similar to: Looping Over Data Frames

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 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

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

Hello! I have a data frame with several rows, for example: 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)) I would like to find y - a data frame that only has the unique rows from x, i.e.: 1,2,3 1,2,2 1,1,1 Thanks a lot for your hints! Dimitri -- Dimitri Liakhovitski gfk.com <http://marketfusionanalytics.com/> [[alternative HTML

Hello, I am trying to wrap my head around the coordinates systems associated with the layout() function ...with the end goal of simply drawing a decorative line in the upper margin of my figure, which is composed of three plots. My output is defined as this: ---------------------------------------------------------------------------- postscript("out.ps", horizontal=FALSE,

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

Hi. Can anyone suggest a simple way to obtain in R a list of vector coordinates of the following form? The code below is Mathematica. In[5]:= Flatten[Table[{i,j,k},{i,3},{j,4},{k,5}], 2] Out[5]= {{1,1,1},{1,1,2},{1,1,3},{1,1,4},{1,1,5},{1,2,1},{1,2,2},{1,2,3},{1 ,2,4},{1,2, 5},{1,3,1},{1,3,2},{1,3,3},{1,3,4},{1,3,5},{1,4,1},{1,4,2},{1,4,3}, {1,4,

Hi all, I would like to run a function with several nested conditions, which are completely factorial. The input data (x1) has two different sample sizes, so: x1 <- dat1 x1 <- dat2 Then a can have 3 different values: a <- 0.15 a <- 0.35 a <- 0.50 Then b can have 2 different values: b <- data.matrix (c(0.5,0.5,0.5)) b <- data.matrix (c(0.2,0.4,0.6)) Then d can have 5

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, 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 List, I have a problem I'm finding it difficult to make headway with. Say I have 6 ordered observations, and I want to find all combinations of splitting these 6 ordered observations in g groups, where g = 1, ..., 6. Groups can only be formed by adjacent observations, so observations 1 and 4 can't be in a group on their own, only if 1,2,3&4 are all in the group. For example,

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 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, I'd like to ask you a question again. It is basically about data frames, NAs and tabulate function. I have this data frame. I already used this in one of the previous questions of mine. It intentionally looks this simple, my real 'df' dataframe is much bigger actually and again, I am not willing to annoy anyone with huge databases... So, my database: id

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

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

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)

Hi everyone, my question might be very trivial, but I could not come up with an answer... I want to find out how often a matrix contains a certain vector as row: x1<-c(1,2,3) x2<-c(1,5,6) x3<-c(7,8,9) A<-matrix(c(rep(x1,5),rep(x2,5),rep(x3,5),rep(x1,5)),nrow=20,ncol=3,byrow=T) How can I find out, how many times x1 is a row of A? Thanks in advance and best regards, Sebastian

On 10/26/2015 1:46 PM, Geert Stappers via Syslinux wrote: > On Mon, Oct 26, 2015 at 12:09:40PM -0600, Alan Sparks via Syslinux wrote: >> I'm still trying fruitlessly to get some sort of local disk boot from >> syslinux EFI to work... using the 6.03 modules. Tried various >> combinations of configurations on Gene's test binaries. > > The test binariers Gene

Hi All, I am interested in aggregating a data frame based on 2 categories--mean effect size (r) for each 'id's' 'mod1'. The 'with' function works well when aggregating on one category (e.g., based on 'id' below) but doesnt work if I try 2 categories. How can this be accomplished? # sample data id<-c(1,1,1,rep(4:12)) n<-c(10,20,13,22,28,12,12,36,19,12,