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I wish to convert Data to Data2. Both sets are provided below. Note that the second set has an arbitrary parameter "t". How can this be done.

Data = {{{2013, 1, 1}, 49.90611}, {{2013, 1, 2}, 
   50.33952}, {{2013, 1, 3}, 50.75688}, {{2013, 1, 4}, 
   50.7649}, {{2013, 1, 7}, 50.24321}, {{2013, 1, 8}, 49.256}, 
     {{2013, 1, 9}, 49.28008}, {{2013, 1, 10}, 
   49.44863}, {{2013, 1, 11}, 49.26403}, {{2013, 1, 14}, 
   49.46468}, {{2013, 1, 15}, 49.56099}, {{2013, 1, 16}, 49.9944}}



Data2 = {{t, 49.90611}, {t + 1, 50.33952}, {t + 2, 50.75688}, {t + 3, 
   50.7649}, {t + 4, 50.24321}, {t + 5, 49.256}, {t + 6, 
   49.28008}, {t + 7, 49.44863}, {t + 8, 49.26403}, 
     {t + 9, 49.46468}, {t + 10, 49.56099}, {t + 11, 49.9944}}
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  • $\begingroup$ Where did you get stuck ? $\endgroup$
    – image_doctor
    Commented Aug 26, 2015 at 1:08

2 Answers 2

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Data2=MapThread[{t + #1, Last@#2} &, {Range[0, Length@Data - 1], Data}]

{{t, 49.9061}, {1 + t, 50.3395}, {2 + t, 50.7569}, {3 + t, 50.7649}, {4 + t, 50.2432}, {5 + t, 49.256}, {6 + t, 49.2801}, {7 + t, 49.4486}, {8 + t, 49.264}, {9 + t, 49.4647}, {10 + t, 49.561}, {11 + t, 49.9944}}

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Works if dates aren't sequential days:

Data2 = Thread[{t + QuantityMagnitude@DateDifference[Data[[1, 1]], #] & /@ Data[[All, 1]], Data[[All, 2]]}]

(Not sure if that's what you were going for, if not...)

Data2 = Thread[{t + Range[0, Length@Data - 1], Data[[All, 2]]}]
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  • $\begingroup$ @kala Yes, these inputs are good alternatives, flexible. $\endgroup$
    – thils
    Commented Aug 26, 2015 at 1:26
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    $\begingroup$ @thils Note that the first option calculates number of DAYS from first data point and adds t. Again, not sure if that's what you were after, but coupled with @image_doctor's answer, you have a couple good choices. $\endgroup$
    – kale
    Commented Aug 26, 2015 at 1:27

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