similar to: If Then Trouble

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

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 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

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

Dear R Users, I am working with gsub for the first time. I am trying to remove some characters from a string. I have hit the problem where the period is the shorthand for 'everything' in the R language when what I want to remove is the actual periods. In the example below, I simply want to remove the periods as I have removed the comma, but instead the complete string is wiped out. I

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

Dear R Helpers, I am trying to isolate a set of characters between two other characters in a long string file. I tried some of the examples on the R help pages and elsewhere, but I am not able to get it. Your help would be much appreciated. require(scrapeR)

Dear R Helpers, I need help with a slightly unusual situation in which I am trying to select some columns from a data frame. I know how to use the subset statement with column names as in: x=as.data.frame(matrix(c(1,2,3, 1,2,3, 1,2,2, 1,2,2, 1,1,1),ncol=3,byrow=T)) all.cols<-colnames(x) to.keep<-all.cols[1:2] Kept<-subset(x,select=to.keep) Kept

Dear R Helpers, I did a search for deleting rows based on conditions but wasn't able to find an example that addressed the error that I am getting. I am hoping that this is a simple syntax phenomenon that somebody else knows off the top of their head. My apologies for not providing a reproducible example but I think that the information given will allow someone to give me a hint. I want to

Dear R Helpers, I am trying to change the name of an object using the assign function. When I use paste on the new object but not the old, everything is fine: The new object is a direct copy of the old object. When I use a paste for both the new and the old object, however, the new object is simply the character representation of the old object name, not the old object itself. The example

Hi, I have been learning the quantmod package over the last several days. I went to check some of my data pulls against other sources and was surprised to find that a few tickers that have single characters do not successfully scrape from Google Finance using getFin(). Particularly require(quantmod) getFin("A") getFin("E") getFin("F") getFin("G")

I am having difficulty with dependencies for the R rpm for Fedora Core 2. In attempting to load R-2.0.0.0.fdr.1.fc2.i386.rpm it fails the libtk8.4.so dependency even though I have this library loaded. The library is located in: /usr/local/lib. How might I tell R where to find this library? I am new to Linux and loading rpms so I am hoping there is a simple answer and that the problem is

Dear R Helpers, First, I apologize for asking for help on the first of my topics. I have been looking at the posts and pages for apply, tapply etc, and I know that the solution to this must be ridiculously easy, but I just can't seem to get my brain around it. If I want to produce a set of tables for all the variables in my data, how can I do that without having to type them into the table

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.

Dear R Helpers, I am trying to do calculations on multiple data frames and do not want to create a list of them to go through each one. I know that lists have many wonderful advantages, but I believe the better thing is to work df by df for my particular situation. For background, I have already received some wonderful help on how to handle some situations, such as removing columns:

Dear R-users, I'm using lattice package and function xyplot for the first time so you will excuse me for my inexperience. I'm facing quite a simple problem but I'm having troubles on how to solve it, I've read tons of old mails in the archives and looked at some slides from?Deepayan Sarkar but still can not get the point. This is the context. I've got data on 9 microRNAs, each

Is it possible to create a stub that returns hash values. For example I would like to convert this: @siteItem = stub(''plmSiteItem'', :one => "uno") To something like this: @siteItem = stub(''plmSiteItem'', {''one'' => ''uno'', ''two'' => ''dos''} ) So that I can do this: