similar to: Matching multiple columns in a data frame

Hi, I'm trying to match peaks between chromatographic runs. I'm able to match peaks when they are chromatographed with the same method, but not when there are different methods are used and spectra comes in to play. While searching I found the ALS package which should be usefull for my application, but I couldn't figure it out. I made some dummy chroms with R, which mimic my actual

Hello all, I have a matrix whose column names look like a1 a2 b1 b2 b3 c1 c2 1 2 3 7 1 3 2 4 6 7 8 1 4 3 Now, I can have any number of a's. not just two as shown above and same goes for b's and c's. I need to extract all the a's columns and put them in another matrix, extract all b's columns and put them in some matrix

Dear all, I have a data frame that looks like this: c1 c2 c3 A B C B C A A A B and so on; I?d like to produce one single vector consisting of the columns c1,c2, c3, such that vector=("A","B","A","B","C","A","C","A","B") I guess it?s easy to do but I don?t know how...Can anyone

Hi, I have a long matrix of the following form which I would like to subset according to the third column: [x y z]: a1 c1 1 a1 c1 2 a2 c1 1 a1 c2 1 a1 c2 2 . . . The first two columns a characters ai and cj. I would like to keep all the rows where there are two entries for z, 1 and 2. That is, I want: a1 c1 1 a1 c1 2 a1 c2 1 a1 c2 2 . . . I try to use something like

Hello, I have a sample data frame which looks like this day od month 1 1 0.1 2 2 3 #VALUE! 1 3 5 0.4 12 4 7 0.8 10 5 11 - 3 6 14 s 7 7 18 -- 12 8 27 19 7 Now i wish to filter all the non numeric values and replace it with "NA". The data frame

Just to repeat: you have NamesWide<-data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Curly")) and you want NamesLong<-data.frame(Names=c("Tom","Dick","Larry","Curly")) There must be something I am missing, because NamesLong <- data.frame(Names = c(NamesWide$Name1, NamesWide$Name2)) appears to

Hello , I am trying to create a histogram in order to compare between two groups and would like it to be similar to the figure attached. How can I generate this using R ? Thank you, Nandini http://r.789695.n4.nabble.com/file/n3582448/5634-15977-1-PB.gif -- View this message in context: http://r.789695.n4.nabble.com/Histogram-tp3582448p3582448.html Sent from the R help mailing list archive at

The data set consists of two sets of matrices, as labelled by the columns, T's and C's. > xy x T1 T2 T3 T4 T5 C1 C2 C3 C4 C5 [1,] 50 0.00 0.00 33.75 0.00 0.00 0.00 36.76 0.00 35.26 0.00 [2,] 13 34.41 0.00 0.00 36.64 32.86 34.11 35.80 37.74 0.00 0.00 [3,] 14 35.85 0.00 33.88 36.68 34.88 34.58 0.00 32.75 37.45 0.00 [4,] 33 34.56

These are supported on GK20A and GM107. Signed-off-by: Ilia Mirkin <imirkin at alum.mit.edu> --- Was a bit torn on where to place the enums... we're about to gut all the xml definitions so this seemed appropriate for now. Tested on GK20A only. src/gallium/drivers/nouveau/nv50/nv50_formats.c | 64 +++++++++++++++++++++++++ src/gallium/drivers/nouveau/nvc0/nvc0_screen.c | 10 ++++

Why not fix the new names instead to be like the old names? Seems like that would be way simpler... On Feb 15, 2016 12:38 AM, "Ben Skeggs" <skeggsb at gmail.com> wrote: > From: Ben Skeggs <bskeggs at redhat.com> > > We've previously had identical naming between vertex and texture > formats, so it mostly made sense to define these together. > > However,

I am trying to extract rows from a data.frame based on the presence/absence of a single value in any column. I've figured out how to do get the positive matches, but the remainder (rows without this value) eludes me. Mining the help pages and archives brought me, frustratingly, very close, as you'll see below. My goal: two data frames, one with -99 in at least one column in each row,

can list names attributes be preserved through S4 class containment? seems to be so but only if the containment relationship is direct ... see below. > setClass("c1", contains="list") [1] "c1" > l1 = list(a=1, b=2) > o1 = new("c1", l1) > names(o1) # pleasant surprise [1] "a" "b" > setClass("c2",

Hi, I observed below behavior with Instruction combiner (visitMul Canonicalization) It tries to convert "(X+C1)*C2" to "X*C2+C1*C2" While transforming if operation is guaranteed to not overflow it does not reflect same property to transformed instructions. Consider following scenarios: 1) If input is ((X+C1)*C2)<nsw> Then post canonicalization output should be

From: Ben Skeggs <bskeggs at redhat.com> Signed-off-by: Ben Skeggs <bskeggs at redhat.com> --- src/gallium/drivers/nouveau/nv50/g80_defs.xml.h | 279 ++++++++++++++++++++++++ 1 file changed, 279 insertions(+) create mode 100644 src/gallium/drivers/nouveau/nv50/g80_defs.xml.h diff --git a/src/gallium/drivers/nouveau/nv50/g80_defs.xml.h

hello gurus, i have a data frame like this HTN HTN_FDR Dyslipidemia CAD t1d_ptype[1:25] 1 Y Y Y T1D 2 T1D 3 Ctrl_FDR 4 T1D 5 Y Ctrl 6 Ctrl 7

Hi all, I have the following problem: I have a data frame (actually it is a prop.table) which I want to print as a list, e.g.: C1 C2 C3 R1 0.0 0.0 1.0 R2 1.0 0.0 0.0 R3 0.0 0.0 0.0 R4 0.0 1.0 0.0 should be printed like C1;R1;0.0 C2;R1;0.0 C3;R1;1.0 C1;R2;1.0 C2;R2;0.0 ..... Is there any existing solution out there or could somebody please give me a hint on how to

Hello R community, I am computing weighted average statistic by using ddply function: My data set is: N1 T1 S1 I1 C1 FY-4 ROE11 EPS11 MKT11 N1 T1 S1 I1 C1 FY-3 ROE12 EPS12 MKT12 N1 T1 S1 I1 C1 FY-2 ROE13 EPS13 MKT13 N1 T1 S1 I1 C1 FY-1 ROE14 EPS14 MKT14 N1 T1 S1 I1 C1 FY0 ROE15 EPS15 MKT15 N1 T1 S1 I1 C1 FY1 ROE16 EPS16 MKT16 N1 T1 S1 I1 C1 FY2 ROE17

Hi dear list, I'm scrambling with this: i have a p-dimensional data frame for which i want to have a pairs() plot, with ellipse added in each lower.panel. I want the ellipse's parameter (mu, sigma) to be extracted from the appropriate entries in a pre-computed matrix ("dab" in the code below). Can it be done ? Thanks in advance, melly<-function (x, y, col =

Dear R-users, For a data frame (say in this example X) I want to look up the corresponding value in a 'look-up data frame' (in this example Y). The for-loop works but is very time-consuming because 'X' in reality is very big. Therefore I would like to have a solution with apply. However, I do not succeed. Any suggestions? Thanks in advance, Hanneke

Thank you Jim Holtman and Mark Leeds for your help. Original question: >How do I do the following more concisely? > Bout[is.na(Bout$bd.n), 'bd.n'] <- 0 > Bout[is.na(Bout$ht.n), 'ht.n'] <- 0 > Bout[is.na(Bout$dbh.n), 'dbh.n'] <- 0 >. . . Solution: for (i in c('bd.n', 'ht.n', 'dbh.n')) Bout[is.na(Bout[[i]]), i] <- 0 Toy