I'd like to create a list with days from the 4th of July to the 4th of October starting in July 1970 to the current date. I tried to use DateRange, but I'm a bit confused cause the documentation is't very clear on this point.

What should I do? Have you ever tried to do such a thing?

  • 1
    $\begingroup$ You mean from the 4th of July to the 4th of October of every year? $\endgroup$
    – Öskå
    Aug 4, 2014 at 11:02
  • 1
    $\begingroup$ Some important things that you left out from your question: how do you want to results to be formatted and how do you plan to use them later? $\endgroup$
    – m_goldberg
    Aug 4, 2014 at 11:06
  • 1
    $\begingroup$ Does my most recent edit of the title characterize your question correctly? $\endgroup$
    – m_goldberg
    Aug 4, 2014 at 11:18
  • 5
    $\begingroup$ Does this produce your desired result: Flatten[DateRange[{#, 7, 4}, {#, 10, 4}, "Day"] & /@ Range[1974, 2013], 1]? $\endgroup$
    – ubpdqn
    Aug 4, 2014 at 11:20
  • 1
    $\begingroup$ @ubpdqn. I'm reading this as the entire summer after the 7/4 up to the current year. if current date of current year is after 4th but before the end of summer, it ends there, otherwise, to end of most recent summer. $\endgroup$
    – m_goldberg
    Aug 4, 2014 at 11:22

3 Answers 3


Borrowing from @ubpdqn, my interpretation of the question (which, hopefully, should also be valid in 2015):

td = Today // Normal;
{cy, cm, cd} = Take[td, 3];
days = Flatten[DateRange[{#, 7, 4}, {#, 10, 4}, "Day"] & /@ Range[1970, cy], 1];

 Today > Interpreter["Date"]["04/October/" <> ToString@cy], days,
 Today < Interpreter["Date"]["04/July/" <> ToString@cy], DeleteCases[days, {cy, __}], True,
 DeleteCases[days, {y_, m_, d_} /; y == cy && m > cm || y == cy && m == cm && d > cd]]

My answer is based on @eldo's and @ubpdqn's answer (thanks guys!). Remove // Flatten[#, 1] & to get the dates grouped into years.

(* Defining 2 allDates functions to prevent counting days
when the date of the current year hasn't reached the starting date of the interval 
i.e. take the "nearest past summer" *)

{{1970,7,4},{1970,7,5},{1970,7,6},{1970,7,7},{1970,7,8},{1970,7,9},{1970,7,10},{1970,7,11},{1970,7,12},{1970,7,13},{1970,7,14},{1970,7,15},{1970,7,16},{1970,7,17}, ...4096... ,{2014,7,22},{2014,7,23},{2014,7,24},{2014,7,25},{2014,7,26},{2014,7,27},{2014,7,28},{2014,7,29},{2014,7,30},{2014,7,31},{2014,8,1},{2014,8,2},{2014,8,3},{2014,8,4}}

{{1970,7,4},{1970,7,5},{1970,7,6},{1970,7,7},{1970,7,8},{1970,7,9},{1970,7,10},{1970,7,11},{1970,7,12},{1970,7,13},{1970,7,14},{1970,7,15},{1970,7,16},{1970,7,17}, ...4157... ,{2014,9,21},{2014,9,22},{2014,9,23},{2014,9,24},{2014,9,25},{2014,9,26},{2014,9,27},{2014,9,28},{2014,9,29},{2014,9,30},{2014,10,1},{2014,10,2},{2014,10,3},{2014,10,4}}

Comparison with @eldo's method

My method seems to not require any Internet connection (as compared to @eldo's method), and it seems to run faster as well.


(* {0.069004,0.776044} *)

Equal[Rest@s1, Rest@eldoDates]
(* True *)

Again with the understanding that an answer has already been accepted. I submit the following only as an exercise:

repeatdaterange[startdate_, enddate_] := Module[{startyear, startmonth, startday, endyear, endmonth,endday, curyear, curmonth, curday, stopyear, stepping, lastmonth, listofdates, rangeofdates},
{startyear, startmonth, startday} = startdate;
{ endyear, endmonth,endday} = enddate;
 {curyear, curmonth, curday} = Take[DateList[],3];
stopyear = If[curmonth >=startmonth, curyear, Subtract[curyear,1] ];
stepping = Subtract[endmonth ,startmonth];
lastmonth = If[curmonth>=endmonth &&  curday >= endday,endmonth,curmonth];
listofdates =Table[{i,j,startday}, {i,startyear,stopyear}, {j,startmonth,If[i == stopyear,lastmonth,endmonth],stepping} ];
listofdates = If[First[Dimensions[Last[listofdates] ]] ==  1, ReplacePart[ listofdates, Length[listofdates] -> {Last[listofdates][[1]],{stopyear,lastmonth,curday } }],listofdates ];
rangeofdates = Apply[DateRange, listofdates,1]

One can add another parameter to the function for final date instead of current runtime in the DateList[{insert final date here}]. There may be small boundary issues but this runs fairly fast.


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