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I have two sets of data. The one is daily temperature data and another is 16-day vegetation observation data (value of Jan 1, Jan 17 etc), throughout a year. But I need daily values of vegetation observation data. How to obtain daily value from 16-day vegetation data?

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  • $\begingroup$ 16 day data comprise data about every day? present a sample $\endgroup$
    – user10016
    Mar 21, 2012 at 12:07
  • $\begingroup$ Mar,06 0.01 Mar,22 0.04 Apr,07 0.24 Apr,23 0.25 May,09 0.19 May,25 0.58 etc until Jan to December. $\endgroup$
    – Bandrush
    Mar 21, 2012 at 12:16
  • $\begingroup$ If you impute daily vegetation data (perhaps via interpolation) and then go on to perform analyses in which those data are compared to temperatures, most assessments of variation and statistical significance will be incorrect because you will appear to have far more observations than you really do. Usually it's better to condense the temperature data to match the vegetation data, perhaps by taking 16-day means (or minima or maxima, depending on what may be scientifically important). $\endgroup$
    – whuber
    Mar 21, 2012 at 16:14

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You might obtain estimates of vegetation observation data by interpolating the data you have. You can choose among the different interpolation methods, linear, polynomial, spline etc.

In the figure below the data you provided are plotted together with a linear interpolation (green line) and a third order polynomial interpolation (red line). Note that with the polynomial interpolation you can get some weird effect, such as negative vegetation observation data points between the first two dates.

interpolation plot

And here are all the values you would get by interpolating the data you provided. In the second column are listed the outputs of the linear interpolation, in the third column, the data of the polynomial interpolation.

table

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  • $\begingroup$ Such interpolations present substantial statistical complications: typically, the interpolated data do not even have the same means, variances, or extrema exhibited by the original data. This can make their use in subsequent analyses (presumably, comparisons to temperature) suspect. $\endgroup$
    – whuber
    Mar 21, 2012 at 16:16
  • $\begingroup$ @whuber I agree that this method shouldn't be used to get data for free. But maybe for just an exploratory analysis can be useful. $\endgroup$
    – VLC
    Mar 21, 2012 at 16:26
  • $\begingroup$ Even for EDA it will be confusing, because (a) the interpolated values have an arbitrary element arising from the interpolation and (b) the amount of numbers is 16 times greater than it really needs to be. Besides, there are so many effective EDA tools available for such datasets that there appears to be no exploratory reason to interpolate. $\endgroup$
    – whuber
    Mar 21, 2012 at 16:28

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