similar to: Create New Column Inside Data Frame for Many Data Frames

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)

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

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 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 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 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

Dear R Helpers, A few weeks ago I asked for some help on how to accomplish modifications to data in a set of data frames. As part of that request I mentioned that I realized that one way to accomplish my goal was to put the data frames together in a list but that I was looking for a way to do it with data frames and a loop because I "believe the better thing is to work df by df for my

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? Quantmod. In the subject line. I agree that they look like dates, I don't know how to determine if they are actually dates. Josh Ulrich usually answers questions along these lines very informatively and quickly. One reasonable course of action is to wait to see if he does the same with this one. --JJS ________________________________ From: Bert Gunter <bgunter.4567 at

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

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)) >

Dear R Helpers, I have run into a problem trying to perform a number of actions on a set of quantmod data objects through a loop and I am hoping that this is an easy problem for someone else as opposed to very difficult for me. The example task is to get the first three objects of the quarterly balance sheet for a number of companies from the getFinancials object and put them together into a

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 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

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 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 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)

On 5 March 2018 at 02:46, Sparks, John wrote: | I agree that they look like dates, I don't know how to determine if they are actually dates. You know options but you are confused about maturity dates, i.e. expiry? In information in that list (ie along the date dimension) is the expiry; at each date you have another list for both puts and calls, and inside each of those a grid given by the

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 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