假设我找回了一个字符串：
"27,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,12,17,17,41,17,17,17,17,17,17,17,17,17,17,17,17,17,26,26,26,26,26,26,26,26,26,29,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,40,48,28,28,28,28,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,34,34,34,34,34,34,36,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,40,40,40,40,40,40,40,40,41,41,41,41,41,41,41,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,43,43,43,43,43,43,43,43,43,43,43,43,43,44,44,44,44,48,49,29,41,6,30,11,29,29,36,29,29,36,29,43,1,29,29,29,1,41"
我通过调用
将其转换为数组str.split(',')
然后通过调用
将其变为哈希arr.compact.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }
我会找回一个看起来像
的哈希{"1"=>2, "6"=>1, "39"=>23, "36"=>23, "34"=>39, "32"=>31, "30"=>18, "3"=>8, "2"=>10, "28"=>36, "29"=>21, "26"=>41, "27"=>48, "49"=>1, "44"=>4, "43"=>14, "42"=>34, "48"=>2, "40"=>9, "41"=>10, "11"=>1, "17"=>15, "12"=>1}
但是，我想按键排序。
我已尝试列出here列出的解决方案。
我相信我的问题与他们的键是字符串的事实有关。
我最接近的是使用
Hash[h.sort_by{|k,v| k.to_i}]
答案 0 :(得分：5)
不应将散列视为已排序的数据结构。它们具有其他优点和用例，以便按顺序返回它们的值。正如MladenJablanović已经指出，当你需要一个有序的键/值对时，一组元组可能是更好的数据结构。
当前版本的Ruby中的 但是实际上存在一种顺序，当您在哈希上调用示例each时，会返回键/值对，这就是插入的顺序。使用此行为，您只需构建一个新哈希，并按所需顺序将所有键/值对插入到新哈希中。但请记住，稍后添加更多条目时，订单将会中断。
string = "27,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,12,17,17,41,17,17,17,17,17,17,17,17,17,17,17,17,17,26,26,26,26,26,26,26,26,26,29,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,40,48,28,28,28,28,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,34,34,34,34,34,34,36,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,40,40,40,40,40,40,40,40,41,41,41,41,41,41,41,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,43,43,43,43,43,43,43,43,43,43,43,43,43,44,44,44,44,48,49,29,41,6,30,11,29,29,36,29,29,36,29,43,1,29,29,29,1,41"
sorted_number_count_tupels = string.split(',').
group_by(&:itself).
map { |k, v| [k, v.size] }.
sort_by { |(k, v)| k.to_i }
#=> [["1",2],["2",10],["3",8],["6",1],["11",1],["12",1],["17",15],["26",41],["27",48],["28",36],["29",21],["30",18],["32",31],["34",39],["36",23],["39",23],["40",9],["41",10],["42",34],["43",14],["44",4],["48",2],["49",1]]
sorted_number_count_hash = sorted_number_count_tupels.to_h
#=> { "1" => 2, "2" => 10, "3" => 8, "6" => 1, "11" => 1, "12" => 1, "17" => 15, "26" => 41, "27" => 48, "28" => 36, "29" => 21, "30" => 18, "32" => 31, "34" => 39, "36" => 23, "39" => 23, "40" => 9, "41" => 10, "42" => 34, "43" => 14, "44" => 4, "48" => 2, "49" => 1}
答案 1 :(得分：3)
假设您从
开始str = "27,2,2,2,41,26,26,26,48,48,41,6,11,1,41"
并创建了以下哈希
h = str.split(',').inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }
#=> {"27"=>1, "2"=>3, "41"=>3, "26"=>3, "48"=>2, "6"=>1, "11"=>1, "1"=>1}
我删除了compact，因为数组str.split(',')仅包含（可能为空）字符串，而不包含nil。
在继续之前，您可能希望将最后一步更改为
h = str.split(/\s*,\s*/).each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
#=> {"27"=>1, "2"=>3, "41"=>3, "26"=>3, "48"=>2, "6"=>1, "11"=>1, "1"=>1}
在正则表达式上拆分允许在每个逗号之前或之后有一个或多个空格，Enumerable#each_with_object可以避免需要那个讨厌的; h。 （注意块变量是相反的。）
然后
h.sort_by { |k,_| k.to_i }.to_h
#=> {"1"=>1, "2"=>3, "6"=>1, "11"=>1, "26"=>3, "27"=>1, "41"=>3, "48"=>2}
创建一个新哈希，其中包含按键的整数表示排序的h键值对。请参阅Hash#sort_by。
请注意，我们创建了两个哈希值。通过修改h，可以通过以下方式实现这一目标。
h.keys.sort_by(&:to_i).each { |k| h[k] = h.delete(k) }
#=> ["1", "2", "6", "11", "26", "27", "41", "48"] (each always returns the receiver)
h #=> {"1"=>1, "2"=>3, "6"=>1, "11"=>1, "26"=>3, "27"=>1, "41"=>3, "48"=>2}
最后，另一种方法是在创建哈希值之前对str.split(',')进行排序。
str.split(',').sort_by(&:to_i).each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
#=> {"1"=>1, "2"=>3, "6"=>1, "11"=>1, "26"=>3, "27"=>1, "41"=>3, "48"=>2}
答案 2 :(得分：2)
String#split无法返回nil元素。 compact在这里没用。但split可能会返回一个空字符串：
p "1,,2,3".split(',')
# ["1", "", "2", "3"]
p "1,,2,3".split(',').compact
# ["1", "", "2", "3"]
p "1,,2,3".split(',').reject(&:empty?)
# ["1", "2", "3"]
如果您必须在inject块中使用两个语句，each_with_object可能是个更好的主意：
arr.compact.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }
可以改写：
arr.compact.each_with_object(Hash.new(0)) { |e, h| h[e] += 1 }
如果需要对结果进行排序，则对数组可能比哈希更适合。
如果您接受将整数作为键，则可能使您的代码更容易编写。
可以重写代码：
str.split(',')
.reject(&:empty?)
.map(&:to_i)
.group_by(&:itself)
.map { |k, v| [k, v.size] }
.sort
输出：
[[1, 2], [2, 10], [3, 8], [6, 1], [11, 1], [12, 1], [17, 15], [26, 41], [27, 48], [28, 36], [29, 21], [30, 18], [32, 31], [34, 39], [36, 23], [39, 23], [40, 9], [41, 10], [42, 34], [43, 14], [44, 4], [48, 2], [49, 1]]
如果您真的想要哈希，可以添加.to_h：
{1=>2, 2=>10, 3=>8, 6=>1, 11=>1, 12=>1, 17=>15, 26=>41, 27=>48, 28=>36, 29=>21, 30=>18, 32=>31, 34=>39, 36=>23, 39=>23, 40=>9, 41=>10, 42=>34, 43=>14, 44=>4, 48=>2, 49=>1}
答案 3 :(得分：0)
您可以将arr.compact.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }分配给变量并按键对其进行排序：
num = arr.compact.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }
num.keys.sort
这会按键对哈希进行排序。
答案 4 :(得分：0)
Ruby哈希将保持添加键的顺序。如果数组足够小以排序我只会改变
str.split(',').
到
str.split(',').sort_by(&:to_i)
为了获取值，因此您也散列排序...