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This question already has an answer here:

temperatures1 = 
  WeatherData["London", "Temperature", {{2013, 8, 14}];

temperatures3 = 
  WeatherData["London", "Temperature", {{2013, 8, 15}]
ListPlot[{temperatures1,temperatures3}] 

How can i do this for for yesterday's "MeanHumidity" compared to today's mean temperature ?

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marked as duplicate by b.gates.you.know.what, ubpdqn, Michael E2, Jens, Mr.Wizard Aug 18 '14 at 18:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ thanks for editing it,can u tell me plz how can i do it? $\endgroup$ – Tigran Aug 18 '14 at 10:15
  • $\begingroup$ I think you should use MeanTemperature. To find out all available properties, type WeatherData["Properties"]. Have a look at mathematica.stackexchange.com/questions/57583/… $\endgroup$ – hieron Aug 18 '14 at 10:56
  • $\begingroup$ Thank lot it help me lot $\endgroup$ – Tigran Aug 18 '14 at 11:08
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{meanhumidity, meantemp} = 
   WeatherData["London", #, {{2007, 1, 1}, {2007, 12, 31}, "Day"}] & /@ { 
      "MeanHumidity", "MeanTemperature"}; 
plotdata = Transpose[{meanhumidity[[;; -2, 2]], meantemp[[2 ;;, 2]]}];
ListPlot[plotdata, Frame -> True, FrameLabel -> {"Mean Humidity [t-1]", "Mean Temp [t]"}]

enter image description here

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