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POHelp me replace a for loop with an "apply" function
 

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  1. POHelp me replace a for loop with an "apply" function
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    <p>...if that is possible</p> <p>My task is to find the longest streak of continuous days a user participated in a game.</p> <p>Instead of writing an sql function, I chose to use the R's rle function, to get the longest streaks and then update my db table with the results.</p> <p>The (attached) dataframe is something like this:</p> <pre><code> day user_id 2008/11/01 2001 2008/11/01 2002 2008/11/01 2003 2008/11/01 2004 2008/11/01 2005 2008/11/02 2001 2008/11/02 2005 2008/11/03 2001 2008/11/03 2003 2008/11/03 2004 2008/11/03 2005 2008/11/04 2001 2008/11/04 2003 2008/11/04 2004 2008/11/04 2005 </code></pre> <p>I tried the following to get per user longest streak</p> <pre><code># turn it to a contingency table my_table &lt;- table(user_id, day) # get the streaks rle_table &lt;- apply(my_table,1,rle) # verify the longest streak of "1"s for user 2001 # as.vector(tapply(rle_table$'2001'$lengths, rle_table$'2001'$values, max)["1"]) # loop to get the results # initiate results matrix res&lt;-matrix(nrow=dim(my_table)[1], ncol=2) for (i in 1:dim(my_table)[1]) { string &lt;- paste("as.vector(tapply(rle_table$'", rownames(my_table)[i], "'$lengths, rle_table$'", rownames(my_table)[i], "'$values, max)['1'])", sep="") res[i,]&lt;-c(as.integer(rownames(my_table)[i]) , eval(parse(text=string))) } </code></pre> <p>Unfortunately this for loop takes too long and I' wondering if there is a way to produce the res matrix using a function from the "apply" family.</p> <p>Thank you in advance </p>
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    1. USGeorge Dontas
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      1. USMatt Weller
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    1. COIf you want to do in sql (or with sqldf in R) there is an excellent discussion in this other SO thread http://stackoverflow.com/questions/1176011/sql-to-determine-minimum-sequential-days-of-access/1176255#1176255
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      1. USEduardo Leoni
    2. COWhy are you using that paste/eval scheme? It seems that that is going to give you a huge performance hit?
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      1. USJonathan Chang
    3. COI agree with Jonathan; and this also makes it really hard to read. Is that what you're trying to do? res.1<-matrix(nrow=dim(my.table)[1], ncol=2) for (i in 1:dim(my.table)[1]) { res.1[i,] <- c(as.integer(rownames(my.table)[i]), sum(as.table(my.table)[i,])) }
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      1. USVince
 

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