similar to: Replacing for loop with tapply!?

Displaying 20 results from an estimated 400 matches similar to: "Replacing for loop with tapply!?"

2009 Nov 13
3
sum(row1==y) if row2=x
Hi to all is there any construct to sum data=data.frame(row1=c(1,1,3,1,2,3,2,2,1,3,4,5,2,3,2,1) , row2=c(2,2,1,1,1,2,1,2,1,1,1,1,2,2,2,1) ) Means I would like to get all y of row1 if in row2 of the data.frame is an x f.e row1=3 and row2=2 so I would like to get 6 And is there another construct to get the count of pairs where row1=3 and row2=2 means the result should be
2008 Dec 16
2
converting a data-frame by a defined rule
Hi, I have a data frame with several columns. Now I want to transfer the data into a new variable (also a data frame), but I only want a part of the data, defined by a rule ... for example; I have following data frame: row1 row2 row3 x 2 3 x 1 4 y 5 3 y 2 3 I know want a data frame, only with lines containing x in row1. I know how to do that for one row (f <-
2008 Dec 16
1
refer to next line within a data-frame an select cases
Hi, I have a problem sorting and selecting entries within a data-frame and I don't know if it is possible to solve it with R ... (probably yes, but I have no idea how). Following Data; row1 row2 a 12 pos NA a 3 neg NA a 5 neg NA a 11 pos NA I want to extract the values in row 2 in the lines with an "a" in row1. But I want to have two vectors: vector x with all
2010 May 11
1
create a data.frame for aov
Hi R-experts, I try to find a way to transfer a matrix to a data.frame that is used as input of aov. can you give me advice for that? >mdat <- matrix(c(1,2,3, 11,12,13), nrow = 2, ncol=3, byrow=TRUE, dimnames = list(c("row1", "row2"), c("Col1", "Col2", "Col3"))) >mdat      Col1 Col2 Col3 row1    1    2    3 row2   11   12   13 ===>
2010 Sep 10
4
for loop help please!
Hi Everyone, I have a 2-dim data.matrix(e.g., table1) in which row1 specifies a range of values. row2 - rown specify the number of times I want to replicate each corresponding value in row1. I can do this with the following function: rep(c(table1[1,]),c(table1[X,])) #where X would go from 2 - n. Now, I can do this manually by changing the values of X and save each resulting array/vector in
2005 Nov 28
7
combine two columns
Hi, I have an R programming problem and I havent found anything in the documentation yet: I have a data matrix, in which two neighbouring columns represent replicates of the same experiment, e.g. something like this: A A B B C C row1 1 1 1 2 2 2 row2 1 1 1 1 1 2 I would like to test, if the values for the two replicates in a row are the same or if they differ and generate a new
2009 Jan 17
2
data.frame: how to extract parts
Hi, I have a problem with the R syntax. It's perhaps pretty simple, but I don't understand it ... I can extract a column from a data.frame with the following code for example ... b$row1[b$row1 == "male"] so I see all male-entries. But I cannot extract all lines of a data.frame depending on this criterium; only.male <- b[b$row1 == "male"] With that, I
2012 Aug 03
3
embedding data frame in R code?
I would like to insert a few modest size data frames directly into my R code. a short illustration example of what I want is d <- read.csv( _END_, row.names=1 ) , "col1", "col2" "row1",1,2 "row2",3,4 __END__ right now, the data sits in external files. I could put each column into its own vector and then combine into a data frame, but this seems
2009 Aug 27
1
generating multiple sequences in subsets of data
I'm running into a problem I can't seem to find a solution for. I'm attempting to add sequences into an existing data set based on subsets of the data. I've done this using a for loop with a small subset of data, but attempting the same process using real data (200k rows) is taking way too long. Here is some sample data and my ultimate goal >
2018 Nov 21
3
Subsetting row in single column matrix drops names in resulting vector
Hi Rui. Thanks for answer, I'm aware of drop = FALSE option. Unfortunately it doesn't resolve the issue - I'm expecting to get a vector, not a matrix . ??, 21 ????. 2018 ?. ? 20:54, Rui Barradas <ruipbarradas at sapo.pt>: > Hello, > > Use drop = FALSE. > > a[1, , drop = FALSE] > # col1 > #row1 1 > > > Hope this helps, > > Rui Barradas
2010 Mar 30
1
Adding RcppFrame to RcppResultSet causes segmentation fault
Hi, I'm a bit puzzled. I uses exactly the same code in RcppExamples package to try adding RcppFrame object to RcppResultSet. When running it gives me segmentation fault problem. I'm using gcc 4.1.2 on redhat 64bit. I'm not sure if this is the cause of the problem. Any advice would be greatly appreciated. Thank you. Rob. int numCol=4; std::vector<std::string>
2006 May 12
7
RJS and page.select collection size
I need to implement a conditional in my RJS template which looks something like: if (page.select(''row1'').first != null) page << "new TableRow.MoveAfter(''row1'', ''newrow'');" else page << "new TableRow.MoveAfter(''row2'', ''newrow'');" end Now, dumb question.. My
2010 May 31
2
accessing a data frame with row names
Readers, I have entered a file into r: ,column1,column2 row1,0.1,0.2 row2,0.3,0.4 using the command: dataframe<-read.table("/path/to/file.csv",header=T,row.names=1) When I try the command: dataframe[,2] I receive the response: NULL I was expecting: row1 0.2 row2 0.4 What is my error with the syntax please? Yours, r251 mandriva2009
2009 Jun 13
3
How Can I insert another column data into the CSV file when I use FasterCSV?
Hi, All, Suppose I have a CSV file, there is data in it. * Column 1 Column2 Column 3 Column 4 Row1 a b c Row2 a2 b2 c2* You know, the column 4 is no data Now, I would like to insert data to Column 4, after save, the CSV file will be: * Column 1
2010 Sep 06
1
Creating named.list from two matrix columns
Hi Friends, I am new to R. On R utility class pages, creating "named.list" is described with this command : new("named.list",a=1,b=2) For large matrix having two columns, such as : "row1" 2334 "row2" 347 "row3" 379 ... I want to create a named.list like : $row1 [1] 2334 $row2 [1] 347 ... Can anyone explain how "named.list"
2012 Feb 06
2
dividing values of each column in a dataframe
Hey Guys I want to divide(numerically) all the columns of a data frame by different numbers. Here is what I am doing but getting a weird error. The values in each column are not getting divided by the corresponding value in the denominator vector instead by the alternative values. I am sure I am messing up something. Appreciate your help -Abhi head(counts) WT_CON WT_RB MU_CON
2018 Nov 21
2
Subsetting row in single column matrix drops names in resulting vector
Hello here. I'm struggling to understand R's subsetting behavior in couple of edge cases - subsetting row in a single column matrix and subsetting column in a single row matrix. I've read R's docs several times and haven't found answer. Consider following example: a = matrix(1:2, nrow = 2, dimnames = list(c("row1", "row2"), c("col1"))) a[1, ] # 1
2018 Nov 27
1
Subsetting row in single column matrix drops names in resulting vector
Dmitriy Selivanov (selivanov.dmitriy at gmail.com) wrote: > Consider following example: > > a = matrix(1:2, nrow = 2, dimnames = list(c("row1", "row2"), c("col1"))) > a[1, ] > # 1 > > It returns *unnamed* vector `1` where I would expect named vector. In fact > it returns named vector when number of columns is > 1. > Same issue applicable
2004 Oct 25
2
Reading sections of data files based on pattern matching
I am about to write general functions to read the output of simulations models. These model generate output files with different sections which I want to analyze plot etc. Since this will be used many people at the department I wanted to make sure that will do this in the best way. For instance I want to read a snippets of data from a text that look like this.
2009 Mar 05
3
Dropping rows conditionally
Dear R-help team, I am getting addicted to using R but keep on getting many challenges on the way especially on data management (data cleaning). I have been wanting to drop all the rows if there values are `NA' or have specific values like 1 or 2 or 3. mdat <- matrix(1:21, nrow = 7, ncol=3, byrow=TRUE, dimnames = list(c("row1",