similar to: Changing multiple instances in data.frame

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: >

Upon closer examination I see that you are not using the split version of df1 as I usually would, so here is a reproducible example: #---- df1 <- read.table( text= "ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA ", header=TRUE, as.is=TRUE ) sdf <- split( df1, df1$ID ) # note the extra [ 1 ]

I don't know. You seem to be posting in HTML so your code is mangled. Can you post plain text and use the reprex package to make sure it produces the errorin a clean R session? -- Sent from my phone. Please excuse my brevity. On January 8, 2018 8:03:45 AM PST, Ek Esawi <esawiek at gmail.com> wrote: >Thank you Jeff. Your code works, as usual , perfectly. I am just >wondering why

Hi With the example, na.locf seems to be the easiest way. > library(zoo) > na.locf(df1) ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 4 b aa TRUE 5 5 b ab FALSE 5 Cheers Petr > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Jeff > Newmiller > Sent: Monday, January

Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use that going forward in your calculations? That is generally the preferred approach in R so you can re-do your calculations easily if you find a mistake later. -- Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com> wrote: >I just came

OPS! Sorry i did indeed posted the code in HTML; should have known better. ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z}) error. unexpected symbol in sdf2 On Mon, Jan 8, 2018 at 11:44 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > I don't know. You seem to be posting in HTML so your code is mangled. Can you post plain text and use the reprex package to

Yes, you are right if the IDs are always sequentially-adjacent and the first non-NA value appears in the first record for each ID. -- Sent from my phone. Please excuse my brevity. On January 8, 2018 2:29:40 AM PST, PIKAL Petr <petr.pikal at precheza.cz> wrote: >Hi > >With the example, na.locf seems to be the easiest way. >> library(zoo) > >> na.locf(df1) > ID

You can enforce these assumptions by sorting on multiple columns, which leads to na.locf(df1[ order(df1$ID,df1$Value), ]) On Mon, Jan 8, 2018 at 4:19 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > Yes, you are right if the IDs are always sequentially-adjacent and the > first non-NA value appears in the first record for each ID. > -- > Sent from my phone. Please

Hi all-- I stumbled on this problem online. I did not like the solution given there which was a long UDF. I thought why cannot split and l/s apply work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2 variables a and b;

"Enforce" is overstating it... results will differ if there are no non-NA values for a given ID, and there is a potential further discrepancy if there are multiple non-NA values. But these issues were not identified by the OP, so may not be relevant in their case. -- Sent from my phone. Please excuse my brevity. On January 8, 2018 6:41:33 AM PST, Eric Berger <ericjberger at

Hi all, probably really simple to solve, but having no background in programming I haven't been able to figure this out: I have two dataframes like df1 <- data.frame(names1=c('aa','ab', 'ac', 'ad'), var1=c(1,5,7,12)) df2 <- data.frame(names2=c('aa', 'ab', 'ac', 'ad', 'ae'), var2=c(3,6,9,12,15)) Now I want merge

I just came up with a solution right after i posted the question, but i figured there must be a better and shorter one.than my solution sdf1[[1]][1,4]<-lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com> wrote: > Hi all-- > > I stumbled on this problem online. I did not like the solution given > there

Dear R'sians, Would really appreciate if you could suggest a more efficient way to order the columns of a dataset. The column names of the dataset contain indices separated by a period. Following are examples of my code and the dataset. oC <- function(tg=x2) { lth <- length(grep("T",names(tg))) thix <-

Hi, Thanks in advance for reading this post. I received some affymetrix genotyping data back recently (250K, Nsp array)…However, in order for me to do any analysis on this data set, I need to add append the annotation file to it. Basically I want to do something that looks like this: Snpfile(tab delimited): SNPID Genotype X Y 123 AA 13.4 1.2

Hi R users, I am trying to extract some attributes (age, sex, area) from dataframe "AB" that has 101,269 observations of 28 variables to dataframe "t2" that has 47 observations of 6 variables. They share a column called "id", which is a factor with 47 levels. I want to end up with a dataframe that has 47 observations of 9 variables (the original 6 variables of t2,

Clearly, I don't understand what order() is doing and as ususl the help for order seems to only confuse me more. For some reason I just don't follow the examples there. I must be missing something about the data frame sort there but what? I originally wanted to reverse-order my data frame df1 (see below) by aa (a factor) but since this was not working I decided to simplify and order by

> On Jul 7, 2017, at 6:03 AM, John Kane via R-help <r-help at r-project.org> wrote: > > This is not serious problem but I just wonder if someone can explain what is happening. > The same command within a dataframe is giving me a factor and as a plain vector is giving me a character. It's probably something simple that I have read and forgotten but I thought I'd ask.

Hi, I am sure there is an easy way to do it, but I can't find it. I have a data frame that has 15 columns and 7000 rows. The only values inside the data.frame are "aa", "ab", "bb" as you can see an example bellow. 1 2 3 1 aa ab ab 2 ab ab ab 3 aa aa aa 4 bb bb bb What I would like to do, is to generate a vector (or another data.frame) with 7000 rows, and 3

Thanks Marc. It never occurred to me that I would need a ""stringsAsFactors" expression in a data.frame.? I could have sworn I never did before when mocking up some data but clearly I was wrong or there has been a change in R v. 3.4.1 which seems unlikely. On Friday, July 7, 2017, 10:37:29 AM EDT, Marc Schwartz <marc_schwartz at me.com> wrote: > On Jul 7, 2017, at 6:03

Hola: Aplicaré lo que dices de usar data.table. Sobre hacer for, etc. Ya lo he hecho, pero pensaba que usar dplyr haría la tarea más rápida por estar este implementado en C (si no me equivoco). Siempre que puedo utilizo estas funciones, porque codificas menos y van más rápido. En este caso el problema es que no he encontrado la manera de hacer lo que quiero con dplyr o similar. La idea era al