similar to: aggregate with cumsum

Displaying 20 results from an estimated 5000 matches similar to: "aggregate with cumsum"

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",
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
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
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,
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 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
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
2007 Apr 20
2
Fastest way to repeatedly subset a data frame?
Hi - I have a data frame with a large number of observations (62,000 rows, but only 2 columns - a character ID and a result list). Sample: > my.df <- data.frame(id=c("ID1", "ID2", "ID3"), result=1:3) > my.df id result 1 ID1 1 2 ID2 2 3 ID3 3 I have a list of ID vectors. This list will have anywhere from 100 to 1000 members, and
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
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
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
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>
2012 Feb 22
1
Lattice and horizontally stacked density plots
Hello, I am try to make a density plot where plots are stacked like the one found here: http://dsarkar.fhcrc.org/lattice/book/images/Figure_14_03_stdBW.png I am facing problems, however. Using the code example below, I'd like to generate a separate panel for each val of id2. Within each panel, I'd like to have individual histograms each on separate lines based on the value of id1. ?Note
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
2008 Sep 25
2
How to order some of my columns (not rows) alphabetically
Hello, I have a dataframe with 9 columns, and I would like to sort (order) the right-most eight of them alphabetiaclly, i.e.: ID1 ID2 F G A B C E D would become ID1 ID2 A B C D E F G Right now, I'm using this code: attach(data) data<-data.frame(ID1,ID2,data[,sort(colnames(data)[3:9])]) detach(data) but that's not very elegant. Ideally I could specify which columns to sort and
2009 Dec 10
1
problem with data processing in R
Hi, I'm stuck with parsing data into R for heatmap representation. The data looks like: 1 id1 x1 x2 x3 .... x20 2 id1 x1 x2 x3 .... x20 3 id1 x1 x2 x3 .... x20 4 id1 x1 x2 x3 .... x20 ......... 348 id2 x1 x2 x3 .... x20 349 id2 x1 x2 x3 .... x20 350 id2 x1 x2 x3 .... x20 351 id2 x1 x2 x3 .... x20 ......... The data is sorted for the IDs (id1,id2 .....id40) and I like to
2011 Apr 20
1
How to check if a value of a variable is in a list
Hi all, I am working with some social network analysis in R and ran into a problem I just cannot solve. Each observation in my data consists of a respondent, some characteristics and up to five friends. The problem is that all of these five friends might no show up later as a respondent (observation). Therefore I might not have characteristics on all the friends listed in the data and I want to
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
2008 Feb 07
1
Help with package reshape, wide to long
Hello, I am having difficulty figuring out how to use functions in the reshape package to perform a wide to long transformation I have a "wide" dataframe whose columns are like this example: id1 id2 subject treat height weight age id1 and id2 are unique for each row subject and treat are not unique for each row height, weight, and age are different types of measurements made on
2012 Jan 27
1
multiple column comparison
Hello, I have a very large content analysis project, which I've just begun to collect training data on. I have three coders, who are entering data on up to 95 measurements. Traditionally, I've used Excel to check coder agreement (e.g., percentage agreement), by lining up each coder's measurements side-by-side, creating a new column with the results using if statements. That is, if