我正在尝试对数组进行排序，但它总是返回相同的原始输入。你知道为什么吗？
puts "Please tell me the array(separated by space, finish by enter twice):"
x = []
input = ' '
while input != ''
input = gets.chomp
x.push input
end
puts x.sort
【问题讨论】：
你需要split 输入和concat 到一个奇异数组：
x = [ ]
loop do
input = gets.chomp
break if (input.empty?)
x += input.split
end
puts x.sort
【讨论】：
split 相当于split(/\s+/)。
split 的不同方式。很有启发性。
使用String#split。看看：
x = '1 3 2 4 55'
x.split.sort
#=> ["1", "2", "3", "4", "55"]
如果x 是一个字符串数组，您可以执行以下操作：
x = ['1 3 100', '2 30 4 10']
x.flat_map { |string| string.split }.sort
#=> ["1", "10", "100", "2", "3", "30", "4"] Note it is sorted alphabetically.
如果您想对输入进行数字排序，请考虑：
x = ['1 3 100', '2 30 4 10']
x.flat_map { |string| string.split.map(&:to_i) }.sort
#=> [1, 2, 3, 4, 10, 30, 100]
虽然在现实生活中我更喜欢 @tadman 的解决方案，因为它会立即对输入进行操作，统一输入。
【讨论】：
x 是一个数组，而不是一个字符串。不过你走在正确的轨道上。
flat_map 或 .flatten 会解决这个问题。
split 等同于split(' ')。
思考这个：
x = [] # => []
input = '1 2 3' # => "1 2 3"
x.push input # => ["1 2 3"]
x.sort # => ["1 2 3"]
基本上，您在代码中输入单个字符串，将其推入数组，然后对单个元素数组进行排序。这不会做任何事情，因为您无法对单个元素数组进行排序并获得不同的顺序。 （嗯，你可以期待它会有所不同，但事实并非如此。）
改为：
x = [] # => []
x.push '1' # => ["1"]
x.push '3' # => ["1", "3"]
x.push '2' # => ["1", "3", "2"]
模拟输入三个不同的字符串。排序x 现在返回一个排序数组：
x.sort # => ["1", "2", "3"]
这里还有一些值得思考的事情：
require 'fruity'
string = ('a'..'z').to_a.join(' ')
string # => "a b c d e f g h i j k l m n o p q r s t u v w x y z"
3.times do
compare do
split1 { string.split }
split2 { string.split(' ') }
split3 { string.split(/\s/) }
split4 { string.split(/\s+/) }
end
end
# >> Running each test 1024 times. Test will take about 1 second.
# >> split1 is faster than split2 by 10.000000000000009% ± 10.0%
# >> split2 is faster than split3 by 3x ± 0.1
# >> split3 is similar to split4
# >> Running each test 2048 times. Test will take about 1 second.
# >> split1 is faster than split2 by 10.000000000000009% ± 10.0%
# >> split2 is faster than split3 by 3x ± 0.1
# >> split3 is similar to split4
# >> Running each test 1024 times. Test will take about 1 second.
# >> split1 is faster than split2 by 10.000000000000009% ± 10.0%
# >> split2 is faster than split3 by 3x ± 0.1
# >> split3 is similar to split4
split1 和 split2 的结果大约有 10% 支持 split1，因此，Fruity 有时会报告 10% 的差异，有时则不会。 split3 和 split4 也是如此。
还有：
string = ('a'..'z').to_a.zip(26.times.map{ ' ' * (rand(4) + 1)}).join
string # => "a b c d e f g h i j k l m n o p q r s t u v w x y z "
compare do
split1 { string.split }
split2 { string.split(' ') }
split3 { string.split(/\s/) }
split4 { string.split(/\s+/) }
end
# >> Running each test 1024 times. Test will take about 1 second.
# >> split1 is similar to split2
# >> split2 is faster than split4 by 3x ± 0.1
# >> split4 is faster than split3 by 2.1x ± 0.1 (results differ: ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"] vs ["a", "", "", "", "b", "", "c", "", "", "", "d", "", "", "", "e", "", "f", "", "", "g", "", "h", "", "", "i", "j", "", "k", "l", "", "m", "", "", "", "n", "", "o", "", "p", "q", "r", "", "", "", "s", "", "", "t", "", "", "u", "", "v", "w", "", "x", "", "", "", "y", "z"])
最后一个结果是不同的，因为split(/\s/) 只寻找单个空白字符，而不是像split(/\s+/) 那样的倍数。我留下这个结果是为了展示 + 对引擎的影响。
【讨论】：
如果您在一行中输入所有数字，那么您只需拨打一次gets。见下文。我们读取单行输入，将输入拆分为任意数量的空格并将值转换为其整数形式。然后我们将它们排序显示。
$ cat foo.rb
puts "Please tell me the array(separated by space, finish by enter twice):"
input = gets.chomp
x = input.split(/\s+/).map(&:to_i)
puts x.sort
$ ruby foo.rb
Please tell me the array(separated by space, finish by enter twice):
2423 888 3333 222
222
888
2423
3333
【讨论】：
您的代码需要在不同的行中输入项目。一个简单的“修复”是调整帮助信息：
puts "Please tell me the array (one item per line, finish by enter twice)"
并按预期提供物品：
$ ruby sort.rb
Please tell me the array (one item per line, finish by enter twice)
b
c
a
a
b
c
请注意，您的代码会将空行添加到数组中。这可以通过检查input来解决：
x = []
input = ' '
while input != ''
input = gets.chomp
x.push input unless input.empty?
end
使用loop 和break 代替while 可以避免必须检查空input 两次（input != '' / input.empty?）和随机初始值（input = ' '）：
x = []
loop do
input = gets.chomp
break if input.empty?
x.push input
end
【讨论】：
如果您希望对数字数组进行排序，请更改此行：
x.push input.to_i
否则输入如下：
4
10
1
将被订购：
1
10
4
如果您需要字符串，那么您的程序应该可以运行，并且应该提供更多详细信息。
【讨论】：