ruby重新排序日期数组，字符串部分保留空字符串

Question

我有一个包含空字符串的日期数组，我想用两种方式重新排序，每种情况下，空字符串应该总是在数组的前面。

[" ", "2018-10-01", "2019", " ",  "2019-06-20", "2019-06", "2019-10", "2019-01-01", "2017", "2018-05", "2018", "2018-05-10",  " "]

第一个结果是用这种方式重新排序。

["", "", "", "2019-10-01", "2019-10", "2019-06-20", "2019-06", "2019", "2018-01-01", "2018-05", "2018-05-10", "2017"]

第二种方式，我想要这样重新订购：

[" ", " ", " ", "2017", "2018-01-01", "2019-06-20", "2019-06", "2019-10-01", "2019-10", "2019"]

我尝试了下面的代码，但没有得到预期的结果。

  ["", "", "", "2019-10-01", "2019-10", "2019-06-20", "2019-06", "2019", "2018-01-01", "2018-05", "2018-05-10", "2017"].sort_by { |date| parts = date.split('-').map(&:to_i) }

更新

提升的排序是否有可能以这个顺序出现。yy，yy，然后仅仅是年份，这意味着排序升序应该以[" ", " ", " ", "2017","2018-05-10", "2018-05", "2018-10-01", "2018", "2019-01-01", "2019-06-20", "2019-06", "2019-10", "2019"]的形式出现。因此，它不是一个上升的规则排序，而是基于导致上述顺序的模式。

同样的模式适用于排序降序。

Answer 1

解决方案1-正常排序

您可以根据字符串是否为空这一事实，首先对列表进行分区。然后对现在的日期进行排序，并将它们连在一起。

dates = [" ", "2018-10-01", "2019", " ",  "2019-06-20", "2019-06", "2019-10", "2019-01-01", "2017", "2018-05", "2018", "2018-05-10",  " "]

asc = ->(a, b) { a <=> b }
desc = ->(a, b) { b <=> a }

blank, present = dates.partition(&:blank?)
result1 = blank + present.sort(&desc)
#=> [" ", " ", " ", "2019-10", "2019-06-20", "2019-06", "2019-01-01", "2019", "2018-10-01", "2018-05-10", "2018-05", "2018", "2017"]
result2 = blank + present.sort(&asc)
#=> [" ", " ", " ", "2017", "2018", "2018-05", "2018-05-10", "2018-10-01", "2019", "2019-01-01", "2019-06", "2019-06-20", "2019-10"]

备注：，这只是根据字母顺序对数组进行排序。只要您使用的是yyyy-mm-dd格式(如果只有一位数，则使用零运算)，这是很好的。如果日期是以另一种格式提供的，则首先要将其转换为日期。

解决方案2-假定丢失月或日的最高值

strings = [" ", "2018-10-01", "2019", " ",  "2019-06-20", "2019-06", "2019-10", "2019-01-01", "2017", "2018-05", "2018", "2018-05-10",  " "]

array_to_date = lambda do |(year, month, day)|
  month ||= 12
  day ||= 31

  begin
    Date.new(year, month, day)
  rescue ArgumentError
    raise unless (1..12).cover? month
    raise unless (1..31).cover? day
    array_to_date.call([year, month, day - 1])
  end
end

date_regex = /\A(\d{4})(?:-(\d{2})(?:-(\d{2}))?)?\z/
yyyy_mm_dd = ->(date_string) { date_regex.match(date_string).captures.compact.map(&:to_i) }
string_to_date = yyyy_mm_dd >> array_to_date

asc = ->(a, b) { string_to_date.call(a) <=> string_to_date.call(b) }
desc = ->(a, b) { string_to_date.call(b) <=> string_to_date.call(a) }

dates, non_dates = strings.partition(&date_regex.method(:match?))
result1 = non_dates + dates.sort(&desc)
#=> [" ", " ", " ", "2019", "2019-10", "2019-06", "2019-06-20", "2019-01-01", "2018", "2018-10-01", "2018-05", "2018-05-10", "2017"]
result2 = non_dates + dates.sort(&asc)
#=> [" ", " ", " ", "2017", "2018-05-10", "2018-05", "2018-10-01", "2018", "2019-01-01", "2019-06-20", "2019-06", "2019-10", "2019"]

注释：--这不是最有效的解决方案，因为每次调用#sort块时都必须转换字符串。如果您正在处理大型数组，您可以首先转换所有的值并将它们保存在一个散列中。整理的时候查一下。

当前正则表达式还允许像"0000-00-00"这样的字符串传递，您可能希望使它更加具体。

Answer 2

arr = [" ", "2018-10-01", "2019", " ",  "2019-06-20", "2019-06", "2019-10",
       "2019-01-01", "2017", "2018-05", "2018", "2018-05-10",  " "]

上升的类型：

def sort_asc(arr)
  arr.sort
end

​

sort_asc(arr)
  #=> [" ", " ", " ",
  #    "2017",
  #    "2018", "2018-05", "2018-05-10", "2018-10-01",
  #    "2019", "2019-01-01", "2019-06", "2019-06-20", "2019-10"]

下降的那一类：

def sort_dsc(arr)
  arr.sort_by { |s| s == ' ' ? '99' : s }.reverse
end

​

sort_dsc(arr)
  #=> [" ", " ", " ",
  #    "2019-10", "2019-06-20", "2019-06", "2019-01-01", "2019",
  #    "2018-10-01", "2018-05-10", "2018-05", "2018",
  #    "2017"]