programming”

和

“my”

左右两侧的单词

现在以比赛的第1组和第2组为例

sentence = 'Programming is my passion'
word     = 'is'
matches  = sentence.match(/\b(\w+) #{word} (\w+)\b/)

matches[1] # => "Programming"
matches[2] # => "my"

---

这个想法是：

• \b

  -单词边界

• （\w+）

  -在一个编号组中包含尽可能多的单词字符

---

您可以使用以下功能：

\b(\w+)\b\s+is\s+\b(\w+)\b
            ^^  

---

您也可以使用此正则表达式：

(\S+)\s+is\s+(\S+)

演示：

带变量：

string = 'Programming is my passion'

matchers = string.match(/(?<before>\w+)?\s?is\s?(?<after>\w*)?/)

matchers[:before] # "Programming"
matchers[:after] # "my"

string = 'Programmingismy passion'
# same tesults

string='编程是我的爱好'
matchers=string.match（/（？\w+）\s？是\s？（？
这里有一个解决方案

str = "Programming is my passion"
word = "is"
words = str.split(/\W/)
index = words.index(word)
before, after = words[index-1], words[index+1] if index > 0
p before 
#=> "Programming"
p after
#=> "my"

一种没有正则表达式但具有

如果句子改为
“编程是我的爱好，因为它非常有趣”
，因为现在句子中有两个“是”的实例，那么这个函数应该如何工作呢
str = "Programming is my passion"
word = "is"
words = str.split(/\W/)
index = words.index(word)
before, after = words[index-1], words[index+1] if index > 0
p before 
#=> "Programming"
p after
#=> "my"

str = "Programming is my passion"

def neighbors(s, w)
  s.split(/\W/)
   .each_cons(3)
   .find   { |_,e,_| e == w }
   .reject { |e| e == w }
end

before, after = neighbors(str, 'is')

def neighbours str, pattern
  str.scan(/(\S*)\s+#{pattern}\s+(\S*)/).flatten
end

neighbours 'Programming is my passion', 'is'
#⇒ ["Programming", "my"]