Sorry possibly very silly question? Couldn't find the answer? How do I load this kind of .dat file in R and stck them in one column? I have been trying
NerveData<-as.vector(read.table("D:/Dropbox/nerve.dat", sep=" ")$value)
The data set looks like
0.21 0.03 0.05 0.11 0.59 0.06
0.18 0.55 0.37 0.09 0.14 0.19
0.02 0.14 0.09 0.05 0.15 0.23
0.15 0.08 0.24 0.16 0.06 0.11
0.15 0.09 0.03 0.21 0.02 0.14
0.24 0.29 0.16 0.07 0.07 0.04
0.02 0.15 0.12 0.26 0.15 0.33
If you want to read all the data in as a single vector, use
src <- "http://www.stat.cmu.edu/~larry/all-of-nonpar/=data/nerve.dat"
NerveData <- scan(src, numeric())
Actually I found a easier solution thanks for the initial helps
Nervedata<-read.table("nerve.dat",sep ="\t")
Nervedata2<-c(t(Nervedata))
Simply use read.table with the correct separator. Which in your case is probably \t, a tab character.
So try:
NerveData = read.table("D:/Dropbox/nerve.dat", sep="\t")
Related
I have the following code:
S = [100 200 500 1000 10000];
H = [0.14 0.15 0.17 0.19 0.28;0.14 0.16 0.18 0.20 0.29;0.15 0.17 0.19 0.21 0.31;0.16 0.17 0.20 0.22 0.32;0.23 0.22 0.28 0.30 0.44;0.23 0.23 0.29 0.3 0.5;0.33 0.32 0.4 0.42 0.63;0.32 0.31 0.39 0.40 0.61;0.23 0.23 0.30 0.30 0.50];
for i = 1:9
hold on
plot(S, H(i,:));
legend('GHM01','GHM02','GHM03','GHM04','GHM05','GHM06','GHM07','GHM08','GHM09'); %legend not correctly
axis([100 10000 0.1 1])
end
set(gca,'xscale','log')
The x-axis looks like this:
Because The S-values are very far from each other, I used a logaritmic x-axis (and linear y-axis).
I have on the axis 5 values (see S), and I only want those 5 values visible on the x-axis with equidistant spacing between the values. How do I do this? Or is there a better alternative to display my x-axis, rather than logaritmic scale?
If you want the X-axis ticks to be equally distant although they are not (neither on a linear nor on a log scale) then you basically treat this axis as categorical, and then it should get and ordinal temporary value (say 1:5) to determine the distance between them.
Here is a quick implementation of your comment above:
S = {'100' '200' '500' '1000' '10000'};
H = [0.14 0.15 0.17 0.19 0.28;...
0.14 0.16 0.18 0.20 0.29;
0.15 0.17 0.19 0.21 0.31;
0.16 0.17 0.20 0.22 0.32;
0.23 0.22 0.28 0.30 0.44;
0.23 0.23 0.29 0.3 0.5;
0.33 0.32 0.4 0.42 0.63;
0.32 0.31 0.39 0.40 0.61;
0.23 0.23 0.30 0.30 0.50];
f = figure;
plot(1:length(S),H);
f.Children.XTick = 1:length(S);
f.Children.XTickLabel = S;
TMHO this is the most straightforward way to solve this problem ;)
I like to find difference between my samples but when I use diff() my first sample miss.
input:
data
XX.3.22 XX.1.2 XX.5.19 XX.2.21 XX.2.16 XX.5.27 XX.3.5 XX.2.12 XX.4.15
0.00 0.12 0.17 0.20 0.21 0.26 0.27 0.27 0.32
diff(data)
output:
XX.1.2 XX.5.19 XX.2.21 XX.2.16 XX.5.27 XX.3.5 XX.2.12 XX.4.15
0.05 0.05 0.03 0.01 0.05 0.01 0.00 0.05
I do not want miss first (XX.3.22) sample.
I expect:
XX.3.22 = 0.12
I need help with interpreting an error message using corrplot.
Here is my script
install.packages("ggplot2")
install.packages("corrplot")
install.packages("xlsx")
library(ggplot2)
library(corrplot)
library(xlsx)
#set working dir
setwd("C:/R")
#read xlsx data into R
df <- read.xlsx("TP_diff_frame.xlsx",1)
#set column as index
rownames(df) <- df$country
#remove column
df2<-subset(df, select = -c(country) )
#round values to to decimals
corrplot(df2, method="shade",shade.col=NA, tl.col="black", tl.srt=45)
My df2:
> df2
a b c d e f g
Sweden 0.09 0.19 0.00 -0.25 -0.04 0.01 0.00
Germany 0.11 0.19 0.01 -0.35 0.01 0.02 0.01
UnitedKingdom 0.14 0.21 0.03 -0.32 -0.05 0.00 0.00
RussianFederation 0.30 0.26 -0.07 -0.41 -0.09 0.00 0.00
Netherlands 0.09 0.16 -0.05 -0.26 0.02 0.02 0.01
Belgium 0.12 0.20 0.01 -0.34 0.01 0.00 0.00
Italy 0.14 0.22 0.01 -0.37 0.00 0.00 0.00
France 0.14 0.24 -0.04 -0.34 0.00 0.00 0.00
Finland 0.16 0.17 0.01 -0.26 -0.08 0.00 0.00
Norway 0.15 0.21 0.10 -0.37 -0.09 0.00 0.00
And the error message:
> corrplot(df2, method="shade",shade.col=NA, tl.col="black", tl.srt=45)
Error in matrix(unlist(value, recursive = FALSE, use.names = FALSE), nrow = nr, :
length of 'dimnames' [2] not equal to array extent
I think the problem is that you are plotting the data frame instead of the correlation matrix. Try to change the last line to this:
corrplot(cor(df2), method="shade",shade.col=NA, tl.col="black", tl.srt=45)
The function cor calculates the correlation matrix, which is what you need to plot
In order to use the corrplot package for heatmap plots you should pass your data.frame to a matrix and also use the is.corr argument.
df2 <- as.matrix(df2)
corrplot(df2, is.corr=FALSE)
Another option is to break it up into two lines of code.
df2 <- cor(df, use = "na.or.complete")
corrplot(df2, method="shade",shade.col=NA, tl.col="black", tl.srt=45)
I'd run a simple corrplot (e.g. corrplot.mixed(df2)) make sure it works, then get into the fine tuning and aesthetics.
I am generating a lot of ftable() crosstabulations for a descriptive report. Example:
AUS BEL BUL EST FRA GEO GER HUN ITA NET NOR ROM RUS
30- primary 0.06 0.03 0.07 0.03 0.02 0.03 0.03 0.02 0.05 0.03 0.05 0.04 0.02
secondary 0.30 0.09 0.16 0.10 0.10 0.14 0.10 0.16 0.11 0.08 0.08 0.09 0.11
tertiary 0.05 0.07 0.04 0.05 0.07 0.06 0.02 0.04 0.02 0.05 0.06 0.02 0.09
30+ primary 0.07 0.16 0.12 0.07 0.16 0.03 0.05 0.11 0.35 0.21 0.09 0.17 0.03
secondary 0.40 0.20 0.30 0.29 0.25 0.35 0.35 0.34 0.27 0.20 0.27 0.34 0.26
tertiary 0.13 0.23 0.13 0.18 0.17 0.17 0.18 0.09 0.09 0.23 0.23 0.06 0.24
60+ primary 0.00 0.12 0.10 0.13 0.14 0.07 0.05 0.12 0.09 0.11 0.06 0.19 0.12
secondary 0.00 0.05 0.05 0.08 0.06 0.10 0.14 0.09 0.02 0.04 0.11 0.07 0.06
tertiary 0.00 0.05 0.03 0.06 0.03 0.04 0.07 0.03 0.01 0.05 0.06 0.02 0.07
I am looking for a function that could take the ftable() or table() output, and highligh values that deviate from the row-mean, or assign an overall gradient to the text of the values, e.g. from 0-100% the values are coloured from red to green.
The output is now processed through knitr, but I'm not sure at which point in the toolchain I could intervene and add colour based on the relative size of the values.
You can use the latex function, in the Hmisc package.
# Example shamelessly copied from http://www.karlin.mff.cuni.cz/~kulich/vyuka/Rdoc/harrell-R-latex.pdf
cat('
\\documentclass{article}
\\usepackage[table]{xcolor}
\\begin{document}
<<results=tex>>=
library(Hmisc)
d <- head(iris)
cellTex <- matrix(rep("", nrow(d) * ncol(d)), nrow=nrow(d))
cellTex[2,2] <- "cellcolor{red}"
cellTex[2,3] <- "color{red}"
cellTex[5,1] <- "rowcolor{yellow}"
latex(d, file = "", cellTexCmds = cellTex, rowname=NULL)
#
\\end{document}',
file="tmp.Rnw" )
Sweave("tmp.Rnw")
library(utils)
texi2pdf("tmp.tex")
To generate latex tables from R objects, you can use the xtable package. It is available on CRAN, take a look at the documentation. To get the color in the table, use the color latex package. Some example code:
library(xtable)
n = 100
cat_country = c("NL","BE","HU")
cat_prim = c("primary","secondary","tertiary")
dat = data.frame(country = sample(cat_country, n, replace = TRUE),
prim = sample(cat_prim, n, replace = TRUE))
ftable_dat = ftable(dat)
## Make latex table:
latex_table = xtable(as.table(ftable_dat))
To get what you want I made the following hack (ugly one). The trick is to print the xtable object and than edit that:
latex_table = within(latex_table, {
# browser()
primary = ifelse(primary > 12, sprintf("\\textbf{%s}", primary), primary)
#primary = sub("\\{", "{", primary)
})
printed_table = print(latex_table)
printed_table = sub("backslash", "\\", printed_table)
printed_table = sub("\\\\}", "}", printed_table)
printed_table = sub("\\\\\\{", "{", printed_table)
printed_table = sub("\\$", "\\", printed_table)
printed_table = sub("\\$", "\\", printed_table)
cat(printed_table)
Which leads to:
% latex table generated in R 2.14.1 by xtable 1.6-0 package
% Thu Feb 16 13:10:55 2012
\begin{table}[ht]
\begin{center}
\begin{tabular}{rrrr}
\hline
& primary & secondary & tertiary \\
\hline
BE & 10 & 5 & 11 \\
HU & \textbf{13} & 13 & 8 \\
NL & 11 & 17 & 12 \\
\hline
\end{tabular}
\end{center}
\end{table}
This example makes a number in the primary category bold, but it can work for colorization just as easily. Maybe someone else has a more elegant solution?
I always transpose by using t(file) command in R.
But i it is not running properly (not running at all) on big data file (250,000 rows and 200 columns). Any ideas.
I need to calculate correlation between 2nd row (PTBP1) with all other rows (except 8 rows including header). In order to do this I transpose rows to columns and then use cor function.
But I struck at transpose fn. Any help would be really appreciated!
I copied example from one of the post in stackoverflow (They are also almost discussing the same problem but seems no answer yet!)
ID A B C D E F G H I [200 columns]
Row0$-1 0.08 0.47 0.94 0.33 0.08 0.93 0.72 0.51 0.55
Row02$1 0.37 0.87 0.72 0.96 0.20 0.55 0.35 0.73 0.44
Row03$ 0.19 0.71 0.52 0.73 0.03 0.18 0.13 0.13 0.30
Row04$- 0.08 0.77 0.89 0.12 0.39 0.18 0.74 0.61 0.57
Row05$- 0.09 0.60 0.73 0.65 0.43 0.21 0.27 0.52 0.60
Row06-$ 0.60 0.54 0.70 0.56 0.49 0.94 0.23 0.80 0.63
Row07$- 0.02 0.33 0.05 0.90 0.48 0.47 0.51 0.36 0.26
Row08$_ 0.34 0.96 0.37 0.06 0.20 0.14 0.84 0.28 0.47
........
250,000 rows
Use a matrix instead. The only advantage of a dataframe over a matrix is the capacity to have different classes in the columns and you clearly do not have that situation, since a transposed dataframe could not support such a result.
I don't get why you want to transpose the data.frame. If you just use cor it doesn't matter if your data is in rows or columns.
Actually, it is one of the major advantages of R that it doen's matter if your data fits in the classical row-column pattern as SPSS and others programs require data to be.
There are numerous ways to correlate the first row with all other rows (I don't get which rows you want to exclude). One is using a loop (here the loop is implicit in the call to one of the *apply family functions):
lapply(2:(dim(fn)[1]), function(x) cor(fn[1,],fn[x,]))
Note that I expect you data.frame to ba called fn. To skip some rows change the 2 to the number you want. Furthermore, I would probably use vapply here.
I hope this answer points you in the correct direction and that is to not use t() if you absolutely don't need it.