similar to: figure margins too large for a barplot in png, pdf ok

Hello! I promise I looked into help files before asking. Still cannot figure it out. I think it's because I am totally confused what packages use lettice, which use trellis, etc. Sections 1 and 2 below produce the data and the data to plot. My question is about barplot in Section 3. I am trying to: 1. add only horizontal gridlines and manipulate the type and color of that line. tck = 1 is not

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 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

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 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

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

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

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

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 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 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

I need to fit a graph into a column of a 2-column paper. I found that just specifying width and height parameters (3.2in x 3.5in) to plot doesn't decrease the fonts of the main title, axis titles, and labeling numbers, and tick sizes. So I have to add cex to all labels and titles and manage ticks. However, I can't decrease the space between axis label and numbers on ticks. Is there a

Sometimes I'm iterating over a list where names are keys into another data structure, e.g. a related list. Then I can't use lapply as it does [[]] and loses the name. Then I do something like this: do.one <- function(ldf) { # list-dataframe item key <- names(ldf) meat <- ldf[[1]] mydf <- some.df[[key]] # related data structure r.df <-

Greetings -- coming from Python/Ruby perspective, I'm wondering about certain features of R as a programming language. Say I have a huge table t of the form run ord unit words new 1 1 6939 1013 641 1 2 275 1001 518 1 3 3314 1008 488 1 4 14154 1018 463 1 5 2982 1006 421 Alternatively, it

Hi All, I need some help in putting text in a stacked barplot. The barplot is filled with 5 levels and now I would like to put text to each level in the stacked barplot. However, it seems that the code that I am using is not placing the text at the correct hight (centered at each fill) in the barplot. Any suggestions to improve the code to make it work? barchart(FREQ ~ VISIT

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,