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...quot; ?"Larry" "Curly" returns a vector, or: > stack(NamesWide)[, 1, drop = FALSE] ? values 1 ? ?Tom 2 ? Dick 3 ?Larry 4 ?Curly which returns a data frame with a single column named 'values'. Regards, Marc Schwartz On April 3, 2023 at 11:08:59 AM, Sparks, John (jspark4 at uic.edu (mailto:jspark4 at uic.edu)) wrote: > 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"),Name...

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

...package that provide date tools. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sun, Mar 4, 2018 at 6:28 PM, Sparks, John <jspark4 at uic.edu> wrote: > 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) > &...

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)

...Tom Name1 2 Dick Name1 3 Larry Name2 4 Curly Name2 If there were multiple columns, you might do > stack(NamesWide[,c("Name1","Name2")])$values [1] "Tom" "Dick" "Larry" "Curly" On Tue, 4 Apr 2023 at 03:09, Sparks, John <jspark4 at uic.edu> wrote: > 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&quot...

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)) >

...package that provide date tools. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sun, Mar 4, 2018 at 6:28 PM, Sparks, John <jspark4 at uic.edu<mailto:jspark4 at uic.edu>> wrote: 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...

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

...26: > unname(unlist(NamesWide)) Why not: NamesWide <- data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Curly")) NamesLong <- data.frame(Names=with(NamesWide, c(Name1, Name2))) > > On April 3, 2023 8:08:59 AM PDT, "Sparks, John" <jspark4 at uic.edu> wrote: >> 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"),Na...

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 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")

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

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 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, 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 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

Hi R Helpers, I recently tried to take advantage of the ability to download all the tickers in the S&P 500 using the functionality of tidyquant, but it threw an error. For summary, the set of commands that I ran was library(tidyquant) tq_index_options() tq_index("SP500") sessionInfo() R feedback including error message and sessionInfo are provided below. Guidance would be