Ruby：如何计算一个字符串在另一个字符串中出现的次数？
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Ruby: How to count the number of times a string appears in another string?
提问by Johnson
I'm trying to count the number of times a string appears in another string.
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。
I know you can count the number of times a letter appears in a string:
我知道你可以计算一个字母出现在字符串中的次数：
string = "aabbccddbb"
string.count('a')
=> 2
But if I search for how many times 'aa' appears in this string, I also get two.
但是如果我搜索 'aa' 在这个字符串中出现的次数，我也会得到两次。
string.count('aa')
=> 2
I don't understand this. I put the value in quotation marks, so I'm searching for the number of times the exact string appears, not just the letters.
我不明白这个。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。
回答by Cary Swoveland
Here are a couple of ways to count the numbers of times a given substring appears in a string (the first being my preference). Note (as confirmed by the OP) the substring 'aa'appears twice in the string 'aaa', and therefore five times in:
这里有几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（由 OP 确认）子字符串'aa'在 string 中出现两次'aaa'，因此在：
string="aaabbccaaaaddbb"
#1
#1
Use String#scanwith a regex that contains a positive lookahead that looks for the substring:
将String#scan与包含查找子字符串的正向前瞻的正则表达式一起使用：
def count_em(string, substring)
string.scan(/(?=#{substring})/).count
end
count_em(string,"aa")
#=> 5
Note:
笔记：
"aaabbccaaaaddbb".scan(/(?=aa)/)
#=> ["", "", "", "", ""]
A positive lookbehind produces the same result:
正向后视产生相同的结果：
"aaabbccaaaaddbb".scan(/(?<=aa)/)
#=> ["", "", "", "", ""]
As well, String#scancan be replaced with String#gsub.
同样，String#scan可以替换为String#gsub。
#2
#2
Convert to an array, apply Enumerable#each_cons, then join and count:
转换为数组，应用Enumerable#each_cons，然后加入并计数：
def count_em(string, substring)
string.each_char.each_cons(substring.size).map(&:join).count(substring)
end
count_em(string,"aa")
#=> 5
We have:
我们有：
enum0 = "aaabbccaaaaddbb".each_char
#=> #<Enumerator: "aaabbccaaaaddbb":each_char>
We can see the elements that will generated by this enumerator by converting it to an array:
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", "b"]
enum1 = enum0.each_cons("aa".size)
#=> #<Enumerator: #<Enumerator: "aaabbccaaaaddbb":each_char>:each_cons(2)>
Convert enum1to an array to see what values the enumerator will pass on to map:
转换enum1为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", "b"]]
c = enum1.map(&:join)
#=> ["aa", "aa", "ab", "bb", "bc", "cc", "ca",
# "aa", "aa", "aa", "ad", "dd", "db", "bb"]
c.count("aa")
#=> 5
回答by tadman
It's because the countcounts characters, not instances of strings. In this case 'aa'means the same thing as 'a', it's considered a set of characters to count.
这是因为count计数字符，而不是字符串的实例。在这种情况下'aa'，与 含义相同'a'，它被认为是一组要计算的字符。
To count the number of times aaappears in the string:
要计算aa字符串中出现的次数：
string = "aabbccddbb"
string.scan(/aa/).length
# => 1
string.scan(/bb/).length
# => 2
string.scan(/ff/).length
# => 0
回答by Georgii Gavrilchik
try to use string.split('a').count - 1
尝试使用 string.split('a').count - 1
