Tag Info

Hot answers tagged

43

The short answer is, yes! There is a whole undocumented package TemporalData` containing some useful functions. The results below are from my own spelunking. Feel free to add/amend as appropriate. Let's set up some simple TemporalData objects to explore them: fakedata = Transpose@{DatePlus[{2001, 1}, {#, "Month"}] & /@ Range[0, 99], ...


23

Just a literal implementation of a formula for the day of the week: Clear[dow]; dow[{year_, month_, day_, _ : 0, _ : 0, _ : 0}] := Module[{Y = If[month == 1 || month == 2, year - 1, year], m = Mod[month + 9, 12] + 1, y, c, s = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}}, y = Mod[Y, 100]; c = Quotient[Y, 100]; ...


21

I will provide one solution which will be using Java and a simple Java reloader I recently introduced. This solution brings to the table up to 100-fold speed-up for large lists of dates. Preparation I will borrow @Mike's functions to generate a random list of dates, from his code in his recent question RandomDateList[] := { RandomInteger[{1800, 2100}], ...


21

Date-picker implementation in Mathematica The following is my implementation of a simple date-picker. The current date is highlighted in LightBlue and the weekends are highlighted in LightGreen. The selected date is always highlighted in LightRed (the default selection is the current date). You can tap into this calendar by using the Dynamic values for ...


18

This site has exactly what you want here, already in Mathematica code. One example here:


17

Since I'm living in Europe I'm sticking to the European definition of week number which is equivalent to the ISO standard. According to this standard, a week starts on Monday and the first week is the week containing 4 January. Taking this into account you could do therefore do something like weekNumberISO[date_] := Module[{day0, year}, With[{days = ...


16

You can use DateDifference to find the time between January 1st and April 17th: DateDifference["Jan. 1", "April 17", "Week"] (* {15.2857, "Week"} *) If you want the "week number" as you've put it, you can just do: Ceiling@First@DateDifference["Jan. 1", "April 17", "Week"] which gives 16. Edit based on Szabolcs's comment: To ensure this works for Jan ...


15

The format specification for DateList is pretty flexible. Since we know that we have <h3> tags wrapped around things, we can just account for them: DateList[{#, {"<h3>", "MonthName", "Day", ",", "Year", "</h3>"}}] & /@ Flatten[dates] (* ==> {{2001, 1, 18, 0, 0, 0.}, {2001, 2, 1, 0, 0, 0.}, {2001, 2, 2, 0, 0, 0.}, {2001, 2, ...


14

It took me quite a while, but finally, here's a visualization of the perigee of Flamsteed's comet: I should first note two things: first, some of the needed data for computing the orbit of comet C/1683 O1 was missing in AstronomicalData["CometC1683O1", "Properties"], and I had to pull information from external sources to supplement the information ...


13

This question might be considered a duplicate. It is closely related to these: Considerations when determining efficiency of Mathematica code Difference between AbsoluteTiming and Timing Benchmarking expressions Profiling from Mathematica However, one simple reading of this question that I do not believe is covered in the answers above is answered with ...


12

I've shown off Larsen's method before (and see this as well), but here it is as a formal answer: larsen[{yr_Integer, mo_Integer, da_Integer, ___}] := Module[{y = yr, m = mo, d = da, q}, If[m < 3, y--; m += 12]; q = d + 2 m + 1 + Quotient[3 (m + 1), 5] + y + Quotient[y, 4] + Quotient[y, 400] - Quotient[y, 100]; {Sunday, Monday, Tuesday, ...


12

data = FinancialData["SPY", "Jan. 1, 2011"] /. {d_List, v_} :> {AbsoluteTime@d, v}; model = a x^4 + b x^3 + c x^2 + d x + e; fit = FindFit[data, model, {a, b, c, d, e}, x] modelf = Function[{x}, Evaluate[model /. fit]] Plot[modelf[x], {x, Min@data[[All, 1]], Max@data[[All, 1]]}, Epilog -> Map[Point, data]] Edit Better (tick labels showing dates) ...


12

Import[(* file *), "Table", "DateStringFormat" -> {"Year", "-", "Month", "-", "Day"}] seems to work... As a test: Export["test.dat", {{"2010-05-19", 17}, {"2010-05-20", 20}, {"2010-05-21", 19}}, "FieldSeparators" -> " "]; Import["test.dat", "Table", "DateStringFormat" -> {"Year", "-", "Month", "-", "Day"}] {{{2010, 5, 19}, ...


12

Using Simon's data: In[6]:= datelist = {"29/02/2008", "15/12/2007", "06/09/2007", "06/10/2008", "05/03/2007", "24/01/2010", "19/06/2009", "03/11/2009", "02/02/2010", "25/12/2009"}; We can just sort the data by the absolute time: In[7]:= SortBy[datelist, AbsoluteTime[{#, {"Day", "Month", "Year"}}] &] Out[7]= {"05/03/2007", "06/09/2007", ...


11

Assuming that the days and times in the gathered data list are strings you could do something like days = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"}; data2 = Reap[ Sow[(AbsoluteTime[#2] - AbsoluteTime["00:00:00"])/3600., #1] & @@@ Flatten[data, 1], days, #2 &][[2, All, 1]]; ...


11

To fit a function and to calculate the moving average you need to convert your dates in absolute time using AbsoluteTime[]. data = FinancialData["IBM", "Jan. 1, 2004"]; newdata = Table[{AbsoluteTime[data[[i, 1]]], data[[i, 2]]}, {i, Length[data]}]; lm = LinearModelFit[newdata, x, x]; movAvg = MovingAverage[newdata, 200]; ...


11

So this generates the heatmap: << Calendar` year = 1990; yearLen = DaysBetween[{year, 1, 1}, {year, 12, 31}] + 1; data = RandomReal[1, yearLen]; days = Map[DayOfWeek[{year, 1, #}] &, Range[3, 9]]; day1 = Position[days, DayOfWeek[{year, 1, 1}]][[1, 1]]; dayn = Position[days, DayOfWeek[{year, 12, 1}]][[1, 1]]; Paddata = Join[ConstantArray[100, day1 ...


11

A simple string-based approach is to swap the order of day/month/year, do the Sort and then swap back again: (* Example data *) datelist = DateString[# + AbsoluteTime[{2007, 01, 01}], {"Day", "/", "Month", "/", "Year"}] & /@ RandomInteger[10^8, 10] {"29/02/2008", "15/12/2007", "06/09/2007", "06/10/2008", "05/03/2007", "24/01/2010", ...


11

For full ranges There is the function DayRange that can be used for this purpose, but not in the same simple way like CharacterRange. For the days: DayName /@ DayRange[Today, Today ~DatePlus~ {{1, "Week"}, {-1, "Day"}}] {Wednesday, Thursday, Friday, Saturday, Sunday, Monday, Tuesday} For the months: DateValue[#, "MonthName"] & /@ DayRange[Today, ...


11

data = Cases[DayRange[{2000, 1, 1}, {2399, 12, 31}, Friday], {y_, m_, d_} :>d] // Tally // Sort; TableForm[data[[ ;; ;; 2]], TableHeadings -> {None, {"Day", "Number of Fri."}}] (Reverse@SortBy[data, Last])[[;; 5]] {{27, 688}, {20, 688}, {13, 688}, {6, 688}, {25, 687}}


10

I'll start with a slightly reformatted version of your data: data = {{"Monday", 3}, {"Tuesday", 4}, {"Wednesday", 6}, {"Thursday", 6}, {"Friday", 10}, {"Saturday", 12}, {"Sunday", 11}, {"Monday", 3}, {"Tuesday", 4}, {"Wednesday", 4}, {"Thursday", 5}, {"Friday", 10}, {"Saturday", 10}, {"Sunday", 9}, {"Monday", 2}, {"Tuesday", 3}, ...


10

Jean Meeus's Astronomical Algorithms (as well as the related book Astronomical Formulæ for Calculators) is what you should start looking at whenever you need to deal with algorithms for quantities of astronomical interest. For instance, here is a translation of Meeus's method for the Julian Date: Options[jd] = {"Calendar" -> "Gregorian"}; ...


10

This recent post reminded me that AbsoluteTime is a fast kernel function. Using the RandomDates function from Leonid's post: dates = RandomDates[500000]; Needs["Calendar`"] rls = Thread[ Range[0, 6] -> {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday} ]; Timing[result1 = DayOfWeek /@ dates;] Timing[result2 = ...


10

I would do this with rule replacement. First, you want to have a list of the days of the week in the appropriate order: days = DayName@{0, 0, #} & /@ Range[5, 11]; Then, you can take tallied results and turn them into a list of rules: tallied = {{Tuesday, 533}, {Sunday, 487}, {Saturday, 481}, {Friday, 422}, {Thursday, 353}, {Wednesday, 371}}; ...


9

Without reading Leonid's answer (which is probably better) I recommend something like this: fillDates[dates_] := Module[{f, all}, all = Part[DateList /@ (Range[##, 24*60^2] & @@ AbsoluteTime /@ dates[[{1, -1}, 1]]), All, {1, 2, 3}]; (f[#[[1]]] = #) & ~Scan~ dates; f[x_] := {x, 0}; f /@ all ] fillDates @ {{{2012, 1, 1}, 1}, {{2012, ...


9

I confess to being allergic to database operations that require data to be in a particular sort order: it's too easy for huge mistakes to creep in. What is needed here is to turn the source list (say, list1) into a lookup table so it reliably returns its value (second element in the list) when given its key (first element in the list). To assure ...


9

You can get about 100 times faster by using Java, without any particular tuning, but you will have to provide the date format explicitly. Here is the solution based on Java reloader. Implementation Load the Java reloader Compile the following class: JCompileLoad@ " import java.text.ParseException; import java.text.SimpleDateFormat; import ...


9

I would do it like this, using AbsoluteTime, which is often much faster than alternatives. Module[{ intv = Interval /@ Map[AbsoluteTime, table1[[2 ;;, {1, 2}]], {-2}], type = table1[[2 ;;, 3]], data = Rest[table2], out }, out = Pick[type, intv ~IntervalMemberQ~ #] & /@ AbsoluteTime /@ data; Join[List /@ data, out /. {} -> {"none"}, 2] ...


8

Options will do it: data = {{{2006, 10, 1}, 10}, {{2006, 10, 15}, 12}, {{2006, 10, 30}, 15}, {{2006, 11, 20}, 20}}; plot = DateListPlot[data]; Options[plot, FrameTicks] For example, modifying the contents of FrameTicks: data = {DateList[{2006, #, 1}], #} & /@ Range[40]; plot = DateListPlot[data]; ticks = FrameTicks /. Options[plot, FrameTicks]; ...


8

Ok, Java solution, by popular demand. Solution Load the Java reloader Compile this class: JCompileLoad@ "import java.util.Calendar; public class SecondOfYearVectorized{ public static double[] secondOfYear(int[][] dates){ Calendar calendar = Calendar.getInstance(); double[] result = new double[dates.length]; for(int ...



Only top voted, non community-wiki answers of a minimum length are eligible