similar to: How to replace match words whith colum name of data frame?Hi--

Dear ?, I'm sure that there are many ways to do what you want; here's one: > cbind(concept_df, category= + ifelse(apply( + sapply(chemical_df$chemical, + function(x) grepl(x, concept_df$concept)), + 1, any), + "chemical", "")) concept category 1 butan

I have two data frame. I want to use "chemical_df" to match "concept_df " concept_df <- data.frame(concept=c("butan acid ", "nano diamond particl", "slurri composit", "composit ph polis", " inorgan particl ", "grind liquid", "liquid formul", "nanoparticl", "size abras particl",

Try the stringr package. This should work chemical=c("basic", "alkalin", "alkali", "acid", " ph ", "hss") chemical_match <- str_c(chemical, collapse = "|") chemical_match concept_df$match[str_detect(concept_df$concept, chemical_match)] <- "chemical" concept_df > concept_df concept match

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

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

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: >

Because you need to separate the instructions with a ; (semi-colon). Hope this helps Rui Barradas Enviado a partir do meu smartphone Samsung Galaxy.-------- Mensagem original --------De: Ek Esawi <esawiek at gmail.com> Data: 08/01/2018 16:03 (GMT+00:00) Para: Jeff Newmiller <jdnewmil at dcn.davis.ca.us>, r-help at r-project.org Assunto: Re: [R] Replace NAs in split lists Thank you

OPS, Sorry i did not read the post carfully. Mine will not work if you have zeros on columns A and B.. But you could modify it to work for specific columns i believe. EK On Wed, Nov 22, 2017 at 8:37 AM, Ek Esawi <esawiek at gmail.com> wrote: > Hi *Massimo,* > > *Try this.* > > *a <- mydf==0mydf[a] <- NAHTHEK* > > On Wed, Nov 22, 2017 at 5:34 AM, Massimo Bressan

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

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 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 Staff This is my file (www.fiscalforecasting.com/data.csv) if you don't download this file, my dataset same as following Year Month A B C D E 2005 July 0 *4* NA NA *1* 2005 July 0 NA NA 0 *9* 2005 July NA *4* 0 *1* 0 2005 July *4* 0 *2* *9* NA I try to count non-zero values which are not NA values for every *column* *Sincerely* *Engin YILMAZ*

In y <- substr(x, i, 1) your third integer needs to be the location not the number of digits, so change it to y <- substr(x, i, i) and you should get what you want. Cheers, Tim > Date: Sun, 21 Jan 2018 10:50:31 -0500 > From: Ek Esawi <esawiek at gmail.com> > To: Luigi Marongiu <marongiu.luigi at gmail.com>, r-help at r-project.org > Subject: Re: [R] substr

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

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

"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

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

I am recoding some data. Many values that should be 1.5 are recorded as 1-2. Some example data and my solution is below. I am curious about better approaches or any other suggestions. Thanks! # example input data myData <- read.table(textConnection("id, v1, v2, v3 a,1,2,3 b,1-2,,3-4 c,,3,4"),header=TRUE,sep=",") closeAllConnections() # the first column is IDs so remove

Full_Name: Steven McKinney Version: 2.9.0 OS: Mac OS X 10.5.6 Submission from: (NULL) (142.103.207.10) A corrupt data frame can be constructed as follows: foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4]

Have you gone through any R tutorials? If not, why not? If so, maybe you need to spend some more time with them. It looks like you want us to do your work for you. We don't do this. See (and follow) the posting guide below for what we might do (we're volunteers, so no guarantees). Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and