similar to: any and all

Hi Avi, As D?nes T?th has rightly diagnosed, you are building an "all or nothing" filter. However, you do not need to explicitly spell out all columns that you want to filter for; the "tidy" way would be to use a helper function like `if_all()` or `if_any()`. Consider this example (I hope I understand your intentions correctly): ``` library(dplyr) data <- tribble(

On 12/04/2024 3:52 p.m., avi.e.gross at gmail.com wrote: > Base R has generic functions called any() and all() that I am having trouble > using. > > It works fine when I play with it in a base R context as in: > >> all(any(TRUE, TRUE), any(TRUE, FALSE)) > [1] TRUE >> all(any(TRUE, TRUE), any(FALSE, FALSE)) > [1] FALSE > > But in a tidyverse/dplyr

Hi Avi, As Duncan already mentioned, a reproducible example would be helpful to assist you better. Having said that, I think you misunderstand how `dplyr::filter` works: it performs row-wise filtering, so the filtering expression shall return a logical vector of the same length as the data.frame, or must be a single boolean value meaning "keep all" (TRUE) or "drop all"

Although there may well be many ways to do what is being asked for with the tidyverse, sometimes things are simple enough to do the old-fashioned way. The request seems to have been to do something to all rows in ONE specific group but was phrased in the sense of wanting to know which group your functionality is being called in. What grouping gains you is more worthwhile if you are interested in

Hi, I have a grouped data set and would like to calculate weighted proportions for a large number of factor variables within each group member. Rather than using dplyr::count() on each of these factors individually, the idea would be to do it for all factors at once. Does anyone know how this would work? Here is a reproducible example: ############################################################

Base R. Regarding code improvements: 1. Personally I find (\(...) ...)() notation hard to read (although by placing (\(x), the body and )() on 3 separate lines it can be improved somewhat). Instead let us use a named function. The name of the function can also serve to self document the code. 2. The use of dat both at the start of the pipeline and then again within a later step of the pipeline

Laurent: As I don't use dplyr, this won't help you, but I hope you and others may find it entertaining anyway. If I understand you correctly (and ignore this if I have not), there are a ton of ways to do this in base R, including using switch() along the lines you noted in your post. However, when the functions get sufficiently complicated or numerous, it may be useful to store them in a

Hi R Helpers, I have been looking for an example of how to execute different dplyr mutate statements on the same dataframe in a single step. I show how to do what I want to do by going from df0 to df1 to df2 to df3 by applying a mutate statement to each dataframe in sequence, but I would like to know if there is a way to execute this in a single step; so simply go from df0 to df1 while executing

Thanks. I found this to be quite informative and a nice example of how useful R-Help can be as a resource for R users. Best, Bert On Mon, Jul 22, 2024 at 4:50?AM Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > Base R. Regarding code improvements: > > 1. Personally I find (\(...) ...)() notation hard to read (although by > placing (\(x), the body and )() on 3

?s 22:50 de 17/10/2024, Sparks, John escreveu: > Hi R Helpers, > > I have been looking for an example of how to execute different dplyr mutate statements on the same dataframe in a single step. I show how to do what I want to do by going from df0 to df1 to df2 to df3 by applying a mutate statement to each dataframe in sequence, but I would like to know if there is a way to execute this

Hi Folks, I have just started using dplyr and could use some help getting unstuck. It could well be that dplyr is not the package to be using, but let me just pose the question and seek your advice. Here is my basic data frame. head(h) subject ageGrp ear hearingGrp sex freq L2 Ldp Phidp NF SNR 1 HALAF032 A L A F 2 0 -23.54459 55.56005 -43.08282

Hi Martin, On 11/29/2017 10:46 PM, Martin Morgan wrote: > On 11/29/2017 04:15 PM, T?th D?nes wrote: >> Hi, >> >> A benchmarking study with an additional (data.table-based) solution. > > I don't think speed is the right benchmark (I do agree that correctness > is!). Well, agree, and sorry for the wording. It was really just an exercise and not a full

Hello. My question is in the subject line. Using R 4.1.3 on Windows 10. Commented MWE below. Thanks. --Chris Ryan library(dplyr) library(lattice) ## fabricate a dataframe dd <- data.frame(agency = sample(LETTERS, size = 5), total = sample(100:200, size = 5), las = sample(20:40, size = 5)) dd <- dd %>% mutate(proportion = las/total, bubble = total/100) ## attempt to make a dotplot

On 11/29/2017 04:15 PM, T?th D?nes wrote: > Hi, > > A benchmarking study with an additional (data.table-based) solution. I don't think speed is the right benchmark (I do agree that correctness is!). For the R-help list, maybe something about least specialized R knowledge required would be appropriate? I'd say there were some 'hard' solutions -- Michael (deep

Hello I have a question. I made an r-script and did a few commands needed to make some new variables. They all work out well, and when I run the commands, the new variables appear in the dataset. I can also work with these new variables to make other new variables from them. Also, when I use summary(dataset), those new variables appear in the summary of the dataset. But, when I do summary(new

I am new to this thread. At the risk of presenting something that has been shown before, below I demonstrate how a column in a data frame can be dropped using a wild card, i.e. a column whose name starts with "th" using nothing more than base r functions and base R syntax. While additions to R such as tidyverse can be very helpful, many things that they do can be accomplished simply

Thank you Jeff. Your code works, as usual , perfectly. I am just wondering why if i put the whole code in one line, i get an error message. sdf2 <- lapply( sdf, function(z){z$Value <-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z}) error. unexpected symbol in sdf2 Thanks again EK On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: >

Hi, I am trying to replicate the R's result for VCV matrix of estimated coefficients from linear model as below data(mtcars) model <- lm(mpg~disp+hp, data=mtcars) model_summ <-summary(model) MSE = mean(model_summ$residuals^2) vcov(model) Now I want to calculate the same thing manually, library(dplyr) X = as.matrix(mtcars[, c('disp', 'hp')] %>% mutate(Intercept =

Hello - I am trying to plot flows in a number of rivers within the same plot, and need to colour the lines differently, using a colour-blind palette. Code beneath works but has colours assigned by the program I have made some simple dummy data: ## code 1 cb8<- c("#000000", "#E69F00", "#56B4E9", "#009E73","#F0E442", "#0072B2",

Dear All, I am trying to reshape the data with some conditions. A small part of the data looks like below. Like this there will be more data with repeating ID. Count id name type 117 335 sally A 19 335 sally A 167 335 sally B 18 340 susan A 56 340 susan A 22 340 susan B 53 340 susan B 135 351 lee A 114 351 lee A 84 351 lee A 80 351 lee A 19 351 lee A 8 351 lee A 21 351 lee A 88 351 lee B 111 351