Question

我有一个包含空字符串的日期数组，我想用两种方式对其重新排序，每种情况下，空字符串应始终位于数组的前面。

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

第一个结果就是以这种方式重新排序

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

我希望能够以这种方式重新排序的第二种方式：

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

我尝试了以下代码，但没有得到期望的结果。

  ["", "", "", "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

升序排序有可能以此顺序出现。 yy-mm-dd，然后是yy-mm，然后是年份，这意味着排序应采用这种形式 [" ", " ", " ", "2017","2018-05-10", "2018-05", "2018-10-01", "2018", "2019-01-01", "2019-06-20", "2019-06", "2019-10", "2019"]。因此，它不是常规的升序排序，而是基于导致上述顺序的模式。

相同的模式适用于降序排列。

Answer 1

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

升序：

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"]

降序排列：

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"]

Answer 2

解决方案＃1-常规排序

您可以首先基于字符串是否为空的事实对列表进行分区。然后对所有当前日期进行排序，并将它们重新连接在一起。

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"]

注意：这只是根据字母顺序对数组进行排序。只要您使用yyyy-mm-dd格式（如果只有一位，则使用零）就可以了。如果以其他格式提供日期，则您要先将其转换为日期。

解决方案2-假定缺少的月份或日期为最高值

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"]

注意：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。如果使用大型数组，则可以先转换所有值，然后将它们保存在哈希中。然后在排序时查找它们。

当前的正则表达式还允许传递"0000-00-00"之类的字符串，您可能想使其更加具体。

红宝石对日期进行重新排序，字符串部分保留空字符串

2 个答案:

解决方案＃1-常规排序

解决方案2-假定缺少的月份或日期为最高值