Get the complement date range

Question

Suppose that I have a date range list

drlist = {{"2015-01-01", "2015-01-05"}, 
          {"2015-01-07", "2015-01-07"}, 
          {"2015-01-12", "2015-01-23"}}

and a range of date

sdate = "2015-01-04";
edate = "2015-01-14";
r = {sdate, edate};

to make things easy to understand, {"2015-01-04", "2015-01-14"} in reality represent something like {"2015-01-04-00-00","2015-01-14-12-00"}, i.e., the date start from 0:00 but end at some time of the day, say "12:00" is enough for me.

Then my question is how can I get the range of date that in r but not in drlist, e.g., it should be output something like:

cdrlist = {{"2015-01-05", "2015-01-06"}, {"2015-01-07", "2015-01-11"}}

and if we set endate = "2015-01-25", then the output should be

cdrlist = {{"2015-01-05", "2015-01-06"}, 
           {"2015-01-07", "2015-01-11"}, 
           {"2015-01-23", "2015-01-25"}}

It should be noted that in this case the day "25" should be contained included.

UPDATE

To clarify the confusing results I were demonstrated, let us just consider the day, the day range we have is union of closed intervals: $$ drlist=[1.0,5.5]\sqcup[7.0,7.5]\sqcup[12.0,23.5] $$ and the day range we set is just a closed interval: $$ r=[4.0,25.5] $$ thus the set $r\setminus drlist$ equals to $$ r\setminus drlist=(5.5,7)\sqcup(7.5,12)\sqcup(23.5,25.5], $$ and translated this back to day range shoulde be $$ 5-6,7-11,23-25. $$ Let me explain why $(5.5,7)$ should be $5-6$? Note that in our notation, $5-6$ represent $[5.0,6.5]$ which is exactly the interval contains $(5.5,7)$, since the $0.5$ should be understand very close to $1.0$. Also $(23.5,25.5]=(24,26)=[23,25]$ in above sense.

Answer 1

I claim it is impossible to remain consistent with your convention of adding hours to second elements. Either we keep track of hours everythwere or we are in trouble.

So now I'm going to ignore that :)

drlist = {{"2015-01-01", "2015-01-05"}, 
          {"2015-01-07", "2015-01-08"}, 
          {"2015-01-12", "2015-01-23"}};
r = {"2015-01-04", "2015-01-14"};

I'm going to work on seconds:

restTime = Interval @@ Partition[Flatten[{0, Map[AbsoluteTime, drlist, {2}], 10^10}],
                                 2]

main = Interval[AbsoluteTime /@ r]

Interval[{0, 3629059200}, {3629404800, 3629577600}, 
         {3629664000, 3630009600}, {3630960000, 10000000000}]
Interval[{3629318400, 3630182400}]

List @@ Map[DateList[#][[;; 3]] &, 
            IntervalIntersection[main, restTime],
            {2}]

{{{2015, 1, 5}, {2015, 1, 7}}, {{2015, 1, 8}, {2015, 1, 12}}}

I'm living formatting to you. Also, if you want to add DatePlus for those second elements, feel free to do so :)

Answer 2

This first part finds the complementary dates.

sdate = "2015-01-04";
edate = "2015-01-14";
r = {sdate, edate};

drlist = {
   {"2015-01-01", "2015-01-05"},
   {"2015-01-07", "2015-01-07"},
   {"2015-01-12", "2015-01-23"}};

If[$VersionNumber == 10,
 fromDate[datelist_List] :=
  QuantityMagnitude@DateDifference[{1899, 12, 30}, datelist],
 fromDate[datelist_List] := DateDifference[{1899, 12, 30}, datelist]] (* Note 1 *)

undoDate[int_Integer] := StringJoin@Riffle[
   ToString /@ DatePlus[{1899, 12, 30}, int] , "-"]

r2 = r /. a_String :> fromDate[Take[DateList[{a, {"Year", "Month", "Day"}}], 3]];
drlist2 = drlist /. a_String :> fromDate[Take[DateList[{a, {"Year", "Month", "Day"}}], 3]];

r3 = Range[0, #2 - #1] + #1 & @@ r2;
drlist3 = Union@Flatten[Range[0, #2 - #1] + #1 & @@@ drlist2];

cdrlist1 = Complement[r3, drlist3];
undoDate /@ cdrlist1

{"2015-1-6", "2015-1-8", "2015-1-9", "2015-1-10", "2015-1-11"}

Now finding the date ranges. (It would be interesting to see better ways of doing this.)

cdrlist2 = Transpose[{cdrlist1,
    Append[Differences@cdrlist1 /. x_ /; x > 1 -> 0, 0]}];

If[cdrlist2[[1, 2]] == 1, output = {},
  output = List@Take[cdrlist2, 1];
  cdrlist2 = Rest@cdrlist2];

While[Length[cdrlist2] > 0,
 a = TakeWhile[cdrlist2, Last[#] == 1 &];
 b = Drop[cdrlist2, Length[a]];
 If[b == {}, AppendTo[output, a],
  AppendTo[a, First[b]];
  AppendTo[output, a];
  cdrlist2 = Rest[b]]]

cdrlist3 = Replace[output, x_List /; Length[x] > 1 :> x[[{1, -1}]], {1}];

cdrlist = cdrlist3 /. {a_Integer, _Integer} :> undoDate[a]

{{"2015-1-6"}, {"2015-1-8", "2015-1-11"}}

Note 1. fromDate[] actually converts a date to an Excel serial date, but valid only from 1st March 1990 due to Excel's compatibility carry-over of the Lotus 1-2-3 leap-year bug. (1990 is not a leap-year.)

Answer 3

I do not follow your logic for why {{"2015-01-05", "2015-01-06"}, {"2015-01-07", "2015-01-11"}} is split into two ranges since "2015-01-06" and "2015-01-07" are adjacent, however overlooking that here is a fairly clean approach.

Helper functions:

toList[x : {_String, _String}, os_: {12, "Hour"}] := 
  MapAt[DatePlus[#, os] &, DateRange @@ DateList /@ x, -1]

toString = DateString[#, {"Year", "-", "Month", "-", "Day"}] &;

And with data:

drlist = {{"2015-01-01", "2015-01-05"},
          {"2015-01-07", "2015-01-07"},
          {"2015-01-12", "2015-01-23"}};

r = {"2015-01-04", "2015-01-25"};

Process:

Complement[toList[r], Join @@ toList /@ drlist];

#[[{1, -1}]] & /@ Split[%, #[[3]] + 1 == #2[[3]] &];

Map[toString, %, {2}]

{{"2015-01-05", "2015-01-11"}, {"2015-01-23", "2015-01-25"}}

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You may also find inspiration from answers to my own question back on Stack Overflow:

• Overlapping strips