similar to: Replace NAs in split lists

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

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

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 ]

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

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

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

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;

Hi, I have two dataframes one with 144 rows and 160 columns (SDF1) and one with 12 rows and 160 columns (SDF2). Now I'm trying to divide rows 1:12 with SDF2, rows 13:24 with SDF2, rows 25:36 with SDF 2, . In S-Plus the following code works fine: DFS = SDF1[1:144,1:60] / as.vector(SDF2[1:12,1:160]) but in R when I try to implement the formula I get the following error: "/

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

Then you need to rethink your data structure. Use a list instead of a data frame. The components of a list can have different lengths, and the "apply" family of functions (lapply(), etc.) can operate on them. Consult any good R tutorial for details. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it."

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

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*

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

Hi All? I was looking at a posting from June-17. I managed to solve it. However, when I changed the example in the posting, my solution will work only once at a time which was mentioned by Jim Lemon on his response to the original posting. This means that my solution will have to be repeated as many times as the maximum number of spaces on each gap; something that may not work well for large

BLS #rsync --version rsync version 3.0.4pre2 protocol version 30 Copyright (C) 1996-2008 by Andrew Tridgell, Wayne Davison, and others. Web site: http://rsync.samba.org/ Capabilities: 64-bit files, 64-bit inums, 32-bit timestamps, 64-bit long ints, socketpairs, hardlinks, symlinks, IPv6, batchfiles, inplace, append, ACLs, xattrs, iconv, symtimes rsync comes with ABSOLUTELY NO

Hi, I have a dataframe SDF1 that looks like this: Char1 Char2 Char 3 W.2007.02 W.2007.09 W.2007.16 W.2008.13 A C1 F1 F2 F3 A C2 F4 B C3 F5 F6 I have another dataframe SDF2 with 163 cols that has the following column names Char1 Char2 Char 3 W.2007.02 W.2007.03 W.2007.04