similar to: ragged array with append

I'd like to control my namespace thoroughly, separated by task. Is there a way, in R session, to introduce namespaces for tasks dynamically and switch them as needed? Or, is there a combination of load/save workspace steps which can facilitate this? Cheers, Alexy

I'm running rle() on a long vector, and get a result which looks like > uc Run Length Encoding lengths: int [1:16753] 1 1 1 1 1 1 1 1 1 1 ... values : int [1:16753] 29462748 22596107 18322820 14323315 12684505 9909036 7296916 6857692 5884755 5883697 ... I can print uc$names or uc$levels separately. Is there any way to print them together as tuples, looking like (29462748, 1)

Greetings -- in order to write back to SQL databases, one needs to create a dataframe with values. I can get column names of an existing table with sqlColumns. Say I have a vector of values (if they're all the same type), or a list (if different). How do I create a dataframe with column names given by my sqlColumns? To make it concrete, how do we make a dataframe A B C 1 2 3

Greetings -- I have a dataframe a with one element a vector, time, of POSIXct values. What's a good way to split the data frame into periods of a$time, e.g. days, and apply a function, e.g. mean, to some other column of the dataframe, e.g. a$value? Cheers, Alexy

Is there an R analog of the Unix command uniq -c: http://en.wikipedia.org/wiki/Uniq Given an array x, uniq -c replaces each contiguous subsequence of identical numbers with a tuple (count, number). E.g. $ cat > usample 10 10 9 8 8 7 7 7 6 3 1 1 1 0 $ uniq -c usample 2 10 1 9 2 8 3 7 1 6 1 3 3 1 1 0 Cheers, Alexy

I'm growing a large dataframe by composing new rows and then doing row <- compute.new.row.somehow(...) d <- rbind(d,row) Is this a fast/preferred way? Cheers, Alexy

Using wk <- with(d, split(word, kind)), I get the following class table: wk$`1` [1] "a" "bra" ... # (*) wk$`10` "ca" "dabra" ... Now I need to export it in the following format: class num_members examples 1 23 a bra ... 10 4 ca dabra For each class C such as `1`, I need to print the

Every time I have to prefix a dataframe column inside the indexing brackets with the dataframe name, e.g. df[df$colname==value,] -- I am wondering, why isn't there an R scoping rule that search starts with the dataframe names, as if we'd said with(df, df[colname==value,]) -- wouldn't that be a reasonable default to prepend to the name search path? Cheers, Alexy

I get tables with millions of rows. For plotting to a screen-size jpg, obviously just about 1000 points are enough. Instead of feeding plot() the original millions of rows, I'd rather shrink the original dataframe, using some kind of the following interpolation: -- split dataframe into chunks of N rows each, e.g. 1000 rows each -- compute average for each column -- issue one new row

I found there's a very good functional set of operations in R, such as apply family, Hadley Wickham's lovely plyr, etc. There's even a Reduce (a.k.a. fold). Now I wonder how can we do pattern-matching? E.g., now I split dimensions like this: m <- dim(V)[1] # R n <- dim(V)[2] # still R While even Matlab allows for [m,n] = size(V) % MATLAB! Ideally I'd be able to

What are idioms for taking a head or a tail of a vector, either up to an index, or from an index to the end? Also -- is it necessary to use length(v) to refer to the last element? E.g., Python has v[:3] # indices 0,1,2 v[3:] # indices 3,4,... v[-1] # the last element of v v[:-1] # all but last Cheers, Alexy

I've had a very long file written out by R with write.table, with fields of time values, converted from POSIXlt as.numeric. Among 2.5 million values, very few had 6 trailing zeroes, and those were output in scientific notation as in the subject. Is this the default behavior for long integers, and how can it be turned off (with all digits for any integer field in write.table)? This

I want to produce a dataframe with integer columns for elements of string pairs: pairs <- c("10 21","23 45") pairs.split <- lapply(pairs,function(x)strsplit(x," ")) pdf <- as.data.frame(pairs.split) names(pdf) <- c("p","q") -- at this point things look good, except the columns are factors, as I didn't change the default

I've used to have a script with a barplot command it in, preceded by a png: png(graph.file,height=H,width=W) barplot(t,names.arg=breaks[2:(length(t)+1)],tck=gridlines) -- worked before R 2.6.2. When I tried it in R 2.6.2, which I have for a while but didn't run with that script, it complained, the margins too large, and I've googled the messages from our list where neither

Greetings -- I'd like to avoid converting a Fortran array of floats into ASCII and back reading it in R. Furthermore it's much faster to dump large arrays in binary, as they take up much less space with full precision -- many decimal points take up many bytes in ASCII versus four or eight per float in raw. Is there a way to read such a Fortran unformatted file back into R?

Did anyone try to write R extensions in OCaml? What would it entail to enable it? Cheers, Alexy

I have a dataframe with many rows like this: > df X1 X2 X3 X4 X5 X6 X7 week d sim1 FALSE TRUE TRUE TRUE TRUE TRUE TRUE 1 0.3064985 sim1 is the rowname, X1..X7,week,d are the column names. X1..X7 are factors, booleans in this case. I need to add another row, represented by the following list: list(rep(T,7),5,0.0) -- i.e, TRUE in all boolean columns,

Certain timeseries I have had outliers, which I removed by assigning NA to their positions. Now acf() refuses to go to work. What's the right way to remove outliers from ts objects, and what are teh standard ways to interpolate NAs in them? Cheers, Alexy

I have two pairs of time intervals: coarse- and fine-grained. They're components of their respective dataframes, looking like, coarse: endtime starttime 1 t1_end t1_start 2 t2_end t2_start ... fine: is the same, except that its intervals presumably fall into the coarse's enclosing ones. The problem is to partition

How do I apply a function to every row of a dataframe most naturally? Specifically, I'd like to filter out any row which contains an Inf in any column. Since all columns are numeric, I guess max should work on a row... -- Alexy