Question

我想知道是否有一种简单的方法可以从字符串中仅获取N个符号，而不会删除整个单词。

例如，我有产品和产品描述信息。描述长度为70到500个符号，但我想只显示前70个符号，如下所示：

可口可乐是最受欢迎和最畅销的软饮料   历史，以及世界上最知名的品牌。

2011年5月8日，可口可乐庆祝其125周年纪念日。创建于   1886年在佐治亚州亚特兰大市，由John S. Pemberton博士，可口可乐公司   首先通过混合在Jacob's Pharmacy作为喷泉饮料提供   可口可乐糖浆与碳酸水。

所以，普通的子字符串方法会给我：

Coca-Cola is the most popular and biggest-selling soft drink in histor

我需要一种方法来获得这个：

Coca-Cola is the most popular and biggest-selling soft drink in ...

Answer 1

只需使用truncate with separator选项：

truncate("Once upon a time in a world far far away", length: 17)
# => "Once upon a ti..."
truncate("Once upon a time in a world far far away", length: 17, separator: ' ')
# => "Once upon a..."

获取更多信息：truncate helper in rails API documentation

Answer 2

此方法使用正则表达式，贪婪地抓取多达70个字符，然后匹配字符串的空格或结尾以实现您的目标

def truncate(s, max=70, elided = ' ...')
  s.match( /(.{1,#{max}})(?:\s|\z)/ )[1].tap do |res|
    res << elided unless res.length == s.length
  end    
end

s = "Coca-Cola is the most popular and biggest-selling soft drink in history, as well as the best-known brand in the world."
truncate(s)
=> "Coca-Cola is the most popular and biggest-selling soft drink in ..."

Answer 3

s = "Coca-Cola is the most popular and biggest-selling soft drink in history, as well as the best-known brand in the world."
s = s.split(" ").each_with_object("") {|x,ob| break ob unless (ob.length + " ".length + x.length <= 70);ob << (" " + x)}.strip
#=> "Coca-Cola is the most popular and biggest-selling soft drink in"

Answer 4

s[0..65].rpartition(" ").first << " ..."

在你的考试中：

s = "Coca-Cola is the most popular and biggest-selling soft drink in history, as well as the best-known brand in the world."    
t = s[0..65].rpartition(" ").first << " ..."
=> "Coca-Cola is the most popular and biggest-selling soft drink in ..."

Answer 5

（受dbenhur的回答启发，但更好地处理第一个最大字符中没有空格或字符串结尾的情况。）

def truncate(s, options = { })
  options.reverse_merge!({
     max: 70,
     elided: ' ...'
  });

  s =~ /\A(.{1,#{options[:max]}})(?:\s|\z)/
  if $1.nil? then s[0, options[:max]] + options[:elided]
  elsif $1.length != s.length then $1 + options[:elided]
  else $1
  end
end

Answer 6

library(data.table)
uber_names <- dt$uber_name
dt[, uber_write_filtered := gsub(
    pattern = paste0("(", paste(uber_names, collapse = "|"), ")"),
    replacement = "", uber_write)]

从字符串中获取前N个字符而不删除整个单词

6 个答案: