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How to skip weekends in DateListPlot, ie, drawing 20 something data points for a month and not leaving 2 days gap between weeks? Currently DateListPlot occupied the actual number of days for a month and drawing a direct line between Friday and next Monday (where no weekend data available).

I want to draw something like this as Excel: Weekends skipped in Excel

While I get this in Mathematica: Weekends included

Sample data:

Dataset1 = {{{2014, 10, 1, 0, 0, 0}, 
326.8442322}, {{2014, 10, 2, 0, 0, 0}, 
335.284444}, {{2014, 10, 3, 0, 0, 0}, 
292.5086867}, {{2014, 10, 6, 0, 0, 0}, 
301.6750855}, {{2014, 10, 7, 0, 0, 0}, 
629.9616564}, {{2014, 10, 8, 0, 0, 0}, 
311.6753747}, {{2014, 10, 9, 0, 0, 0}, 
281.5024327}, {{2014, 10, 10, 0, 0, 0}, 
285.9985823}, {{2014, 10, 13, 0, 0, 0}, 
304.0455324}, {{2014, 10, 14, 0, 0, 0}, 
702.0810431}, {{2014, 10, 15, 0, 0, 0}, 
362.4479495}, {{2014, 10, 16, 0, 0, 0}, 
305.3800495}, {{2014, 10, 17, 0, 0, 0}, 
289.6995727}, {{2014, 10, 20, 0, 0, 0}, 
320.0506679}, {{2014, 10, 21, 0, 0, 0}, 
904.8753329}, {{2014, 10, 22, 0, 0, 0}, 
337.5214556}, {{2014, 10, 23, 0, 0, 0}, 
278.8865421}, {{2014, 10, 24, 0, 0, 0}, 
300.4519167}, {{2014, 10, 27, 0, 0, 0}, 
314.3109259}, {{2014, 10, 28, 0, 0, 0}, 
796.3015273}, {{2014, 10, 29, 0, 0, 0}, 
368.7108282}, {{2014, 10, 30, 0, 0, 0}, 
305.3858455}, {{2014, 10, 31, 0, 0, 0}, 321.1488558}};

g1 = DateListPlot[Dataset1, PlotRange -> {0, 1050}]
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  • $\begingroup$ Please provide some data and the code you used to get the plot. $\endgroup$
    – Edmund
    Commented Mar 7, 2016 at 3:36
  • $\begingroup$ I've updated the sample graphs with data $\endgroup$
    – Leonx
    Commented Mar 7, 2016 at 4:16

2 Answers 2

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I don't think this is logic to be done using DateListPlot because that will contradict the main idea of DateListPlot (as far as I understand DateListPlot).

In my opinion the easy way is to use the normal ListLinePlot.

data2 = MapAt[
   DateString[#, {"Day", "  ", "DayNameShort", "  ", "MonthNameShort",
       "/", "Year"}] &, Dataset1, {;; , 1}];
tiks = Transpose[{Range@Length[data2],Rotate[#, Pi/2] & /@ data2[[;; , 1]]}];
ListLinePlot[data2[[;; , 2]], PlotRange -> All,Ticks -> {tiks, Automatic}]

enter image description here

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  • $\begingroup$ I see... thanks! I suppose same goes to DateListBarChart (ie, using BarChart instead)? $\endgroup$
    – Leonx
    Commented Mar 7, 2016 at 5:52
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    $\begingroup$ @Leonx - DateListBarChart is not a built-in Mathematica function. It is a custom function that was written by someone in your organisation. $\endgroup$
    – Verbeia
    Commented Mar 7, 2016 at 6:45
  • $\begingroup$ @Verbeia, I am sorry, I didn't read carefully enough. $\endgroup$
    – Leonx
    Commented Mar 7, 2016 at 22:52
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ListLinePlot[d1[[All, 2]], PlotRange -> All, 
 Ticks -> {Transpose[{Range@Length@d1, Rotate[#, -Pi/2] & /@ (DateString[#, "DateShort"] 
           & /@ d1[[All, 1]])}], Automatic}]

Mathematica graphics

if your list contains "holes", you'll need to fill them up for the plot to be meaningful:

d2 = Join[d1[[;; 5]], d1[[9 ;;]]];
r = DateRange[d2[[All, 1]][[1, ;; 3]], d2[[All, 1]][[-1, ;; 3]]]
d3 = SortBy[Join[d2, {#, 0} & /@ Select[Complement[r, d2[[All, 1, ;; 3]]], 
     DayName@# != Sunday && DayName@# != Saturday &]], AbsoluteTime[#[[1]]] &]
ListLinePlot[d3[[All, 2]], PlotRange -> All, 
 Ticks -> {Transpose[{Range@Length@d3, Rotate[#, -Pi/2] & /@ (DateString[#, "DateShort"] & /@ 
        d3[[All, 1]])}], Automatic}]

Mathematica graphics

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