similar to: Interpret List Label as Date from Quantmod getOptionChain

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

On 5 March 2018 at 03:13, Sparks, John wrote: | library(quantmod) | #in fairness, I did not include this last time and my example was therefore not reproducible. Apologies to Bert and everyone else #for not following the posting guidelines. | aapl_total<-getOptionChain("AAPL", NULL)> | | How could I then get the subset of the entire list which only has expiry dates in 2019, or

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 Dirk, Thanks for your note. I understand that expiry dates are the dates that the option expires, so I don't think that I am confused about that (although the upper limits of one's confusion is difficult to accurately estimate). My lack of clarity come from treating those "dates" as actual dates as opposed to strings, which one could reasonably interpret them to be from

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)

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 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, 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 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 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 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, 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

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

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, 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 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)