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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.gatessucks, ubpdqn, Michael E2, Jens, Mr.Wizard Aug 18 at 18:07

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thanks for editing it,can u tell me plz how can i do it? –  Tigran Aug 18 at 10:15
    
I think you should use MeanTemperature. To find out all available properties, type WeatherData["Properties"]. Have a look at mathematica.stackexchange.com/questions/57583/… –  hieron Aug 18 at 10:56
    
Thank lot it help me lot –  Tigran Aug 18 at 11:08

1 Answer 1

{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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thank u lot :) but this bring somth wrong :( –  Tigran Aug 18 at 10:40
    
thank u i did it –  Tigran Aug 18 at 10:43

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