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    <p>Problem is that the ksmooth function will return a list, and those lists are saved as that by rollaplly. By the way, I don't think you even want to use rollaplly, as that does not do this for each date but "rolls" over the dataframe. I believe from your explanation that is not the desired behaviour.</p> <p>I couldn't really work it out using a zoo object, as that one is quite restrictive. Maybe somebody else will show you that. You can construct that dataframe using the <code>ddply</code> function from the <code>plyr</code> package :</p> <pre><code>tt &lt;- ddply(test,.(Date), function(x) { as.data.frame(ksmooth(x$return,x$price,"normal",bandwidth=2,n.points=50)) }) </code></pre> <p>tt can then be transformed to a zoo object, using</p> <pre><code>tt2 &lt;- zoo(tt, as.POSIXct(tt$Date) + 1:50) </code></pre> <p>Alternatively, you could do it by hand using a bit of list manipulation. again, the resulting tt can be converted by the line above to a zoo object.</p> <pre><code>tt &lt;- split(test,test$Date) tt &lt;- lapply(tt,function(x){ as.data.frame(ksmooth(x$return,x$price,"normal",bandwidth=2,n.points=50)) }) tt &lt;- do.call(rbind,tt) names(tt) &lt;- c("return","price") tt$Date &lt;- as.Date(gsub("\\.\\d+","",rownames(tt))) </code></pre> <p>Mind you, I used <code>read.table()</code> to construct test :</p> <pre><code>zz &lt;- textConnection( "Date , return , price , vol 20100902 , 0.3 , 15 , 8.5 20100902 , 0.4 , 17 , 8.6 20100902 , 0.6 , 19 , 8.7 20100903 , 0.2 , 13 , 8.2 20100903 , 0.4 , 17 , 8.6 20100903 , 0.8 , 21 , 9.0" ) test &lt;- read.table(zz,header=T,sep=",") test$Date &lt;- as.Date(as.character(test$Date),format="%Y%m%d") close(zz) </code></pre>
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    1. COThank you very much Joris, that looks very promising! I will give my best to make it easier to read for you next time =)
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