similar to: Fastest way to repeatedly subset a data frame?

Displaying 20 results from an estimated 1000 matches similar to: "Fastest way to repeatedly subset a data frame?"

2005 Jul 11
2
building packages on Windows
Hi, all, I just recently upgraded my computer though I'm using the same OS (XP). But now I'm having difficulty building packages and I cannot seem to solve the problem. I'm using R-2.1.1pat on Windows XP. Here is what I tried: D:\Users\sundard\slib\sundar\R>R CMD CHECK sundar * checking for working latex ... OK * using log directory
2012 Feb 21
5
[LLVMdev] buildbot failure in LLVM on clang-x86_64-debian-fnt
All, This buildbot is getting lots of assertion failures in the test suite. They were probably caused by my commit: ------------------------------------------------------------------------ r151049 | foad | 2012-02-21 09:25:52 +0000 (Tue, 21 Feb 2012) | 6 lines Changed paths: M /llvm/trunk/lib/VMCore/LLVMContextImpl.h M /llvm/trunk/lib/VMCore/Type.cpp PR1210: make uniquing of struct and
2004 Apr 30
2
festival and gcc 3.3.2 (Fedora Core 1)
Can someone tell me how to build festival on a machine with gcc 3.3.2? I've searched all around and even found a reference or two that the problem exists but I'm not seeing the fix. thanks! -reed Symtoms are -- ./configure, then.... [root@telephone speech_tools]# make Check system type Remake modincludes.inc NATIVE_AUDIO ok EDITLINE
2006 Sep 13
3
group bunch of lines in a data.frame, an additional requirement
Thanks for pointing me out "aggregate", that works fine! There is one complication though: I have mixed types (numerical and character), So the matrix is of the form: A 1.0 200 ID1 A 3.0 800 ID1 A 2.0 200 ID1 B 0.5 20 ID2 B 0.9 50 ID2 C 5.0 70 ID1 One letter always has the same ID but one ID can be shared by many letters (like ID1) I just want to keep track of the ID, and get
2012 Apr 14
3
Choose between duplicated rows
Dear r experts, Sorry for this basic question, but I can't seem to find a solution? I have this data frame: df <- data.frame(id = c("id1", "id1", "id1", "id2", "id2", "id2"), A = c(11905, 11907, 11907, 11829, 11829, 11829), v1 = c(NA, 3, NA,1,2,NA), v2 = c(NA,2,NA, 2, NA,NA), v3 = c(NA,1,NA,1,NA,NA), v4 = c("N",
2011 Apr 25
2
Problem with ddply in the plyr-package: surprising output of a date-column
Hi Together, I have a problem with the plyr package - more precisely with the ddply function - and would be very grateful for any help. I hope the example here is precise enough for someone to identify the problem. Basically, in this step I want to identify observations that are identical in terms of certain identifiers (ID1, ID2, ID3) and just want to save those observations (in this step,
2008 Jul 09
2
Parsing
Dear R users, I have a big text file formatted like this: x x_string y y_string id1 id1_string id2 id2_string z z_string w w_string stuff stuff stuff stuff stuff stuff stuff stuff stuff // x x_string1 y y_string1 z z_string1 w w_string1 stuff stuff stuff stuff stuff stuff stuff stuff stuff // x x_string2 y y_string2 id1
2006 Jan 14
3
In place editing and external control
Dear all, First I''d like to thank authors for so nice Scriptaculous and Prototype libraries, which helped me already a lot! I have question regarding externalControl parameter in InPlaceEditor. If I understand correctly, I can use that to have one image as a trigger to enter edit mode? I tried with below code but without success: <span id="id1">My text</span>
2005 Nov 09
3
dataframe without repetition
Hello, with a data.frame like this : > toto <- data.frame(id=c("id1","id1","id2","id3","id3","id3"),dpt=c("13","13","34","30","30","30")) > toto id dpt 1 id1 13 2 id1 13 3 id2 34 4 id3 30 5 id3 30 6 id3 30 what is the most efficient ways to obtain : id
2011 Aug 02
2
Data frame to matrix - revisited
Hi, I've tried to look through all the previous related Threads/posts but can't find a solution to what's probably a simple question. ? I have a data frame comprised of three columns e.g.: ? ID1?ID2?Value a?b?1 b?d?1 c?a?2 c?e?1 d?a?1 e?d?2 ? I'd like to convert the data to a matrix i.e.: ? ?a b c d e a n/a 1 2 1 n/a b 1 n/a n/a 1 n/a? c 2 n/a n/a n/a 1 d 1 1 n/a n/a 2 e n/a n/a 1
2010 Jun 03
2
deduplication
Colleagues, I am trying to de-duplicate a large (long) database (approx 1mil records) of diagnostic tests. Individuals in the database can have up-to 25 observations, but most will have only one. IDs for de-duplication (names, sex, lab number...) are patchy. In a first step, I am using Andreas Borg's excellent record linkage package (), that leaves me with a list of 'pairs' looking
2010 Sep 07
1
average columns of data frame corresponding to replicates
Hi Group, I have a data frame below. Within this data frame there are samples (columns) that are measured more than once. Samples are indicated by "idx". So "id1" is present in columns 1, 3, and 5. Not every id is repeated. I would like to create a new data frame so that the repeated ids are averaged. For example, in the new data frame, columns 1, 3, and 5 of the original
2006 Jan 24
9
Number of replications of a term
Hello, Is there a simple and fast function that returns a vector of the number of replications for each object of a vector ? For example : I have a vector of IDs : ids <- c( "ID1", "ID2", "ID2", "ID3", "ID3","ID3", "ID5") I want the function returns the following vector where each term is the number of replicates for the
2013 Apr 12
1
Removing rows that are duplicates but column values are in reversed order
Hi, From your example data, dat1<- read.table(text=" id1?? id2?? value a????? b?????? 10 c????? d??????? 11 b???? a???????? 10 c????? e???????? 12 ",sep="",header=TRUE,stringsAsFactors=FALSE) #it is easier to get the output you wanted dat1[!duplicated(dat1$value),] #? id1 id2 value #1?? a?? b??? 10 #2?? c?? d??? 11 #4?? c?? e??? 12 But, if you have cases like the one
2010 Oct 12
5
aggregate with cumsum
Hello everybody, Data is myd <- data.frame(id1=rep(c("a","b","c"),each=3),id2=rep(1:3,3),val=rnorm(9)) I want to get a cumulative sum over each of id1. trying aggregate does not work myd$pcum <- aggregate(myd[,c("val")],list(orig=myd$id1),cumsum) Please suggest a solution. In real the dataframe is huge so looping with for and subsetting is not a
2008 Jan 10
1
data.frame manipulation: Unbinding strings in a row
Hi all, I have a data.frame I received with data that look like this (comma separated strings in last row): ID Shop Items ID1 A1 item1, item2, item3 ID2 A2 item4, item5 ID3 A1 item1, item3, item4 But I would like to unbind the strings in col(2) items so that it will look like this: ID Shop Items ID1 A1 item1 ID1 A1 item2 ID1 A1 item3 ID2 A2 item4 ID2 A2 item5 ID3 A1 item1 ID3 A1 item3 ID3 A1
2005 Aug 10
1
Why only a "" string for heading for row.names with write.csv with a matrix?
Consider: > x <- matrix(1:6, 2,3) > rownames(x) <- c("ID1", "ID2") > colnames(x) <- c("Attr1", "Attr2", "Attr3") > x Attr1 Attr2 Attr3 ID1 1 3 5 ID2 2 4 6 > write.csv(x,file="x.csv") "","Attr1","Attr2","Attr3" "ID1",1,3,5
2011 May 25
1
Subtracting rows by id
Dear R users, I have two datasets: id1 <- c(rep(1,10), rep(2,10), rep(3,10)) value1 <- sample(1:100, 30, replace=TRUE) dataset1 <- cbind(id1,value1) id2 <- c(1,2,3) subtract.value <- c(1,3,5) dataset2 <- cbind(id2, subtract.value) I want to subtract the number of rows in the subtract.value that corresponds to the id value in dataset1. So for the 1 in id1, I want to
2012 Sep 15
5
create new variable with ifelse? (reproducible example)
Dear R users, I have a reproducible data and try to create new variable "clo" is 1 if know variable is equal to "very well" or "fairly well" and getalong is 4 or 5 otherwise it is 0. rep_data<- read.table(header=TRUE, text=" id1 id2 know getalong 100000016_a1 100000016_a2 very well 4 100000035_a1 100000035_a2 fairly
2006 Feb 09
1
List Conversion
Hello, I have a list (mode and class are list) in R that is many elements long and of the form: >length(list) [1] 5778 >list[1:4] $ID1 [1] "num1" $ID2 [1] "num2" "num3" $ID3 [1] "num4" $ID4 [1] NA I'd like to convert the $ID2 value to be in one element rather than in two.?? It shows up as c(\"num2\", \"num3\") if I try to use