# string operation

Posted 8 years ago
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 How to plot this when such a strange data file. What will make a good list so that eaisly on ecan call them. Attachments:
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Posted 8 years ago
 sorry i do not understand this time stuff. I should read about this. One more question how to find here curve derivatives?
Posted 8 years ago
 its good but time data looks so big number.
Posted 8 years ago
 Your spreadsheet has December 31 year 1899 with times over 1/5 of one second.Your text file has no month or day or year with times over 1/9 of one second.If your month day year and time were so useful number then graph is better.Or subtract date and time of first row from all other rows and get graph that starts at zero.
Posted 8 years ago
 . In[1]:= raw = Import["2015-06-25_TemperatureValues.xls"]; data = Map[Flatten[{AbsoluteTime[#[[1]]], Rest[#]}] &, Drop[First[raw], 7]] Out[2]= {{-86399.96300000000000000, 32.17, 34.96}, {-86399.92600000000000000, 32.17, 34.96}, {-86399.8890000000000000, 32.17, 34.96}, {-86399.8520000000000000, 32.17, 34.96}, {-86399.8150000000000000, 32.17, 34.96}, {-86399.7780000000000000, 32.17, 34.96}, {-86399.7410000000000000, 32.17, 34.96}, {-86399.7040000000000000, 32.17, 34.96}, {-86399.6670000000000000, 32.17, 34.96}, {-86399.6300000000000000, 32.17, 34.96}, {-86399.5930000000000000, 32.17, 34.96}, {-86399.5560000000000000, 32.17, 34.96}} In[3]:= g1 = Show[ListPlot[Map[{#[[1]], #[[2]]} &, data]], ListPlot[Map[{#[[1]], #[[3]]} &, data]]]  In[4]:= raw = Import["testdata.txt"]; data = Map[Flatten[{AbsoluteTime[#[[1]]], ToExpression[Rest[#]]}] &, Map[StringSplit[#, "\t"] &, StringReplace[Rest[StringSplit[raw, "\n"]], "," -> "."]]] Out[5]= {{3629059200, 37, 32.17, 34.96}, {3629059200, 74, 32.17, 34.96}, {3629059200, 111, 32.17, 34.96}, {3629059200, 148, 32.17, 34.96}, {3629059200, 37, 32.17, 34.96}, {3629059200, 74, 32.17, 34.96}, {3629059200, 111, 32.17, 34.96}, {3629059200, 148, 32.17, 34.96}, {3629059200, 185, 32.17, 34.96}, {3629059200, 222, 32.17, 34.96}, {3629059200, 259, 32.17, 34.96}, {3629059202, 333, 32.17, 34.96}, {3629059203, 370, 32.17, 34.96}} In[6]:= g2 = Show[ListPlot[Map[{#[[1]], #[[3]]} &, data]], ListPlot[Map[{#[[1]], #[[4]]} &, data]]] 
Posted 8 years ago
 please see the second data file also Attachments:
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