Loops, dataframes and ggplot 
Below you find an alternative: using the melt function from reshape2
and then faceting with facet_wrap.
require(ggplot2)
require(reshape2)
data.melt < melt(rawdata1, id.var='Period')
ggplot(data.melt, aes(Period, value)) +
geom_line() +
facet_wrap(~variable, scales='free_y')
If you want to use multiplot instead, you could do the following:
out < lapply(names(rawdata1)[1],

R install package RevoScaleR 
I am not sure what R version you are currently using but the latest
stable version is 3.1.2. If the package installation warning is
telling that the package you're trying to install is not available,
that basically means that the package developer has not yet made a
version of the package that is compatible with the R version you're
currently running.
If you really need to use this package (if y

Simple lapply task 
You can use sapply to get a vector
> sapply(dir, function(x) paste("year", x, sep="_"))
[1] "year_2013" "year_2012" "year_2011"
Use lapply to get a list
> lapply(dir, function(x) paste("year", x, sep="_"))
[[1]]
[1] "year_2013"
[[2]]
[1] "year_2012"
[[3]]
[1] "year_2011"

R, data.table, group by column *numbers* AND sum a column 
Here's how I would approach this:
f < function(columns) {
Get < if (!is.numeric(columns)) match(columns, names(DT)) else
columns
columns < names(DT)[Get]
DT[, sum(N), by = columns]
}
The first line (Get..) keeps "columns" as numeric if it's already
numeric or it converts it from characters to numeric if they are not.
Test it out with some sample data:
set.seed(1)
DT <

R: Creating colorized barplot from segments() 
You can use rect for this:
lapply(ind, function(x) {
plot(unlist(x[, c(3, 5)]), unlist(x[, c(4, 6)]), type='n',
xlab='Time [Years]', ylab='Value [mm]', main=x[1, 1])
apply(x, 1, function(y) {
rect(y[3], min(y[4], 0), y[5], max(y[4], 0),
col=if(as.numeric(y[4]) < 0) 'red' else 'blue')
abline(h=0)
})
})

How do I create 3 or four classes of diameter from a range of diameters in R? 
As @Vincent Guillemot notes, all you need is ?cut.
my.data < read.table(text="LOGID DIAMETER
1 10
2 30
3 20
4 10
5 8
6 5", header=TRUE)
my.data$CLASS < cut(my.data$DIAMETER, breaks=c(Inf, 10, 20, 30,
Inf),
labels=c(1,2,3,4))
my.data
# LOGID DIAMETER CLASS
# 1 1 10 1
# 2 2 30 3
# 3 3

How to make this graph in R 
this chart is called a 'Sankey diagram'. I don't really know how to do
in R but I am using d3 to create this at work. An example here:
http://bost.ocks.org/mike/sankey/

RStudio Shiny dynamic selectize 
You need to add the css dynamically also. To target the selectize
input you need to target select#dataset + .selectizecontrol rather
then select#dataset
library(shiny)
runApp(list(
ui = bootstrapPage(
radioButtons("viz_multiple", "Select Type:",
c("Select From List (can use Up/Down + Enter)" =
"multiple",
"Search One (Delete then type keyword)" = "sing

strptime not recognizing %b/%B? 
I am a bit confused by R's behaviour in the case of missing values in
strptime. The documentation is clear enough:
For strptime the input string need not specify the date completely:
it
is assumed that unspecified seconds, minutes or hours are zero, and
an
unspecified year, month or day is the current one. Some components
may
be returned as NA (but an unknown tzone component is represent

constrained nonlinear minimization with many variables 
Try this:
co < coef(lm.fit(ui, ci))
co[is.na(co)] < 0
res < nloptr( x0=co, eval_f=objective, eval_g_eq = constraints,
opts=list(algorithm = "NLOPT_LD_SLSQP"))
giving:
> res
Call:
nloptr(x0 = co, eval_f = objective, eval_g_eq = constraints,
opts = list(algorithm = "NLOPT_LD_SLSQP"))
Minimization using NLopt version 2.4.0
NLopt solver status: 4 ( NLOPT_XTOL_REACHED: O

removing elements of the environment using a loop 
for(i in 1:16) {
rm(list=paste0('Factor',i))
}
although rm(list=paste0('Factor',1:16)) or
rm(list=ls(pattern="Factor"))would be more appropriate...

Optimize the performance of a function 
A couple thoughts: If you have a gigundo dataset, well, it's going to
take time; or you'll need to learn to use the parallel package.
I don't think you want to redefine newdata$ALTMISTA every time thru
the loop, as you're just overwriting the values.
You could remove the i loop by using a vectorized operation with
ifelse :
set.seed(1)
foo<sample(c(1,1),10,rep=T)
foo
[1] 1 1 1

density histogram by intervals in R 
If I understand you right, you have prebinned data:
interval count
[ 0, 10) 1
[ 10, 30) 15
[ 30, 200) 44
[200, 400) 40
total 100
Your code does not work, because hist tries to the bin the values in x
by itself. 1 goes to the first interval, 15 to the second and 44 and
40 to the third one.
I don't know how to do this with hist. With ggplot2, you can u

How to give a "/" in a column name to a dataframe in R? 
You can use check.names=FALSE in the data.frame. By default, it is
TRUE. And when it is TRUE, the function make.names changes the
colnames. ie.
make.names('Cost/Day')
#[1] "Cost.Day"
So, try
dat < data.frame("Cost/Day"=1:10,"Days"=11:20,
check.names=FALSE)
head(dat,2)
# Cost/Day Days
#1 1 11
#2 2 12
The specific lines in data.frame function changing the

Define a number line in R 
I guess you mean (1)^n 1/n ? Because n 1/n would just give you an
infimum of infinity.
Although for my answer it does not matter. To make a line up to given
n.lim, I would do:
nlim<10000
n < 1:nlim
myline < (1)^n1/n
Which gives for the first 15 elements:
myline[1:15]
[1] 2.0000000 0.5000000 1.3333333 0.7500000 1.2000000 0.8333333
[7] 1.1428571 0.8750000

Tabular data to matrix in R 
I would suggest to change from letters to numbers (which is straight
forward using the ASCII codes) and then use the linearized indices of
R matrices to access each pair in a vectorwise manner.
Minimal example:
N < 3
d < data.frame(x = c(1,2), y = c(2,3), v = c(0.1, 0.2))
m < matrix(0, N, N)
m[(d$y1)*N+d$x] = d$v
The output is:
[,1] [,2] [,3]
[1,] 0 0.1 0.0
[2,] 0

List all onedigit differences from a vector 
You could try the following:
a < c(0, 0, 0, 0, 1) # your input vector
m < expand.grid(rep(list(0:1), length(a)) # all combinations of
0/1 of length a
temp < sapply(seq_along(a), function(i) m[,i] == a[i]) # check
the differences
m[rowSums(temp) == (length(a)1),] # use the index to subset
# Var1 Var2 Var3 Var4 Var5
#1 0 0 0 0 0
#18 1 0 0 0 1
#19

Including Rnw files within a package 
To answer this question:
Include the .Rnw files in ./pkgname/inst/latex then when you build the
package, the ./latex folder will go to the root level of the package.
You can then extract the .Rnw files using system.file("latex",
"mytemplate.Rnw", package = "pkgname").

ggplot2: How to plot an orthogonal regression line? 
Caveat: not familiar with this method
I think you should be able to just pass the slope and intercept to
geom_abline to produce the fitted line. Alternatively, you could
define your own method to pass to stat_smooth (as shown at the link
smooth.Pspline wrapper for stat_smooth (in ggplot2)). I used the
Deming function from the MethComp package as suggested at link How to
calculate Total least squa

Spatial interpolation in R gstat without the message 
Various ways exist of stopping output  the nicest being if the
function has an option to suppress it. But krige doesn't seem to have
that.
capture.output works here:
> rm(zn.tr1)
> zn.tr1 # there is no zn.tr1
Error: object 'zn.tr1' not found
> z = capture.output(zn.tr1 < krige(log(zinc) ~ x + y , meuse,
meuse.grid))
> str(zn.tr1) # there is now
Formal class 'SpatialPixelsDataFr

How to draw a graph for erdos.renyi.game method in R 
For a line plot you need to use:
plot(d,type="l")
The default for the plot() function is the dotplot.
If all you want is to get a graph of the degree distribution, you can
use hist(degree(graph)). The function degree.distribution() gives you
a vector of the relative frequencies of degrees. You can get a
histogram showing the distribution of degree values by calling
degree() and feeding it int

Sorting by a slice of the text string 
Assuming chrPos is in ColumnA, please try in a helper column:
=IF(FIND("_",A1)=5,CHAR(64+MID(A1,4,1)),CHAR(64+MID(A1,4,2)))&REPT("0",8LEN(A1)+FIND("_",A1))&MID(A1,FIND("_",A1)+1,8)
OR, for additional requirements as mentioned in comments:
=IF(MID(A1,4,1)="M","W",IF(MID(A1,4,1)="X","X",IF(MID(A1,4,1)="Y","Y",IF(FIND("_",A1)=5,CHAR(64+MID(A1,4,1)),CHAR(64+MID(A1,4,2))))))&REPT(

igraph plot graph with vertex size dependent on betweenness centrality 
Using igraph, and following the example in ?betweenness:
g < random.graph.game(10, 3/10)
plot(g, vertex.size=betweenness(g))
(note numbers are node numbers, not betweenness value)
You may want to rescale your vertex size if you have lots of large
values or otherwise improve the visualisation.
g = graph.lattice(c(10,4))
plot(g,vertex.size=betweenness(g)/10)
without the /10 the vertic

Error in executing knit HTML in R Studio 
When you run "Knit HTML", the code is trying to find the file you're
reading in the same directory where .Rmd is located. As far as I see
you have two options.
Try specifying the absolute path to the file (not very robust, but
handy in some cases).
Figure out the relative path to the file. If you have your .Rmd file
in / and data in /data, relative path should be, e.g.,
read.table("./data/myfile

R Graphics Legends Commands 
You can use expression
plot(0)
legend(x = "topleft", legend = expression(alpha[a], mu[b], gamma[c]))

How to get chunk name in knitr? 
I think this solution is imperfect because it requires a bit of care
in making sure the correct chunks are evaluated, but it gets around
the problem that chunk options are evaluated before hooks are called.
In short, it doesn't use a hook, but instead uses the fact that chunk
options can be R expressions. In this case, a function e() is used
that relies on a global counter variable to dictate whet

How eliminate codes with format 00.00.00A 
You first have to generate the codes :
codes<sapply(180001:200000,
function(x){
decomp<unlist(strsplit(as.character(x),""))
y<paste(paste(decomp[seq(1,6,2)],decomp[seq(2,6,2)],sep=""),collapse=".")
return(y)})
Then you can do :
toMatch<paste(rep(codes,e=3),c('U','B','R'),sep="")
toMatch[1:5]
[1] "18.00.01U" "18.00.01

Extract data for all days for last 30 days from R data frame 
You don't need an iteration. What you could do is, assuming your
data.frame is called X, and the date column, DATE, you could write:
X$DATE=as.Date(X$DATE, format='%d%B%Y')
the 'format' argument is to match your date format you specify in you
question. Then, to get the lines you are interested in, something
like:
X[X$DATE>=as.Date(today(),format='%d%B%Y')30)]
which is all the lines

Error in sorting column of a data frame 
I'm guessing data[[1]] is only one element, not a column or row and
that might explain why your sorting is kind of funny.
You can try :
data<data[order(data[,2],data[,1]),]

R (and R Studio) ignoring Environment variables (default package library) 
The best solution is to not use "User Variables", but "System
Variables", as the latter are profileindependant.
Changing "User Variables" for a given profile seem to require
administrator rights for that profile (at least in our network setup).

vectorize stringr str_match to remove a for loop 
You have to use [,2] instead of [2] because the output is a 2 column
matrix and by indexing [2], you are getting only the 2nd element i.e.
"http://www.slate.fr/story/" instead of the 2nd column`.
str_match(url, "http://.*slate.fr/(.*?)/")[,2]
#[1] "france" "story" "grandformat"
From the description of ?str_match
Vectorised over ‘string’. ‘pattern’ should be a single p

how to write multiple dataframe to multiple sheet of one csv excel file in R? 
This is not possible. That is the functionality of csv to be just in
one sheet so that you can view it either from notepad or any other
such software. If you still try to write it would get over ridden.
Just try to open a csv and open a new sheet and just write some values
and save it. The values which were already there is erased. one excel
file in csv format can have only one sheet.

Clustering connected set of points (longitude,latitude) using R 
Maybe this. First make some coordinates:
> x=c(1.000000, 1.055672, 1.038712, 1.094459, 1.133179, 1.116241,
1.126053, 1.181824 ,1.377892, 5.869881, 5.925270)
> y=c(1.333368, 1.304790, 1.347332, 1.318743, 1.332676, 1.375229,
1.572287, 1.544174, 2.371105 ,2.337032, 2.383415)
Make into a data frame
> xy = data.frame(x=x,y=y)
Now use outer to loop over all pairs of rows and columns to co

Replicate a bunch of syntax (multiple lines) to 20 times with various objects 
Something like this:
#dummy list
mylist <
list(a=data.frame(x=runif(10)),
b=data.frame(x=runif(10)),
c=data.frame(x=runif(10)))
#make temporary Rmd file
write.table(
paste0(
"```{r summary_table",names(mylist),", results='asis'}
kable(mylist['",names(mylist),"'], digits=1, align='r' )
```"),
"temp.Rmd",row.names=FALSE,col.names=FALSE,quote=FALSE)
Output  temp.Rmd

Filling in each of the 9 areas in circle using R 
You can do this with a combination of pie and draw.circle:
library(plotrix)
library(RColorBrewer)
cols < brewer.pal(9, 'Set3')
pie(rep(1, 4), col=cols[1:4], init.angle=45, radius=1, labels='')
par(new=TRUE)
pie(rep(1, 4), col=cols[5:8], init.angle=45, radius=0.5, labels='')
draw.circle(0, 0, 0.166, col=cols[9])
Set col as necessary to fill (or not) the individual sections, e.g.:
pie(re

How to find the smallest number of several variables and return the column name 
One simple approach would be (assuming your data called df)
df[c("min", "minCol")] < t(apply(df, 1, function(x) c(min(x),
names(x[which.min(x)]))))
df
# A B C D min minCol
# 1 23 23 12 34 12 C
# 2 12 13 11 9 9 D
# 3 70 80 67 76 67 C
# 4 43 23 25 40 23 B
Another approach (offered by @akrun) would be a combination of pmin,
do.call and max.col
val < do.call

how to convert mtw minitab files to readable mtp files in R 
If you want to convert your mtw worksheet to mtp extension, simply
open the mtw file on minitab and from File menu select "Save Current
Worksheet As" and save it to mtp extension. It will save it as Text
not Encrypted
You would be able then to use foreign

How do I fix encoding in a data frame regardless of its row or column in R(using dplyr)? 
You could correct this in a couple of ways.
Read the file with the proper encoding (UTF8)
read.csv2(file("filename.csv", encoding="UTF8"))
After reading the file, apply functions to convert to the UTF8
encoding
library(stringi)
df[] < lapply(df, function(x) stri_encode(x, "", "UTF8"))

why can't get the root of equation in R function solve? 
That's because the determinant of a is 0. You've got to specify the
tolerance in your call :
solve(a,b,tol=1e20)
[1] 0.1995899 1.3991798 2.1995899
> a%*%c(0.1995899,1.3991798,2.1995899)
[,1]
[1,] 4
[2,] 7
[3,] 10
But, of course, that is one solution among an infinity of it...
