如何计算角色连续出现的次数
我的代码适用于常规字符数
count = Hash.new(0) str.each_char do |char| count[char] += 1 unless char == " " end count 例如， "aaabbaaaaacccbbdddd"等于’a’= 8，’b’= 4，’c’= 3，’d’= 4。
我想连续发生多少次。 我想要的结果是：’a’= 3，’b’= 2，’a’= 5’c’= 3，’b’= 2，’d’= 4.我该怎么做？
"aaabbaaaaacccbbdddd".each_char.chunk(&:itself).map{|k, v| [k, v.length]} # => [["a", 3], ["b", 2], ["a", 5], ["c", 3], ["b", 2], ["d", 4]]
我对sawa和spickermann的解决方案进行了基准测试：
require 'benchmark/ips' def sawa(string) string.each_char.chunk(&:itself).map{|k, v| [k, v.length] } end def spickermann(string) string.split(//).slice_when { |a, b| a != b }.map { |group| [group.first, group.size] } end Benchmark.ips do |x| string = "aaabbaaaaacccbbdddd" x.report("sawa") { sawa string } x.report("spickerman") { spickermann string } x.compare! end # Calculating ------------------------------------- # sawa 6.293ki/100ms # spickermann 4.447ki/100ms # ------------------------------------------------- # sawa 75.353k (±10.4%) i/s - 371.287k # spickermann 48.661k (±12.0%) i/s - 240.138k # # Comparison: # sawa: 75353.5 i/s # spickermann: 48660.7 i/s - 1.55x slower
关于什么：
string.split(//).slice_when { |a, b| a != b }. map { |group| [group.first, group.size] } #=> [['a', 3], ['b', 2], ['a', 5], ['c', 3], ['b', 2], ['d', 4]]
而不是哈希，使用数组来存储对，如你所见。
str = "aaabbaaaaacccbbdddd" counts = [] str.each_char do |char| # Get the last seen character and count pair last_pair = counts[-1] || [] if last_pair[0] == char # This character is the same as the last one, increment its count last_pair[1] += 1 else # New character, push a new pair onto the list counts.push([char, 1]) end end counts.each { |c| puts "#{c[0]} = #{c[1]}" }
这可以使用chunk更简洁地编写。
str = "aaabbaaaaacccbbdddd" counts = [] str.chars.chunk(&:itself).each { |char, chars| counts << [char, chars.length] } puts counts.inspect
chunk将列表拆分为块。 它通过调用每个元素上的块来决定这一点。 只要块返回与前一个值相同的值，它就会添加到当前块。 一旦它改变，它就会产生一个新的块。 这类似于我们之前在循环中通过存储最后看到的字符所做的事情。
if last_seen == char # it's the same chunk else # it's a new chunk last_seen = char end
itself返回角色。 所以chunk(&:itself)会将字符串拆分成字符块。
新列表是chunk(&:itself)的返回值（在我们的例子中是此块中的字符）加上实际的块（例如字符串“aaa”）。
我更喜欢这种问题的正则表达式：
str = "aaabbaaaaacccbbdddd" counts = str.scan(/(?(?\w)\k +)/).inject([]) do |occurs, match| occurs << [match[1], match[0].size] occurs end puts counts.inspect #=>[["a", 3], ["b", 2], ["a", 5], ["c", 3], ["b", 2], ["d", 4]]
编辑：
我用@sawa运行了相同的基准测试，并添加了正则表达式方式。 看起来好一点。 此外， #itself不适用于ruby < 2.2.x
require 'benchmark/ips' def sawa(string) string.each_char.chunk(&:itself).map{|k, v| [k, v.length] } end def spickermann(string) string.split(//).slice_when { |a, b| a != b }.map { |group| [group.first, group.size] } end def stathopa(string) string.scan(/(?(?\w)\k +)/).inject([]) do |occurs, match| occurs << [match[1], match[0].size] occurs end end Benchmark.ips do |x| string = "aaabbaaaaacccbbdddd" x.report("sawa") { sawa string } x.report("spickerman") { spickermann string } x.report("stathopa") { stathopa string } x.compare! end # Calculating ------------------------------------- # sawa 6.730ki/100ms # spickerman 4.061ki/100ms # stathopa 11.969ki/100ms # ------------------------------------------------- # sawa 70.072k (± 8.9%) i/s - 349.960k # spickerman 43.652k (± 9.5%) i/s - 219.294k # stathopa 132.992k (± 8.8%) i/s - 670.264k # # Comparison: # stathopa: 132992.1 i/s # sawa: 70072.4 i/s - 1.90x slower # spickerman: 43651.6 i/s - 3.05x slower #
要计算每个字符的最大长度序列：
count = Hash.new(0) last_char = nil occurred = 0 str.each_char do |char| if char != last_char occurred = 1 else occurred += 1 end last_char = char count[char] = occurred if (count[char]||0) < occurred end count
或者得到像[['a'，3]，['b'，2]，['a'，5]，['c'，3]，['b'，2]，['d'的结果”，4]]：
count = [] last_char = nil occurred = 0 str.each_char do |char| if char != last_char count.push([last_char, occurred]) occurred = 1 else occurred += 1 end last_char = char end count.push([last_char, occurred]) count
这是一种方法：
s = "aaabbaaaaacccbbdddd" s.chars.uniq.map do |c| p [c, s.split(/[^#{c}]+/).reject(&:empty?).map(&:size)] end.to_h #=> {"a"=>[3, 5], "b"=>[2, 2], "c"=>[3], "d"=>[4]}