search for: shvorob

> 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:

Hello, I am new to R and I am trying to figure out how to print text output from an operation like table() to a pdf file. Is there a simple command to do this, or do you need to work with connections at a rudimentary level? Thanks you, Jesse [[alternative HTML version deleted]]

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 =

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

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...

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

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,

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

murdoch at stats.uwo.ca wrote: > 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 =C3=A2=E2=82=AC= =CB=9Cscipen=C3=A2=E2=82=AC=E2=84=A2 digits >>> wider. >> However,=20 >> >> options(scipen =3D...

...le density in rgl (Mike Marchywka) 50. Re: data acquisition with R? (B.-Markus Schuller) 51. Is it possible to make a matrix to start at row 0? (bogdanno) 52. Re: Rexcel (csrabak) 53. Re: question about constraint minimization (dhacademic at gmail.com) 54. Re: Lost in POSIX (Dimitri Shvorob) 55. Re: Find in R and R books (Georg Otto) 56. Re: using rpart with a tree misclassification condition (meytar) 57. Re: memory profiling (Patrick Leyshock) 58. Some questione about plot (romzero) 59. how do remove those predictor which have p value greater than 0.05 in GLM? (shubha...

Is there a better alternative to x = c(1e7, 2e7) x.lb = c(0,1e7,2e7) s.lb = format(x.lb, scientific = FALSE, big.mark = ",") barplot(x, yaxt = "n", ylab = "") axis(side = 2, at = x.lb, labels = s.lb) (I am sure there is a better alternative to line 2 :)). Thank you. -- View this message in context:

Two problems with the code below. A. It produces empty JPEGs. When the 'bwplot' line alone is submitted, the plot duly shows up. B. When the 'bwplot' line alone is submitted, y labels are values 1 to 6, not actual distinct values of y$maxthreads. (C. I would, of course, prefer to produce plots for all distinct values of x$maxthreads in a single swoop, on a single figure). Can