similar to: Replacing Period in String

Dear All,   I have the following code set up:   x <-2000 y <-8 z <-3   I would need to use these numbers to show up in my plot title "mixed" with text. The x,y,z numbers would need to change, the text would not. So my title should look like this   "x txt1 y txt2 z txt3"   so if: txt1=hours txt2=minutes txt3=seconds   then my title of the plot should read:   2000 hours

Hi all... I?m making a chart dealing with frequencies of isotopes of various elements. For instance, I'd like the following text to appear on a chart with the "35" and "37" as superscripts: Based upon: 35Cl: 75% 37Cl: 25% I am having problems properly parsing the superscript that preceeds the "Cl", since there is no character ahead of the superscript (I saw

Dear R-ers! I have x as a variable in a data frame x. x<-data.frame(x=c("aa.bb","cc.dd.ee")) x$x<-as.character(x$x) x I am sorry for such a simple question - but how can I replace all periods in x$x with spaces? sub('.', ' ', x$x) - removes all letters to the left of each period... Thanks a lot for your advice! -- Dimitri Liakhovitski Ninah.com

hello could you help in solving the following problem I want to replace same consecutive words by a single word in a sentence.. for example --- my name name name is micky so I want the output like this--my name is micky I want this solution for a text file can you tell me the code for it?? thanking you in anticipation -- Shilpa Rai MSc.(2011-2013) Applied Statistics and Informatics Indian

Hello I have a sting of the form "12.084.547,17" which I would like R to understand as a number which has "," as the decimal separator, does anybody know how to do this? thank you Felipe Parra [[alternative HTML version deleted]]

Buenas tardes, grupo. Estoy tratando de hacer la comparación de dos archivos de una misma organización para encontrar las diferencias entre su informe del tema edl año 2005 y el del año 2013: Todos los comandos van bien, a exepción del último "colnames", como se ve en la siguiente secuencia: > pdf1<-"./PLAN de INSPECCIONES/05_seguridad_ciudadana.pdf" >

Dear R Helpers, I am still having trouble with the getOptionChain command in quantmod. I have the latest version of quantmod, etc. so I was under the impression that the problem was solved with updates to the package. If someone could let me know what I need to install in order to make this work, I would really appreciate it. My error message as session info are shown below. Thanks a bunch.

Hi R-Helpers, Sorry to bother you, but I have a simple task that I can't figure out how to do. For example, I have some names in two columns NamesWide<-data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Curly")) and I simply want to get a single column

Could somebody advise me about importing a txt file as a frame? I am using the command: test <- read.delim ("~/docs/perl/expr_ctx.txt2", header=T, sep = "\t", row.names = 1) This gives me an error because there are duplicate rows. In the txt file, the columns are unique subjects and the rows are variables, so I had planned to transform the file after importing. The first

Dear R Helpers, I am trying to write a character value to the row of a data frame and am running into a problem that I don't have when I do this for numeric arguments. For example, the following works just fine: > test<-data.frame(number=numeric(1)) > test[1,]<-.5 > test number 1 0.5 But the following bombs out: > hold<-data.frame(symbol=character(1)) >

Hi R Helpers, Is it possible to interpret the top level of the list as a date after downloading all the option chain data for a ticker? For example, after I run aapl_total<-getOptionChain("AAPL", NULL) the top descriptor of the lists is a date (Mar.09.2018, Mar.23.2018, etc.). So if want to subset down to those parts of the list that correspond to say, (expiration)

Dear R Helpers, I have another one of those problems involving a very simple step, but due to my inexperience I can't find a way to solve it. I had a look at a number of on-line references, but they don't speak to this problem. I have a variable with 20 values > table (testY2$redgroups) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Dear R Helpers, I have a large number of data frames and I need to create a new column inside each data frame. Because there is a large number, I need to "loop" through this, but I don't know the syntax of assigning a new column name dynamically. Below is a simple example of what I need to do. Assume that I have to do this for all 26 letters and you should see the form of the

Dear R Helpers, I have about 20 data frames that I need to do a series of data scrubbing steps to. I have the list of data frames in a list so that I can use lapply. I am trying to build a function that will do the data scrubbing that I need. However, I am new to functions and there is something fundamental that I am not understanding. I use the return function at the end of the function and

Hi, You were on the right track using stack(), but you just pass the entire data frame as a single object, not the separate columns: > stack(NamesWide) ? values ? ind 1 ? ?Tom Name1 2 ? Dick Name1 3 ?Larry Name2 4 ?Curly Name2 Note that stack also returns the index (second column of 'ind' values), which tells you which column in the source data frame the stacked values originated

Basic question: when I use names() to extract the name of a dataframe element, why does it have "." instead of " " between words? Context: I'm importing a CSV file of survey results for analysis. I read them like this: df <- read.csv("surveydata.csv",nrows=40,header=TRUE,

Package? The **names** of the top levels of your lists, "Mar.09.2018", "Mar.23.2018" certainly look like dates and if they are -- I have no idea what package/context is -- they certainly could be formatted as such. See e.g. "date-time" . There are also several package that provide date tools. Cheers, Bert Bert Gunter "The trouble with having an open mind is

Hello, I have several strings where I am trying to eliminate the period and everything after the period, using a regular expression. However, I am having trouble getting this to work. > x = "wa.w" > gsub(x, "\..*", "", perl=TRUE) [1] "" Warning messages: 1: '\.' is an unrecognized escape in a character string 2: unrecognized escape removed

Hi, I am reading a large space-delimited text file into R (41 columns and many rows) and need to do run each row's values through another R object and then write to another text file. So, far using readLines and writeLines seems to be the best bet. I've gotten the data exchange working except each row is read in as one 'chunk', meaning the row has all values between two quotes

Just to repeat: you have NamesWide<-data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Curly")) and you want NamesLong<-data.frame(Names=c("Tom","Dick","Larry","Curly")) There must be something I am missing, because NamesLong <- data.frame(Names = c(NamesWide$Name1, NamesWide$Name2)) appears to