similar to: exporting text output to pdf

Greetings, I need to compare the ratios of vector sizes like this: length(object1) / length(object2) I have many vector objects to compare, so I would like to do it in a loop. I created a loop like this: mat1 <- matrix() for (i in 1:6) { for (j in 1:6) { mat1[i,j] <- length( paste("object",i,sep="")) /

> grep("f[0-9]+=", "f1=5,f22=3,", value = T) [1] "f1=5,f22=3," How do I make the line output c("f1", "f22") instead? (Actually, c(1,22) would be even better). Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Quick-GREP-challenge-tp2339486p2339486.html Sent from the R help mailing list archive at Nabble.com.

I have a (very big - 1.5 rows) dataframe with a (POSIXt" "POSIXlt") column h (hour). Surprisingly, I cannot calculate a simple aggregate over the dataframe. > n.h1 = sqldf("select distinct h, count(*) from x group by h") Error in sqliteExecStatement(con, statement, bind.data) : RS-DBI driver: (error in statement: no such table: x) In addition: Warning message: In

If I want to drop columns x, y, z from dataframe df, is there a better alternative to df$x = NULL df$y = NULL df$z = NULL There are sufficiently many columns remaining to make df = subset(df, select = c(a,b,c,d[etc])) cumbersome. Thank you. -- View this message in context: http://n4.nabble.com/Is-there-a-quicker-way-to-drop-a-data-frame-column-than-setting-it-to-NULL-tp1288617p1288617.html

... Being an R newbie, I can only think of extracting distinct x values with unique, looping over them, extracting matching rows from the original data frame, applying table, and recording the size of table's output alongside the x value being checked. Is there a more elegant way? Thank you. -- View this message in context:

Greetings, I need to conduct a merge on two databases containing information on organizations, but the organization names are often non-identical and there is no common unique identifier. Does anyone know a good way to calculate a similarity measure on two names, or even better is there a natural language matching function in an R package? I did some searches on this but must not know the right

Dataframe cust has Date-type column open.date. I wish to set up another column, with (first day of) the quarter of open.date. To be comprehensive (of course, improvement suggestions are welcome), month = function(date) { return(as.numeric(format(date,"%m"))) } first.day.of.month = function(date) { return(date + 1 - as.numeric(format(date,"%d"))) } first.day.of.quarter =

Greetings, I am not familiar with processing text in R. Can someone tell me how to read each line of words as separate elements in a list? FE, I would like to turn: word1 word2 word3 word2 word4 into a list of length two with three character elements in the first list and two elements in the second. I know that this should be easy, but I am a little confused by the text functions. Thanks in

Hello - I need to read in some tables that are embedded within data files like this: line 1 line 2 data table 01000 10110 00011 end table line 3 line 4 Is there any way to read just the data by telling an input device to start reading when it encounters the keyword "data table" and stop reading at "end table"? Thanks in advance, Jesse -- View this message in context:

How can one simplify the folowing? t$aum[is.na(t$aum)] = 0; t$aum.core[is.na(t$aum.core)] = 0 t$num[is.na(t$num)] = 0; t$num.core[is.na(t$num.core)] = 0 Thank you. -- View this message in context: http://n4.nabble.com/Applying-a-transformation-to-multiple-data-frame-columns-tp1457641p1457641.html Sent from the R help mailing list archive at Nabble.com.

... Both readLines() and scan() produce a number_of_lines x 1 vector; trying paste(s, collapse = NULL) leaves it unaffected. How can I concatenate vector elements (lines) into a single string? Thank you. -- View this message in context: http://r.789695.n4.nabble.com/How-to-read-contents-of-a-text-file-into-a-single-string-tp2069303p2069303.html Sent from the R help mailing list archive at

Dear useRs, I have a large dataset (600 to 2000 observations) that I need to cluster with a hierarchical technique, and then examine so that I can use cutree to extract various groups. I would like to visualize the results of hclust in a dendrogram, but this is obviously too much data for the normal 6 x 6 figure. Does anyone know how to make a large graph so that I can print it and view the

Dataframe closed contains balances of closed accounts: each row has month of closure (Date-type column month) and latest balance. I would like to plot by-month distributions of balances. A qplot call below produces several warnings and no output. Can anyone help? Thank you. PS. A really basic task, very similar to the examples on p. 71 of the ggplot2 book, apart from a Date grouping column; I

Greetings, I would like to use some of the spatial statistics functions in R, but I am having trouble entering data. My data is already in a distance matrix format, not an X Y coordinate format (each Xij cell in the matrix represens the distance from point i to j). Does anyone know of a way to convert a distance matrix to a ppp object in spatstat, or an X,Y coordinate system for other

x = structure(list(date = structure(list(sec = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), min = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), hour = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), mday = c(1L, 2L, 3L, 4L, 5L, 8L, 9L, 10L, 11L, 12L, 15L, 16L, 17L, 18L, 19L, 22L, 23L, 24L,

On 02/02/2010 6:20 AM, Dimitri Shvorob wrote: > Ruben Roa has kindly suggested using 'scipen' option - cf. > >> fixed notation will be preferred unless it is more than ???scipen??? digits >> wider. > > However, > > options(scipen = 50) > x = c(1e7, 2e7) > barplot(x) > > still does not produce the desired result. This is strange. I see what

You must create a output filter. Edit /etc/printcap and add your client printer like this: ... PC-PRINTER:\ :sd=/var/spool/lpd/PP01:\ :mx#0:\ :lp=/dev/null:\ :if=/var/spool/lpd/Filter/PP01:\ :sh: ... Adjust the printcap parameters as needed (spool directory, directory to find the filter script). Make sure the lpd can access the directory. Consult your manual. Now create the filter

I have a simple barchart with horizontal bars and horizontal tick labels, produced with barplot(x, horiz = T, names.arg = c, las = 1) The labels are longish strings, truncated on the plot. I wish to leave more space for the left margin, and experiment with mar parameter, barplot(x, horiz = T, names.arg = c, las = 1, mar = c(5, 15, 4, 2)) trying various values for the second vector element, but

> x = structure(list(time = structure(c(1020232904.818, 1020232904.818 ), class = c("POSIXt", "POSIXct"), tzone = ""), price = c(321, 323.5)), .Names = c("time", "price"), row.names = 1:2, class = "data.frame") > x1 = x[,c("price")] > dput(x1) c(321, 323.5) Is there similar syntax that gets "price" as a

i have a time Series of IBM closing px from 1/1/2000 to today I want to graph the time serie by dividing the graph by year and month all the monthly graphs with the same year will go to one page. so from 1/1/2000 to 11/19/2010. i will have 11 pages, and each page will have 12 graphs (jan to dec) except for 2010. I am able to do it in R, but when i use sweave, I can only print the last page.