search for: l_ply

Dear All, to produce output of several columns of a data frame, I tried to use lapply and also l_ply. In both cases, I would like to print a header line containing also the name of the respective column in the data frame. For example, I would like the following lapply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x)))) to produce: [1] "a" [1] "b" and not,...

...al order that I would like. How can I fix this? Thank you! ************************* Code: A_var_df <- data.frame(index=1:length(seq(1.0, -0.9, by= -0.1)), from=seq(1.0, -0.9, by= -0.1), to=seq(0.9, -1.0, by= -0.1)) for(i in 1:length(A_var_df[,1])) { library(plyr) max.col <- ncol(Dchr1) l_ply(seq(1, max.col, by=3573), function(k) write.table(as.data.frame(Dchr1[,k:min(k+3572, max.col)]), paste("./A_", A_var_df[i,1], "/k.csv", sep=""), sep=",", row.names=F, quote=F) ) } ************************** -- Thanks, CC [[alternative HTML version delete...

I am estimating a large model by groups. How do you save the results and?returns the associated quantiles? For this example I need a data frame n?? ?xi??????? mu????????beta 1?? 0.1033614? 2.5389580 0.9092611 2? ?0.3401922? 0.5192882 1.5290615 3?? 0.5130798? 0.5668308 1.2105666 I also want to apply gevrlevelPlot() for each "n" or group. ? #Example n <- c(1, 1, 1, 1, 1, 1, 2, 2, 2,

Dear List, I have a wrapper function that draws a graph that I'd like to use in a vector-like manner. The for-loop version I currently use is below. library(ggplot2) data(economics) h <- 600 w <- 800 #---------------------------------------------------------- draw_metric_by_date <- function( df, i, smooth=FALSE, BASEPATH ) { mlabel <- names(df)[i] qmetric

...rding to the type of object they input (first letter) and output (second letter): * llply = from a list to a list * alply = from an array (or vector, or matrix) to a list * ldply = from a list to a data.frame * d_ply = from a data.frame, ignore output * and so on for llply, laply, ldply, l_ply, alply, aaply, adply, a_ply, dlply, daply, dply, d_ply plyr also provides: * m*ply which works in a similar way to mapply * r*ply which works in a similar way to replicate You can find out more at http://had.co.nz/plyr/, including a 20 page introductory guide, http://had.co.nz/plyr/plyr-intr...

...rding to the type of object they input (first letter) and output (second letter): * llply = from a list to a list * alply = from an array (or vector, or matrix) to a list * ldply = from a list to a data.frame * d_ply = from a data.frame, ignore output * and so on for llply, laply, ldply, l_ply, alply, aaply, adply, a_ply, dlply, daply, dply, d_ply plyr also provides: * m*ply which works in a similar way to mapply * r*ply which works in a similar way to replicate You can find out more at http://had.co.nz/plyr/, including a 20 page introductory guide, http://had.co.nz/plyr/plyr-intr...

There are a few version of apply() (e.g., lapply(), sapply()). I'm wondering if there is one that does not return anything but just silently apply a function to the list argument. For example, the plot function is applied to each element in 'alist'. It is redundant to return anything from apply. apply(alist,function(x){ plot each element of alist})

Hi, Working on a report that is going to have a large number of graphs and summaries. We have 80 "groups" with 20 variables each. Ideally, I'd like to produce ONE page for each group. It would have two columns of 10 graphs and then the 5 number summary of the variables at the bottom. So, perhaps the top 2/3 of the page has the graphs and the bottom third has 20 rows of data

I have a simple 1 way anova coded like summary(ANOVA1way <- aov(Value ~ WellID, data = welldata)) How can I use the BY function to do this ANOVA for each group using another variable in the dataset?? I tried coding it like this, but it doesn't seem to work. summary(ANOVA1way <- by(welldata, Analyte, function(x) aov(Value ~ WellID, data = welldata))) In SAS I would code it like this:

I'm running a long MCMC chain that is generating samples for 22 variables. I have each run of the chain as a row in a matrix. So: Chain[,1] is the column with all the samples for variable one. Chain[,2] is the column with all the samples for variable 2, etc. I'd like to fit all 22 on a single page to print a nice summary. It is OK if the graphs are small, I just need to show the

Hi, I have been examining large data and need to do simple linear regression with the data which is grouped based on the values of a particular attribute. For instance, consider three columns : ID, x, y, and I need to regress x on y for each distinct value of ID. Specifically, for the set of data corresponding to each of the 4 values of ID (76,111,121,168) in the below data, I should invoke