Ruby：如何计算一个字符串在另一个字符串中出现的次数？

我正在尝试计算一个字符串在另一个字符串中出现的次数。

我知道你可以计算一个字母在字符串中出现的次数：

string = "aabbccddbb"
string.count('a')
=> 2

但是，如果我搜索“aa”在此字符串中出现的次数，我也会得到两次。

string.count('aa')
=> 2

我不明白这一点。我将值放在引号中，因此我正在搜索确切字符串出现的次数，而不仅仅是字母。

这里有两种方法来计算给定子字符串在字符串中出现的次数（第一种是我的偏好）。注意（经OP确认）子字符串'aa'在字符串中出现两次'aaa'，因此五次：

str = "aaabbccaaaaddbab"

1. Use 字符串#scan http://www.ruby-doc.org/core-2.1.0/String.html#method-i-scan使用包含查找给定子字符串的正向前瞻的正则表达式

def count_em(str, substr)
  str.scan(/(?=#{substr})/).count
end

count_em(str,"aa")
  #=> 5
count_em(str,"ab")
  #=> 2

Note:

"aaabbccaaaaddbab".scan(/(?=aa)/)
  #=> ["", "", "", "", ""]

正向回顾会产生相同的结果：

"aaabbccaaaaddbab".scan(/(?<=aa)/)
  #=> ["", "", "", "", ""]

还有，String#scan可以替换为以下形式字符串#gsub http://ruby-doc.org/core-2.5.1/String.html#method-i-gsub它接受一个参数（这里是相同的正则表达式）并且没有块，并返回一个枚举器。那种形式的gsub不寻常之处在于与角色替换无关；它只是生成正则表达式的匹配项。

2. 将给定字符串转换为字符数组，应用字符串#each_char https://ruby-doc.org/core-2.7.0/String.html#method-i-each_char then 可枚举#each_cons http://www.ruby-doc.org/core-2.1.1/Enumerable.html#method-i-each_cons, then 可枚举#count https://ruby-doc.org/core-2.7.0/Enumerable.html#method-i-count

def count_em(str, substr)
  subarr = substr.chars
  str.each_char
     .each_cons(substr.size)
     .count(subarr)
end

count_em(str,"aa")
  #=> 5
count_em(str,"ab")
  #=> 2

We have:

subarr = "aa".chars
  #=> ["a", "a"]
enum0 = "aaabbccaaaaddbab".each_char
  #=> #<Enumerator: "aaabbccaaaaddbab":each_char>

我们可以通过将其转换为数组来查看此枚举器将生成​​的元素：

enum0.to_a
  #=> ["a", "a", "a", "b", "b", "c", "c", "a", "a", "a",
  #    "a", "d", "d", "b", "a", "b"]

enum1 = enum0.each_cons("aa".size)
  #=> #<Enumerator: #<Enumerator:
  #      "aaabbccaaaaddbab":each_char>:each_cons(2)> 

Convert enum1到一个数组以查看枚举器将传递给哪些值map:

enum1.to_a
  #=> [["a", "a"], ["a", "a"], ["a", "b"], ["b", "b"], ["b", "c"],
  #    ["c", "c"], ["c", "a"], ["a", "a"], ["a", "a"], ["a", "a"], 
  #    ["a", "d"], ["d", "d"], ["d", "b"], ["b", "a"],
  #    ["a", "b"]]
 
enum1.count(subarr)
  #=> enum1.count(["a", "a"])
  #=> 5