similar to: Averaging within a range of values

Dear all, I have a matrix, called newdata1, > dim(newdata1) [1] 34176 83 It looks like: EntrezID Name S1 S2 S3 S4 S5..... 1 4076 CAPRIN1 0.1 0.2 0.3... 2 139170 WDR40B 0.4 0.5 0.6... 3 5505 PPP1R2P1 0.3 0.3 0.7... 4 4076 CAPRIN1 0.7 0.3 0.2... 5 139170 WDR40B null 0.8

Folks, The contour() function wants x and y to be in increasing order. I have a situation where I have a grid in x and y, and associated z values, which looks like this: x y z [1,] 0.00 20 1.000 [2,] 0.00 30 1.000 [3,] 0.00 40 1.000 [4,] 0.00 50 1.000 [5,] 0.00 60 1.000 [6,] 0.00 70 1.000 [7,] 0.00 80 0.000 [8,] 0.00 90

Hi, Use ?split() #dat1 is the dataset: lst1<- split(dat1,dat1$BaseProd) lst1 #$`2231` ?# BaseProd? CF OSA #1???? 2231 0.5 0.7 #2???? 2231 0.8 0.6 #3???? 2231 0.4 0.8 # #$`2232` ?# BaseProd CF OSA #4???? 2232? 1?? 2 #5???? 2232? 3?? 1 # #$`2233` ?# BaseProd? CF OSA #6???? 2233 0.9 0.5 #7???? 2233 0.7 0.5 #8???? 2233 4.0 5.0 #9???? 2233 5.0 7.0 lst1[[1]] #? BaseProd? CF OSA #1???? 2231 0.5 0.7

Dear R programmers, is there a sensible explanation for the following behaviour? The second command seems not to be interpreted correctly. > seq(0.6, 0.9, by=0.1) == 0.8 [1] FALSE FALSE TRUE FALSE > seq(0.7, 0.9, by=0.1) == 0.8 [1] FALSE FALSE FALSE > c(0.7, 0.8, 0.9) == 0.8 [1] FALSE TRUE FALSE > seq(0.9, 0.7, by=-0.1) == 0.8 [1] FALSE TRUE FALSE I am running R version 1.7.1 on

a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9) b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1) cor(a,b)= -1 a'=qbinom(a, 1, 0.5) b'=qbinom(b, 1, 0.5) why cor(a',b') becomes -0.5 ? -- View this message in context: http://r.789695.n4.nabble.com/qbinom-tp4651460.html Sent from the R help mailing list archive at Nabble.com.

Hi everyone, I have a question on selecting and grouping elements of a data frame. For example: A.df<- [ a c 0.9 b x 0.8 b z 0.5 c y 0.9 c x 0.7 c z 0.6] I want to create a list of a data frame that gives me the unique values of column 1 of A.df so that i can create intersects. That is: B[a]<- [ c 0.9] B[b]<- [ x 0.8

Dear all, I have a dataset as below: id code1 code2 p 1 4 8 0.1 1 5 7 0.9 2 1 8 0.4 2 6 2 0.2 2 4 3 0.6 3 5 6 0.7 3 7 5 0.9 I just want to rewrite it as this (vertical to horizontal): id var1 var2 var3

Hi, I have a Dataframe. A B C D 0.1 0.7 0.9 0.8 0.20 0.60 0.80 0.70 0.40 0.80 0.70 0.76 I need a resultant dataframe (A-B) (C-D) -0.6 0.1 -0.40 0.1 -0.40 -0.06 Any suggestion would be of a great help Thanks Ramya -- View this message in context: http://r.789695.n4.nabble.com/Finding-a-Diff-within-a-Dataframe-columns-tp3247943p3247943.html Sent from

Dear R forum I have a following vector of random no.s x = runif(100, 0.01, 0.99)  [1] 0.47212037 0.77867992 0.33947474 0.93369035   [5] 0.03720073 0.79307831 0.81801835 0.92710688 ................................................. I need to count the random no. falling in the range (0 - 0.10), (0.10 - 0.20), (0.20 - 0.30)..upto (0.90 - 1) Thus, I need to have a data frame as

Is the following developed in my console output a recognized bug or am I using the cat function incorrectly? Thanks, Stan > ifelse(class(data[[n]])!="factor",{print("yes")},{print("no")}) [1] "yes" [1] "yes" > ifelse(class(data[[n]])!="factor",{cat("yes")},{cat("no")}) yesError in ans[test & !nas] <-

hi i try to do a graph of a time series which shows in red the values > -0.05, in blue the values >0.05 and in white the values between -0.05 and 0.05 for un exemple :http://www.appinsys.com/globalwarming/enso.htm thanks !!!!! denisse -- View this message in context: http://r.789695.n4.nabble.com/how-to-do-a-graph-with-tree-different-colors-tp4635206.html Sent from the R help mailing

Can someone show me some code to do normalization by the median of some control genes for the example below? Many Many Thanks in advance This strategy selects a subset of genes (called ?control genes?) and makes the median of their data distribution similar across arrays. ??? ??? id1??? id2??? id3 control1??? 0.8??? 0.7??? 0.6 control2??? 0.6??? 0.2??? 0.4 probe1??? ??? 0.3??? 0.2??? 0.5

Dear R Users, Could you please help me on the following issue? I have a real large yearly data set. For each year I have 365 flow values. Some of the flow values are not known and that’s why you will see NA written in those slots. I wanted to know, is there a way that I can estimate those values? I tried approx command but it seems least helpful for the kind of issue I am up against.

Dear R Helpers, I am trying to extract the modulus from divisions by a sequence of fractions. I noticed that %% seems to behave inconsistently (to my untutored eye), thus: > 0.1%%0.1 [1] 0 > 0.2%%0.1 [1] 0 > 0.3%%0.1 [1] 0.1 > 0.4%%0.1 [1] 0 > 0.5%%0.1 [1] 0.1 > 0.6%%0.1 [1] 0.1 > 0.7%%0.1 [1] 0.1 > 0.8%%0.1 [1] 0 > 0.9%%0.1 The modulus for 0.1, 0.2, 0.4 and 0.8 is

Dear R People: I am getting some odd results when using logical operators: > x <- seq(from=-1,to=1,by=0.1) > > x [1] -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 [16] 0.5 0.6 0.7 0.8 0.9 1.0 > x == -1 [1] TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [13] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE > x

Hi, I'm looking the best way to fill irregular polygons with patterns, Something like the function grid.pattern do, but my case is with irregular polygons. Whit this script I can get it, but I'm looking for an "elegant" solution.. library(grid) grid.polygon(x=c(0.2, 0.8, 0.6, 0.6, 0.8, 0.2), y=c(0.2, 0.2, 0.3, 0.5, 0.7,0.7), gp=gpar(fill="grey",

Hello there, I am not quite sure how to interpret the output of Rprof (in the following the output I was staring at). I was poking around the web a little bit for documentation but without much success. I guess if I want to figure out what takes so long in my code the 2nd table $by.total and the total.pct column (pct = percent) is the most helpful. What does it mean that [ or [.data.frame is

I have a problem with R. I try to compute the confidence interval for my df. When I want to create the plot I have this problem: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ. I try this code: library(dplR) df.rwi <- detrend(rwl = df, method = "Spline",nyrs=NULL) write.table(df.rwi,file="rwi.txt",quote=FALSE,row.names=TRUE)

Dear all, I?m having some problems getting optim with method="Nelder-Mead" to work properly. It seems like there is no way of controlling the step size, and the step size seems to depend on the *difference* between the initial values, which makes no sense. Example: f=function(xy, mu1, mu2) { print(xy) dnorm(xy[1]-mu1)*dnorm(xy[2]-mu2) } f1=function(xy) -f(xy, 0,

Hello, i wrote this function guete and now i want to plot it: but i get this error message. i hope someone can help me. Error in dim(robj) <- c(dX, dY) : dims [product 16] do not match the length of object [1] p_11=seq(0,0.3,0.1) p_12=seq(0.1,0.4,0.1) guete = function(p_11,p_12) { set.seed(1000) S_vek=matrix(0,nrow=N,ncol=1) for(i in 1:N) { X_0=rmultinom(q-1,size=1,prob=p_0)