Displaying 1 result from an estimated 1 matches for "rle_table".
2009 Oct 01
1
Help me replace a for loop with an "apply" function
...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
--- R code follows
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# turn it to a contingency table
my_table <- table(user_id, day)
# get the streaks
rle_table <- 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<-matrix(nrow=dim(my_table)[1], ncol=2)
for...