Re: [R] Removing polygons from shapefile of Scotland and Islands

I believe mapshaper has functionality for removing small 'islands'. There is a webinterface for mapshaper, but I see there is also an R-package (see https://search.r-project.org/CRAN/refmans/rmapshaper/html/ms_filter_islands.html for island removal). If you want to manually select which

[R] Removing polygons from shapefile of Scotland and Islands

Hello I have a shapefile of Scotland, including the islands. The river flow data I am using is only for the mainland and for a clearer and larger map I would like to not plot Orkney and Shetland to the north of the mainland, as I don't need them. The map I have I got from

Re: [R] Removing variables from data frame with a wile card

, February 12, 2023 5:19 PM To: Andrew Simmons Cc: R-help Mailing List Subject: Re: [R] Removing variables from data frame with a wile card In the line suggested by Andrew Simmons, mydata <- mydata[, !grepl("^yr", colnames(mydata)), drop = FALSE] what does drop=FALSE do? Thanks. On 1/14

Re: [R] Removing variables from data frame with a wile card

Great, Thanks. Now I have many options. Steven from iPhone > On Feb 13, 2023, at 10:52 AM, Andrew Simmons wrote: > > What I meant is that that > > mydata[, !grepl("^yr", colnames(mydata)), drop = FALSE] > > and > > mydata[!grepl("^yr", colnames(mydata))] > > should be identical. Some

Re: [R] Removing variables from data frame with a wile card

What I meant is that that mydata[, !grepl("^yr", colnames(mydata)), drop = FALSE] and mydata[!grepl("^yr", colnames(mydata))] should be identical. Some people would prefer the first because the indexing looks the same as matrix indexing, whereas some people would prefer the second because it

Re: [R] Removing variables from data frame with a wile card

Complain, complain... x[ names( x ) != "V2" ] or x[ ! names( x ) %in% c( "V2", "V3" ) ] or any other character or logical or integer expression that selects columns you want... On February 12, 2023 6:38:00 PM PST, Steven Yen wrote: >x[“V2”] would retain columns of x headed by V2. What I

Re: [R] Removing variables from data frame with a wile card

x[“V2”] would retain columns of x headed by V2. What I need is the opposite——I need a data grime with those columns excluded. Steven from iPhone > On Feb 13, 2023, at 9:33 AM, Rolf Turner wrote: > >  >> On Sun, 12 Feb 2023 14:57:36 -0800 >> Jeff Newmiller wrote: >> >> x["V2"] >> >> is

Re: [R] Removing variables from data frame with a wile card

On Sun, 12 Feb 2023 14:57:36 -0800 Jeff Newmiller wrote: > x["V2"] > > is more efficient than using drop=FALSE, and perfectly normal syntax > (data frames are lists of columns). I never cease to be amazed by the sagacity and perspicacity of the designers of R. I would have worried that

Re: [R] Removing variables from data frame with a wile card

Thanks Jeff and Andrew. My initial file, mydata, is a data frame with 92 columns (variables). After the operation (trimming), it remains a data frame with 72 variables. So yes indeed, I do not need the drop=FALSE. > is.data.frame(mydata) [1] TRUE > ncol(mydata) [1] 92 >

Re: [R] Removing variables from data frame with a wile card

x["V2"] is more efficient than using drop=FALSE, and perfectly normal syntax (data frames are lists of columns). I would ignore the naysayers, or put a comment in if you want to accelerate their uptake. As I understand it, one of the main reasons tibbles exist is because of drop=TRUE.

Re: [R] Removing variables from data frame with a wile card

drop = FALSE means that should the indexing select exactly one column, then return a data frame with one column, instead of the object in the column. It's usually not necessary, but I've messed up some data before by assuming the indexing always returns a data frame when it doesn't, so drop =

Re: [R] Removing variables from data frame with a wile card

In the line suggested by Andrew Simmons, mydata <- mydata[, !grepl("^yr", colnames(mydata)), drop = FALSE] what does drop=FALSE do? Thanks. On 1/14/2023 8:48 PM, Steven Yen wrote: > Thanks to all. Very helpful. > > Steven from iPhone > >> On Jan 14, 2023, at 3:08 PM, Andrew Simmons wrote: >>

Re: [R] Removing variables from data frame with a wile card

From: R-help on behalf of Valentin Petzel Sent: Saturday, January 14, 2023 1:21 PM To: avi.e.gr...@gmail.com Cc: 'R-help Mailing List' Subject: Re: [R] Removing variables from data frame with a wile card Hello Avi, while something like d$something <- ... may s

Re: [R] Removing variables from data frame with a wile card

, John Sent: Sunday, January 15, 2023 11:55 AM To: Valentin Petzel ; avi.e.gr...@gmail.com Cc: 'R-help Mailing List' Subject: Re: [R] Removing variables from data frame with a wile card I am new to this thread. At the risk of presenting something that has been shown before, below I demonstrate

Re: [R] Removing variables from data frame with a wile card

lp on behalf of Valentin Petzel Sent: Saturday, January 14, 2023 1:21 PM To: avi.e.gr...@gmail.com Cc: 'R-help Mailing List' Subject: Re: [R] Removing variables from data frame with a wile card Hello Avi, while something like d$something <- ... may seem like you're directly modifying the da

Re: [R] Removing variables from data frame with a wile card

hange. > > For those who like to use the tidyverse, it comes with lots of tools that let > you select columns that start with or end with or contain some pattern and I > find that way easier. > > > > -Original Message- > From: R-help On Behalf Of Steven Yen >

Re: [R] Removing variables from data frame with a wile card

do not want, simply melt away. The part about selecting or deselecting columns can often be used in many of the verbs. From: John Kane Sent: Saturday, January 14, 2023 4:07 PM To: avi.e.gr...@gmail.com Cc: R-help Mailing List Subject: Re: [R] Removing variables from data frame with a wile

Re: [R] Removing variables from data frame with a wile card

I prefer to avoid religious wars. -Original Message- From: Valentin Petzel Sent: Saturday, January 14, 2023 1:21 PM To: avi.e.gr...@gmail.com Cc: 'R-help Mailing List' Subject: Re: [R] Removing variables from data frame with a wile card Hello Avi, while something like d$something <- ... may

Re: [R] Removing variables from data frame with a wile card

ier. > > > > -Original Message- > From: R-help On Behalf Of Steven Yen > Sent: Saturday, January 14, 2023 7:49 AM > To: Andrew Simmons > Cc: R-help Mailing List > Subject: Re: [R] Removing variables from data frame with a wile card > > Thanks to all. Ver

Re: [R] Removing variables from data frame with a wile card

find that way easier. -Original Message- From: R-help On Behalf Of Steven Yen Sent: Saturday, January 14, 2023 7:49 AM To: Andrew Simmons Cc: R-help Mailing List Subject: Re: [R] Removing variables from data frame with a wile card Thanks to all. Very helpful. Steven from iPhone >

Re: [R] Removing variables from data frame with a wile card

The -grep(pattern,colnames) as a subscript is a bit dangerous. If no colname matches the pattern then all columns will be omitted (because -0 is the same as 0, which means no column). !grepl(pattern,colnames) avoids this problem. > mydata <- data.frame(A=1:3,B=11:13) > mydata[, -grep("^yr",

Re: [R] Removing variables from data frame with a wile card

Thanks to all. Very helpful. Steven from iPhone > On Jan 14, 2023, at 3:08 PM, Andrew Simmons wrote: > > You'll want to use grep() or grepl(). By default, grep() uses extended > regular expressions to find matches, but you can also use perl regular > expressions and globbing (after converting

Re: [R] Removing variables from data frame with a wile card

You'll want to use grep() or grepl(). By default, grep() uses extended regular expressions to find matches, but you can also use perl regular expressions and globbing (after converting to a regular expression). For example: grepl("^yr", colnames(mydata)) will tell you which 'colnames' start with

Re: [R] Removing variables from data frame with a wile card

mydata[, -grep("^yr",colnames(mydata))] On Sat, Jan 14, 2023 at 8:57 AM Steven T. Yen wrote: > I have a data frame containing variables "yr3",...,"yr28". > > How do I remove them with a wild cardsomething similar to "del yr*" > in Windows/doc? Thank you. > > > colnames(mydata) >[1]

[R] Removing variables from data frame with a wile card

I have a data frame containing variables "yr3",...,"yr28". How do I remove them with a wild cardsomething similar to "del yr*" in Windows/doc? Thank you. > colnames(mydata)   [1] "year"   "weight" "confeduc"   "confothr" "college"   [6] ...  [41] "yr3"    "yr4"    "yr5"

Re: [R] removing non-table lines

omment="string" in the tidyverse function read_csv(). > > And, of course you can skip lines if that makes sense albeit it can be tricky > with header lines. > > -Original Message- > From: R-help On Behalf Of Rui Barradas > Sent: Sunday, September 18, 2022 6:19

Re: [R] removing non-table lines

v(). And, of course you can skip lines if that makes sense albeit it can be tricky with header lines. -Original Message- From: R-help On Behalf Of Rui Barradas Sent: Sunday, September 18, 2022 6:19 PM To: Nick Wray ; r-help@r-project.org Subject: Re: [R] removing non-table lines H

Re: [R] removing non-table lines

Helo, Unfortunatelly there are many files with a non tabular data section followed by the data. R's read.table has a skip argument: skip integer: the number of lines of the data file to skip before beginning to read data. If you do not know how many lines to skip because it's not

Re: [R] removing non-table lines

Use the skip parameter if the number of header lines is always the same. On September 18, 2022 12:39:50 PM PDT, Nick Wray wrote: >Hello - I am having to download lots of rainfall and temperature data in >csv form from the UK Met Office. The data isn't a problem - it's in nice >columns and can

Re: [R] removing non-table lines

Can you provide a sample of say the first 3 rows then the last 2 rows before the CSV starts. Are there always the same number of lines at the top? Or can it vary depending what non-sense the Met Office decided to contaminate it with? This should be solvable with some sample data. Base R or

[R] removing non-table lines

Hello - I am having to download lots of rainfall and temperature data in csv form from the UK Met Office. The data isn't a problem - it's in nice columns and can be read into R easily - the problem is that in each csv there are 60 or so lines of information first which are not part of the

Re: [R] removing dates from xts object..

Dear Joshua, Thanks a lot... Yours sincrely, AKSHAY M KULKARNI From: Joshua Ulrich Sent: Tuesday, July 26, 2022 2:20 AM To: akshay kulkarni Cc: R help Mailing list Subject: Re: [R] removing dates from xts object.. On Mon, Jul 25, 2022 at 3

Re: [R] removing dates from xts object..

ich Sent: Tuesday, July 26, 2022 1:56 AM To: akshay kulkarni Cc: R help Mailing list Subject: Re: [R] removing dates from xts object.. This is the most straightforward and general way I can think of quickly: library(xts) data(sample_matrix) x <- as.xts(sample_matrix, dateFormat = "Da

[R] removing dates from xts object..

Dear members, I have: > head(testOHLC) HINDUNILVR.NS.Open HINDUNILVR.NS.High HINDUNILVR.NS.Low HINDUNILVR.NS.Close 2007-01-02 217.80 219.00215.45 216.40 2007-01-03 217.00 217.65

Re: [R] Removing a space from a string

Richard Per your requests: 1. Plain text: no spaces 2. read_docx: spaces 3. read_rtf: no spaces 4. Not requested by you: copying from the Word document, then pasting into “vim”: no spaces The Word document was created by hand but #1, #3, and #4 confirm that

Re: [R] Removing a space from a string

Hi! How about this? --- snip --- > x <- c("STRING 01. Remainder of the string","STR ING 01. Remainder of the string","STRIN G 01. Remainder of the string","STR IN G 01. Remainder of the string") > x1<-unlist(strsplit(x,"\\.")) > for (i in seq(1,length(x1),2)) { x[(i+1) %/%

Re: [R] Removing a space from a string

The spaces may not have been VISIBLE in the Word document, but that does not mean that there wasn't anything THERE. - What happens if you open the document in Word and save it as plain text? - What happens if you open the document in Word and save it as RTF, then read that using read_rtf? -

Re: [R] Removing a space from a string

Richard In reply to your “first response”, the text was originally in a Word document and it did NOT contain the errant spaces. I used read_docx in the textreadr package to access the text. The spaces were added during that step. I am copying the maintainer of that package to see if he has

Re: [R] Removing a space from a string

The first response has to be "how did the spaces get there in the first place?" Can you fix the process that creates the data? If the process sometimes generates one extra space, are you sure it never generates two? But let's treat this purely as a regular expression problem, where if there is

Re: [R] Removing a space from a string

Note that my previous strategy can be expressed slightly more clearly as: x <- c("STRING 01. Remainder of the string", "STR ING 01. Remainder of the string", "STRIN G 01. Remainder of the string", "STR IN G 01. Remainder of the string") ## more spaces in this last example entry rx <-

Re: [R] Removing a space from a string

1. Thanks for the nice reprex. 2. However, I thought there was still a bit of ambiguity. I interpreted your specification to mean: "any number of spaces could occur in the beginning alphabetic part of the strings before one or more digits occur followed by a '.' (a period) and then more stuff

Re: [R] Removing a space from a string

On 2020-07-28 23:00 +0200, Rasmus Liland wrote: | | Perhaps by using gregexpr to look for | dots, remove spaces from the substring until the first | finding, then pasting it back. | | strings <- | c("STRING 01. Remainder of the string.", | "STR ING 01. Remainder of the

Re: [R] Removing a space from a string

This RegEx would do it I think: \s(?=.*\s\d*\.) Looks for space - \s Before any strings followed by space, numbers, period text <- "STR ING 01. Remainder of the string" stringr::str_replace_all(text, "\\s(?=.*\\s\\d*\\.)", "") Should do it I think! On 2020-07-28 21:34, Dennis Fisher wrote:

Re: [R] Removing a space from a string

Dear Dennis, On 2020-07-28 13:20 -0700, Dennis Fisher wrote: | Colleagues | | I have strings that contain a space in | an unexpected location. The intended | string is: | “STRING 01. Remainder of the string" | However, variants are: | “STR ING 01. Remainder of the string" |

Re: [R] Removing a space from a string

Only the spaces in STRING. However, if I inadvertently delete the space between STRING and NN, I can add it back in. Dennis Fisher MD P < (The "P Less Than" Company) Phone / Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com > On Jul 28, 2020, at 1:29 PM,

Re: [R] Removing a space from a string

It is possible that there will be > 1 space. But, most likely only one (i.e., a solution for one space will suffice; a solution for > 1 space would be even better) Dennis Fisher MD P < (The "P Less Than" Company) Phone / Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com

Re: [R] Removing a space from a string

On 2020-07-28 21:31, Dennis Fisher wrote: Only the spaces in STRING. However, if I inadvertently delete the space between STRING and NN, I can add it back in. Can there only be one space in STR ING or is ST RI NG possible? Dennis Fisher MD P < (The "P Less Than" Company) Phone / Fax:

Re: [R] Removing a space from a string

On 2020-07-28 21:20, Dennis Fisher wrote: R 4.0.2 OS X Colleagues I have strings that contain a space in an unexpected location. The intended string is: “STRING 01. Remainder of the string" However, variants are: “STR ING 01. Remainder of the string" “STRIN G 01.

[R] Removing a space from a string

R 4.0.2 OS X Colleagues I have strings that contain a space in an unexpected location. The intended string is: “STRING 01. Remainder of the string" However, variants are: “STR ING 01. Remainder of the string" “STRIN G 01. Remainder of the string" I would like a

Re: [R] removing part of a string

Hello, Ulrik's way is simpler therefore better. I would use it, not mine. As for an explanation of mine, here it goes. 1) Parenthesis are meta-characters and must be escaped if you want to match them: \\( and \\) 2) You want to keep them so I used groups, i.e., put what you want between

Re: [R] removing part of a string

I would use sub("\\(.*\\)", "()", s) It is essentially the same as Rui's suggestion, but I find the purpose to be more clear. It might also be a little more efficient. HTH Ulrik On Mon, 21 May 2018, 15:38 Rui Barradas, wrote: > Hello, > > Try this. > > > ss1 <- "z:f(5,

Re: [R] removing part of a string

Hello, Try this. ss1 <- "z:f(5, a=3, b=4, c='1:4', d=2)" ss2 <- "f(5, a=3, b=4, c=\"1:4\", d=2)*z" fun <- function(s) sub("(\\().*(\\))", "\\1\\2", s) fun(ss1) #[1] "z:f()" fun(ss2) #[1] "f()*z" Hope this helps, Rui Barradas On 5/21/2018 2:33 PM, Vito M. R. Muggeo wrote: dear all, I am

[R] removing part of a string

dear all, I am stuck on the following problem. Give a string like ss1<- "z:f(5, a=3, b=4, c='1:4', d=2)" or ss2<- "f(5, a=3, b=4, c=\"1:4\", d=2)*z" I would like to remove all entries within parentheses.. Namely, I aim to obtain respectively "z:f()" or "f()*z" I played with sub() and

Re: [R] Removing columns from big.matrix which have only one value

help-boun...@r-project.org] On Behalf Of Jack Arnestad > Sent: Saturday, April 21, 2018 9:17 PM > To: r-help@r-project.org > Subject: [R] Removing columns from big.matrix which have only one value > > I have a very large binary matrix, stored as a big.matrix to conserve memory > (

Re: [R] Removing columns from big.matrix which have only one value

Maybe base R's unique() function might be useful? It uses hashing I believe. Bert On Sat, Apr 21, 2018, 12:17 PM Jack Arnestad wrote: > I have a very large binary matrix, stored as a big.matrix to conserve > memory (it is over 2 gb otherwise - 5 million columns and 100

[R] Removing columns from big.matrix which have only one value

I have a very large binary matrix, stored as a big.matrix to conserve memory (it is over 2 gb otherwise - 5 million columns and 100 rows). r <- 100 c <- 1 m4 <- matrix(sample(0:1,r*c, replace=TRUE),r,c) m4 <- cbind(m4, 1) m4 <- as.big.matrix(m4) I need to remove every column which has only

Re: [R] Removing a data subset

Thank Rainer and Jim!! The end result was: #Makes a new data set name "MyData_Minus_MW01" that contains all the data where the Location Column is not equal to MW01 and the comma after that ensures that all columns are copied into the amended data.frame. MyData_Minus_MW01 <- MyData[

Re: [R] Removing a data subset

Hi David, You "just" need to learn how to subset your data.frame, see functions like ?subset and ?"[", as well as a good guide to understand the subtleties! Some graphic functions also have a built-in argument to subset within the function (e.g. argument 'subset' in 'plot.formula'), although

Re: [R] Removing a data subset

Reading in the data from the file x <- read.csv( "ExampleData.csv", header = TRUE, stringsAsFactors = FALSE ) Subsetting as you want x <- x[ x$Location != "MW01", ] This selects all rows where the value in column 'Location' is not equal to "MW01". The comma after that ensures that

[R] Removing a data subset

Say I have a dataset that looks like LocationYear GW_Elv MW011999 546.63 MW021999 474.21 MW031999 471.94 MW041999466.80 MW012000545.90 MW022000546.10 The whole dataset is at

Re: [R] removing dropouts from survival analysis

Sorry. You may get private replies, but this *is* way OT on this list. Try stats.stackexchange.com instead for statistical queries. Or, better yet, find local consulting help. Non-random dropouts are a difficult issue. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people

[R] removing dropouts from survival analysis

Dear All. Apologies for posting a question regarding survival analysis, and not R, to the R-help list. In the past I received the best advices from the R community. The random censorship model (the censoring times independent of the failure times and vice versa) is one of the fundamental

Re: [R] removing all non-numeric characters from a string, but not "."

Thank you very much, gentlemen! On Tue, Jul 26, 2016 at 5:48 PM, peter dalgaard wrote: > >> On 26 Jul 2016, at 23:28 , Dimitri Liakhovitski >> wrote: >> >> Hello! >> >> I have a string x: >> x <- c("x - 84", "y - 293.04", "z = 12.5") >> >> I

Re: [R] removing all non-numeric characters from a string, but not "."

> On 26 Jul 2016, at 23:28 , Dimitri Liakhovitski > wrote: > > Hello! > > I have a string x: > x <- c("x - 84", "y - 293.04", "z = 12.5") > > I want to remove all the non-numeric stuff from it. The following works: > gsub("[^0-9]", "", x) > > However, it

Re: [R] removing all non-numeric characters from a string, but not "."

> On Jul 26, 2016, at 4:39 PM, Marc Schwartz wrote: > > >> On Jul 26, 2016, at 4:28 PM, Dimitri Liakhovitski >> wrote: >> >> Hello! >> >> I have a string x: >> x <- c("x - 84", "y - 293.04", "z = 12.5") >> >> I want to remove all the

Re: [R] removing all non-numeric characters from a string, but not "."

> On Jul 26, 2016, at 2:28 PM, Dimitri Liakhovitski > wrote: > > gsub("[^0-9]", "", x) ?regex I think you might be bit embarrassed because it seems pretty obvious once you know that character class elements like "." don't need to be escaped so it's just

Re: [R] removing all non-numeric characters from a string, but not "."

> On Jul 26, 2016, at 4:28 PM, Dimitri Liakhovitski > wrote: > > Hello! > > I have a string x: > x <- c("x - 84", "y - 293.04", "z = 12.5") > > I want to remove all the non-numeric stuff from it. The following works: > gsub("[^0-9]", "", x) > > However, it

[R] removing all non-numeric characters from a string, but not "."

Hello! I have a string x: x <- c("x - 84", "y - 293.04", "z = 12.5") I want to remove all the non-numeric stuff from it. The following works: gsub("[^0-9]", "", x) However, it strips my numbers of "." Help - how could I do the same but leave the "." in? Thanks a lot! -- Dimitri Liakhovitski

Re: [R] Removing NAs from dataframe (for use in Vioplot)

> Bert Gunter > on Mon, 2 May 2016 06:20:52 -0700 writes: > Martin et. al.: > na.omit(frame) will remove all rows/cases in which an NA occurs. I'm > not sure that this is what the OP wanted, which seemed to be to > separately remove NA's from

Re: [R] Removing NAs from dataframe (for use in Vioplot)

Martin et. al.: na.omit(frame) will remove all rows/cases in which an NA occurs. I'm not sure that this is what the OP wanted, which seemed to be to separately remove NA's from each column and plot the resulting column. This is what the lapply (and the OP's provided code) does, anyway. Also,

Re: [R] Removing NAs from dataframe (for use in Vioplot)

> Mike Smith > on Sun, 1 May 2016 08:15:44 +0100 writes: On Apr 30, 2016, at 12:58 PM, Mike Smith wrote: Hi First post and a relative R newbie I am using the vioplot library to produce some violin

Re: [R] Removing NAs from dataframe (for use in Vioplot)

> On May 1, 2016, at 12:15 AM, Mike Smith wrote: > On Apr 30, 2016, at 12:58 PM, Mike Smith wrote: > Hi > First post and a relative R newbie > I am using the vioplot library to produce some violin plots. > > DW> It's a package,

Re: [R] Removing NAs from dataframe (for use in Vioplot)

>>> On Apr 30, 2016, at 12:58 PM, Mike Smith wrote: >>> Hi >>> First post and a relative R newbie >>> I am using the vioplot library to produce some violin plots. DW> It's a package, not a library. >>> I have an input CSV with columns off irregular length that

Re: [R] Removing NAs from dataframe (for use in Vioplot)

But require() should not be used interchangeably with library()... the return value from require() should always be tested. -- Sent from my phone. Please excuse my brevity. On May 1, 2016 3:03:59 AM GMT+01:00, Tom Wright wrote: >Never let it be said there's only one way to

Re: [R] Removing NAs from dataframe (for use in Vioplot)

Never let it be said there's only one way to do a thing: require(ggplot2) require(dplyr) #create a sample dataset dat <- data.frame(y1=sample(c(1:10,NA),20,replace=TRUE), y2=sample(c(1:10,NA),20,replace=TRUE),

Re: [R] Removing NAs from dataframe (for use in Vioplot)

> On Apr 30, 2016, at 4:16 PM, David Winsemius wrote: > > >> On Apr 30, 2016, at 12:58 PM, Mike Smith wrote: >> >> Hi >> >> First post and a relative R newbie >> >> I am using the vioplot library to produce some violin plots. It's a package,

Re: [R] Removing NAs from dataframe (for use in Vioplot)

> On Apr 30, 2016, at 12:58 PM, Mike Smith wrote: > > Hi > > First post and a relative R newbie > > I am using the vioplot library to produce some violin plots. I have an input > CSV with columns off irregular length that contain NAs. I want to strip the > NAs out and

[R] Removing NAs from dataframe (for use in Vioplot)

Hi First post and a relative R newbie I am using the vioplot library to produce some violin plots. I have an input CSV with columns off irregular length that contain NAs. I want to strip the NAs out and produce a multiple violin plot automatically labelled using the headers. At the moment

Re: [R] removing factor values in the main data frame

> On 01 Mar 2016, at 18:33 , David L Carlson wrote: > > You need to learn how to send emails in plain text since html gets mangled on > r-help. See your message below. If I understand your question, it has to do > with what happens to factor levels when you subset your

Re: [R] removing factor values in the main data frame

Without more information it is hard to see where things go wrong. To judge from your text, I am not sure you do things correct. You are doing it along the lines of: str(mydata) submydata <- subset(mydata, ... ) submydata <- droplevels(submydata) str(submydata) right? On Tue, 1 Mar 2016 at

Re: [R] removing factor values in the main data frame

Hi, I think droplevels(df) is what you are looking for. Best wishes, Ulrik On Tue, 1 Mar 2016 at 17:33 hoda rahmati via R-help wrote: > Hi all,I have the following main data frame:(mydata)$ TE : num 40 > 40 20 20 20 20 20 20 20 40 ...$ TR : num 49 49 28

Re: [R] removing factor values in the main data frame

I also tried droplevels but after getting str(submydata) again I see no changes in COUNTRY  On Tuesday, March 1, 2016 8:38 AM, Ulrik Stervbo wrote: Hi, I think droplevels(df) is what you are looking for. Best wishes, Ulrik On Tue, 1 Mar 2016 at 17:33 hoda

[R] removing factor values in the main data frame

Hi all,I have the following main data frame:(mydata)        $ TE : num 40 40 20 20 20 20 20 20 20 40 ...        $ TR : num 49 49 28 28 28 28 28 28 28 49 ...        $ COUNTRY : Factor w/ 27 levels "","AU","BA","BE",..: 8 8 8 8 8 ...among the COUNTRY I just need US and AU,first I get a subset to

Re: [R] removing data based on date pairs in a separate data frame

If your start/end pairs are not overlapping you can use findInterval() to do this pretty quickly. E.g., isInABound <- function (x, low, high) { stopifnot(length(low) == length(high)) bounds <- rep(low, each = 2) bounds[seq(2, length(bounds), by = 2)] <- high

Re: [R] removing data based on date pairs in a separate data frame

What is the format of your date columns? -- character, factor, POSIXxx,... ?? See ?str to find out. (Reply to the list, not just me; others are far more facile at dates than I am). Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking

[R] removing data based on date pairs in a separate data frame

Dear R users, I have two data frames. The first contains a date/time column and the concentration of a species: head(mydata) datespecies 1 2016-01-31 23:59:53 -559.17 2 2016-02-01 00:00:53 -556.68 3

Re: [R] Removing a dollar sign from a character vector

The "end of string" special meaning only applies when the dollar sign is at the right end of the string (as it was in the OP attempt). That is, it is NOT generally necessary to wrap it in brackets to remove the special meaning unless it would otherwise be at the end of the pattern string. --

Re: [R] Removing a dollar sign from a character vector

I should have said that R-3.2.3 requires the $ to be backslashed even when it is not at the end of the pattern: > gsub("$[[:digit:]]*", "", c("$VAR", "$20/oz.")) [1] "$VAR""$20/oz." > gsub("\\$[[:digit:]]*", "", c("$VAR", "$20/oz.")) [1] "VAR" "/oz." Modern Linuxen's tools like sed

Re: [R] Removing a dollar sign from a character vector

In certain programs (not current R), a pattern with stuff after a naked dollar sign would not match anything because dollar meant end-of-string. In any case I prefer simple rules like 'backslash a dollar sign' instead of 'backslash a dollar sign at the end of the pattern but not elsewhere'.

[R] Removing a dollar sign from a character vector

What I’ve got: # sessionInfo() R version 3.2.3 (2015-12-10) Platform: x86_64-apple-darwin13.4.0 (64-bit) Running under: OS X 10.11.3 (El Capitan) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets

Re: [R] Removing a dollar sign from a character vector

> y [1] "$1,000.00 " "$1,000.00 " "$1,000.00 " "$2,600.00 " "$2,600.00 " > gsub("$", "", y) [1] "$1,000.00 " "$1,000.00 " "$1,000.00 " "$2,600.00 " "$2,600.00 “ # no change. Why? "$" as a regular expression means "end of string", which has zero length - replacing "end of string" with

Re: [R] Removing a dollar sign from a character vector

y <- as.numeric( gsub( "[$, ]", "", y ) ) -- Sent from my phone. Please excuse my brevity. On February 10, 2016 9:39:16 PM PST, James Plante wrote: >What I’ve got: ># sessionInfo() >R version 3.2.3 (2015-12-10) >Platform: x86_64-apple-darwin13.4.0 (64-bit) >Running under: OS X

Re: [R] removing outlier --> use robust regression !

> Juli > on Sat, 12 Sep 2015 02:32:39 -0700 writes: > Hi Jim, thank you for your help. :) > My point is, that there are outlier and I don´t really > know how to deal with that. > I need the dataframe for a regression and read often that

Re: [R] removing outlier

... and this, of course, is a nice example of how statistics contributes to the "irreproducibility crisis" now roiling Science. Cheers, Bert (Quote from a long ago engineering colleague: "Whenever I see an outlier, I never know whether to throw it away or patent it.") Bert Gunter "Data is not

Re: [R] removing outlier

If this mailing list accepted formatted submissions I would have used the trèsModernSarcastic font for my first sentence. Failing the availability of that mode of communication I am (top) posting through Nabble (perhaps) in "Comic Sans". On Sat, Sep 12, 2015 at 9:52 AM, David Winsemius

Re: [R] removing outlier

On Sep 12, 2015, at 2:32 AM, Juli wrote: > Hi Jim, > > thank you for your help. :) > > My point is, that there are outlier and I don´t really know how to deal with > that. > > I need the dataframe for a regression and read often that only a few outlier > can change your results very much.

Re: [R] removing outlier

Hi Jim, thank you for your help. :) My point is, that there are outlier and I don´t really know how to deal with that. I need the dataframe for a regression and read often that only a few outlier can change your results very much. In addition, regression diacnostics didn´t indcate me the best

Re: [R] removing outlier

Hi Juli, What you can do is to make your outlier remover into a function like this: remove_outlier_by_sd<-function(x,nsd=3) { meanx<-mean(x,na.rm=TRUE) sdx<-sd(x,na.rm=TRUE) return(x[abs(x-xmean) < nsd*sdx]) } Then apply the function to your data frame ("table")

[R] removing outlier

Hey, i want to remove outliers so I tried do do this: # 1 define mean and sd sd.AT_ZU_SPAET <- sd(AT_ZU_SPAET) mitt.AT_ZU_SPAET <- mean(AT_ZU_SPAET) # sd.Anzahl_BAF <- sd(Anzahl_BAF) mitt.Anzahl_BAF <- mean(Anzahl_BAF) # sd.Änderungsintervall <- sd(Änderungsintervall) mitt.Änderungsintervall <-

Re: [R] Removing display of R row names from list.

To: jrkrid...@inbox.com Subject: Re: [R] Removing display of R row names from list. Here is a small example. it is not the real data I am working on but this can help #Chick weight example names(ChickWeight) - tolower(names(ChickWeight)) chick_m - melt(ChickWeight, id=2:4, na.rm

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