# Determining the dates of days

I'm trying to use WeatherData[] to give me (not sure how) the temperature, humidity, wind-speed, cloud cover... at a certain location.

I only want the information for weekdays (e.g. Monday-Friday) for the last 83 years.

How do I make a list of all these dates in a format (DateList or DateString) that WeatherData[] can use?

• Search for DayOfWeek in the documentation. Jun 12, 2012 at 13:28

For variation, here is method that doesn't require the Calendar package and uses a function which returns days of the week ranging from 0 for Sunday to 6 for Saturday:

GaussDay[y_Integer, m_Integer, d_Integer] :=
With[{yDigits = IntegerDigits[y - Boole[m < 3]},
With[{ y1 = FromDigits[yDigits[[1 ;; 2]]],
y2 = FromDigits[yDigits[[3 ;; 4]]]},
Mod[(d + Floor[2.6 ( Mod[m + 9, 12] + 1) - 0.2] + y2 +
Quotient[y2, 4] + Quotient[y1, 4] - 2 y1), 7]]]


And in play:

Select[WeatherData["Chicago",
"Temperature", {{2011, 1, 1}, {2011, 12, 11}}],
MemberQ[Range@5, GaussDay @@ #[[1, 1 ;; 3]]] &]


For comparison, this took about 8.5 seconds to select approx 130,000 measurements when asked for 83 years of data.

• Floor[y2/4] is better expressed as Quotient[y2, 4]. Jun 12, 2012 at 15:53
• This is a good solution. As a rule of thumb, never use Mathematica's built-in date functions unless you absolutely have to. By my tests, they are two orders of magnitude slower than Visual Basic's comparable functions, and even farther behind compiled languages. Jun 12, 2012 at 15:53
• Also, I prefer Larsen's method myself for day-of-week computations: larsen[{yr_Integer, mo_Integer, da_Integer}] := Module[{y = yr, m = mo, d = da, f, q}, If[m < 3, y--; m += 12]; f = If[y >= 1752 && m >= 9 && d >= 14, Quotient[y, 400] - Quotient[y, 100], 5]; q = d + 2 m + 1 + Quotient[3 (m + 1), 5] + y + Quotient[y, 4] + f; Mod[q, 7] + 1]. No digit twiddling necessary; just integer arithmetic. Jun 12, 2012 at 15:59
• @J.M. Thank you. Jun 12, 2012 at 16:00
• If[m < 3, IntegerDigits[y - 1], IntegerDigits[y]] might be better written as IntegerDigits[y - Boole[m < 3]]. Jun 12, 2012 at 16:01

After testing the performance of my first answer, along with J.M.'s suggestion, I don't think it will be fast enough. Here is another approach: filtering after acquisition.

This gives all "Temperature" data for Chicago in 2011, filtering out all Saturdays and Sundays.

Select[
WeatherData["Chicago", "Temperature", {{2011, 1, 1}, {2011, 12, 31}}],
! MatchQ[DateString[#[[1]], "DayName"], "Saturday" | "Sunday"] &
]


If you only want one point for each day, use: {{2011, 1, 1}, {2011, 12, 31}, "Day"}

• No need for the Calendar package. The built-in DateString[{2011, 1, 1}, {"DayName"}] will do (and is slightly faster too). Note that it returns day names as strings instead of symbols. Jun 12, 2012 at 21:03
• @Sjoerd thanks! Updated. Jun 12, 2012 at 21:27

Needs["Calendar"]


Find a Sunday:

DayOfWeek[{1950, 1, 1}] (* Sunday *)


Create a list of offsets from that day for weekdays:

weeks = 10;
offsets = Join @@ Array[Range@5 + 7 # &, weeks, 0];


Generate resolved days from offsets:

weekdays = DaysPlus[{1950, 1, 1}, #]& /@ offsets;


Confirm:

Tally[DayOfWeek /@ weekdays]

{{Monday, 10}, {Tuesday, 10}, {Wednesday, 10}, {Thursday, 10}, {Friday, 10}}

• After which, OP can do something like Outer[WeatherData["Chicago", #1, #2] &, {"Temperature", "Humidity", "WindSpeed", "CloudCoverFraction"}, weekdays, 1]. Jun 12, 2012 at 13:43
• @J.M. I'll try this, thanks! Jun 12, 2012 at 13:44
• @J.M. okay; I was leaving something for the OP to figure out. ;-) Jun 12, 2012 at 13:44
• @Mr.Wizard I'm sure in the long run I'd appreciate it, but right now... No. (I'm just in a hurry.) Jun 12, 2012 at 13:46
• Wizard: I figured that use of Outer[]` is not too well-known among beginners, so I figured it should be pointed out... :) Jun 12, 2012 at 13:46