similar to: Applying a transformation to multiple data frame columns

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 =

> 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]]

My data frame consists of character variables, factors, and proportions, something like c1 <- c("A", "B", "C", "C") c2 <- factor(c(1, 1, 2, 2), labels = c("Y","N")) x <- c(0.5234, 0.6919, 0.2307, 0.1160) y <- c(0.9251, 0.7616, 0.3624, 0.4462) df <- data.frame(c1, c2, x, y) pct <- function(x) round(100*x, 1) I want to

... 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

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

On Mon, Aug 17, 2015 at 8:39 AM, Phelps, Matthew <mphelps at cfa.harvard.edu> wrote: > > > On Tue, Aug 11, 2015 at 10:55 AM, Ralf Aum?ller < > Ralf.Aumueller at informatik.uni-stuttgart.de> wrote: > >> Hello, >> >> after an update from 6.6 to 6.7 the following error message is logged to >> /var/log/messages when I login (per ssh): >>

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,

Hi, I am confused by results from: > ddply(aa, names(aa), colwise(sum)) I thought ddply was just calling colwise(sum)() with each column. However ddply() returns a 13 x 5 result !! The general result I expected is similar to that of apply() , or using colwise(sum)() alone. Shouldn't ddply() produce the same ? Thanks in advance for your help, - Stuart Andrews >

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

I want to apply a more complicated function than what I use in my example, but the idea is the same: Suppose you have a data frame named x and you want to a function applied to each variable, we'll just use the quantile function for this example. I'm trying all sorts of apply functions, but not having luck. My best guess would be: sapply(x, FUN=quantile) -- View this message in

Hello, after an update from 6.6 to 6.7 the following error message is logged to /var/log/messages when I login (per ssh): Aug 11 16:31:21 a1234 automount[1598]: set_tsd_user_vars: failed to get passwd info from getpwuid_r Checked all log-files of my systems running 6.6 with same configuration -- never got such a message (We use NFS/autofs for home-directories, NIS and tcsh (login shell)).

Hello all R gurus, I have a following problem which I hope someone will help me to solve. I have a data.frame in form similar to below. > testframe<-data.frame("Name"=c("aa","aa","aa","aa","aa","bb","bb","bb","bb","bb"),"Value"=c(1,100,1,1,1,100,100,100,100,1))

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

Dear all, Without a loop, I would like transform 3 numeric vectors empty of 0/1 of same length Vec1 : transform 1 to A and 0 to "" Vec2 : transform 1 to B and 0 to "" Vec3 : transform 1 to C and 0 to "" to obtain only 1 vector Vec who is the paste of the 3 vectors (Ex : ABC, BC, AC, AB,...) Any idea ? Thank you for your help -- Michel ARNAUD

Hello, today I updated a CentOS 7.6 ppc64le machine to CentOS 7.7. After reboot to the new kernel (4.18.0-80.7.2.el7.ppc64le) dkms could not build the nvidia-module. Error-message from dkms: Compiler version check failed: The major and minor number of the compiler used to compile the kernel: gcc version 8.3.1 20190311 (Red Hat 8.3.1-3) (GCC) does not match the compiler used here: cc (GCC)