Hi!
Jean-Baptiste Combes wrote:
Hello,
I use R 2.10, and I am new in R (I used to use SAS and lately Stata), I am
using XP.
I have a data which has a data.frame format called x.df (read from a csv
file). I want to take from this data observations for which the variable
Code starts with
Hi!
Do you mean something like this (df is your original data frame):
--- cut here ---
df1-df
df1[[1]]-paste(R,df[[1]],sep=_)
colnames(df1)-c(SERIES,YEAR,value)
df1$value[ df1$YEAR==2009 ]-5
for (i in c(2009:2007)) { df1$value[ df1$YEAR==(i-1) ]-( df1$value[
df1$YEAR==i ]-df$DELTA[ df$year==i ]
Hi!
Let's suppose the values for the x axis are stored in 'values'.
barplot(values, col=c(rep(Red,3),rep(1,length(values)-8),rep(Blue,5)))
HTH,
Kimmo
vikrant kirjoitti:
Suppose I need to draw a Grouped bar plot with 100 values on the X axis. Now
my question is If I need to highlight suppose
Hi!
29.01.2010 12:49, soeren.vo...@eawag.ch wrote:
Hello,
I read the help as well as the examples, but I can not figure out why
the following code does not produce the *given* row names, x and y:
x - 1:20
y - 21:40
rbind(
x=cbind(N=length(x), M=mean(x), SD=sd(x)),
Dear Anna,
19.02.2010 08:17, Anna Carter wrote:
(1) If the dataset contains some variables having all the entries = 0
and while analysing I want to delete those pericular columns, how do
acheive this. i.e.
Let's suppose 'df' is your data frame, then:
subset(df, select=which(colSums(df)!=0))
Hi!
Right, my solution did not take into accound paired negative values
summing up to zero.
This should work in all cases:
df[, which(colSums(df!=0)!=0)]
Kind regards,
Kimmo
__
R-help@r-project.org mailing list
Hi!
22.02.2010 17:45, xin wei wrote:
thank you for reply. I just type: hist(x) from SSH terminal, expecting a
histogram to pop up like what i got under windows.instead I got the
following error msg:
Error in X11(d$display, d$width, d$height, d$pointsize, d$gamma,
d$colortype, :
Hi!
22.02.2010 19:53, xin wei wrote:
hi, Kevin and K.Elo:
thank you for the suggestion. Can you be more specific on these? (like how
exactly get into x-switch or man ssh). I am totally ignorant about linux and
SSH:( Memory limitation forces me to switch from windows to Linux
Hi Mohamed,
mohamed nur anisah wrote (8.2.2008):
Dear lists,
I'm in my process of learning of writing a function. I tried to
write a simple functions of a matrix and a vector. Here are the
codes:
mm-function(m,n){ #matrix function
w-matrix(nrow=m, ncol=n)
for(i in
Hi,
joseph wrote (15.2.2008):
Thanks. I have another question:
In the following data frame df, I want to replace all values in col1
that are higher than 3 with NA. df= data.frame(col1=c(1:5, NA),col2=
c(2,NA,4:7))
My suggestion:
x-df$col1; x[ x3 ]-NA; df$col1-x; rm(x)
-Kimmo
Hi,
Hans Ekbrand wrote (19.2.2008):
I tried the following small code snippet which I copied from the
Introduction to R:
for (i in 2:length(meriter)) { table(meriter[[1]], meriter[[i]]) }
Try:
for (i in 2:length(meriter)) { print(table(meriter[[1]],
meriter[[i]])) }
Kind regards,
Kimmo
Hi,
this might also work for You:
points(example.df$StartDate[ (row(example.df)%%5)==0 ],
example.df$DSR2[ (row(example.df)%%5)==0 ], type=p, pch=3)
points(example.df$StartDate[ (row(example.df)%%5)==0 ],
example.df$DSR2[ (row(example.df)%%5)==0 ], type=p, pch=3)
Kind regads,
Kimmo
Jessi
Hi,
sorry, the correct commands should look like this:
plot(example.df$StartDate[ (row(example.df)%%5)==0 ], example.df$DSR1[
(row(example.df)%%5)==0 ], type=p, ylim=c(0.3,0.9))
points(example.df$StartDate[ (row(example.df)%%5)==0 ],
example.df$DSR2[ (row(example.df)%%5)==0 ], type=p, pch=3)
Hi,
if the columns always follow the same order (indx,var#, var#_lab ...),
then You could use the column numbers and do the following:
1) CN-colnames(df)[#] (#=colnum of 'var#')
2) df$NEW-(here the expression to create the new variable)
3) colnames(df)[ncol(df)]-c(paste(CN,new,sep=_))
This can
Dear all,
I have tried to import a SPSS file in R, but always get the following
message:
--- cut here ---
Error in read.spss(spss-data.sav, :
error reading system-file header
In addition: Warning message:
In read.spss(spss-data.sav, :
spss-data.sav: Variable Y6B_A indicates variable label
Hi again,
many thanks for Your answers. So, if I understood Wei right:
Zhao, Wei (Cancer Center) wrote (12.3.2008):
I had a similar problem when read one of my spss.sav with long
variable label. But when I read another spss.sav with short label the
same way, I don't have problem.
Seems like
Hi,
Albrecht Kauffmann wrote (18.03.2008):
Dear R-helpers,
if I want to read a .dta-file generated by stata 9.0 with read.dta
(foreign), I get the message
not a stata version 5-8 .dta-file. I'm using R-2.6.2 and the latest
version of the foreign package. Has someone any hint?
Seems quite
Hi,
the problem were a couple of overlength labels, indeed. After having
removed them, I was able to import the data without any problems.
Thanks anyway for Your help.
Happy Easter,
Kimmo
__
R-help@r-project.org mailing list
Henrique Dallazuanna wrote (1.4.2008):
You can try this:
x - data.frame()
for(i in LETTERS[1:5]) x[1:10, i] - rnorm(10)
x
Or this:
--- cut here ---
df-data.frame(0) [obsolet, if df already exists]
for (i in 1:10) { df-data.frame(cbind(df,0)); names(df)
[ncol(df)]-as.character(i) }
--- cut
Dear R-helpers,
I am desperately looking for a solution for how to print out the console
output to a standard printer. For example, I would like to print out the
summary.lm() output, the output of different ftable-functions etc. I use
R on a linux machine.
The only ways so far have been to
0.3746872961
[7] 0.6108254879 0.6617410595 0.6694177436 0.4650380281 0.0414420397
0.2307212995
[13] 0.5338913775 0.9186298891 0.0006410333 0.8046684864 0.6205502201
0.5352788521
[19] 0.4255279053 0.7711444888
On Wed, Aug 27, 2008 at 7:44 AM, K. Elo [EMAIL PROTECTED] wrote:
Dear R-helpers,
I am
Hi,
as mentioned in my previous posting, I run R on a linux machine. So a
possible function for printing (in linux) could look like this:
copy2lpr-function(..., PRINTER=lpr) {
LPR-pipe(PRINTER,w)
capture.output(..., file=LPR)
close(LPR)
}
This seems to work... An allows the user to choose
Hi,
is there a command or parameter for reducing the plotting area with
lattice? What I am looking for is an option similar to 'mai' or 'mar'
from the graphs package.
Background: I have plotted several charts with horizontal stacked bars
and now I would like to add info about percentages of each
Dear John,
thanks for Your quick reply.
John Fox wrote:
Dear Kimmo,
MCA is a rather old name (introduced, I think, in the 1960s by
Songuist and Morgan in the OSIRIS package) for a linear model
consisting entirely of factors and with only additive effects --
i.e., an ANOVA model will no
Hi,
cimfasy_rwl[ is.na(cimfasy_rwl) ] -0
Or did I understood Your right?
HTH,
Kimmo
Alfredo Alessandrini wrote:
I'm trying to replace NA with 0 value...
I've write a loop, but don't work...
Where's the problem?
cimfasy_rwl
1991 0.92 0.72 0.50 1.29 0.54 1.22
Dear all,
I use the barchart-function (lattice) for plotting stacked barcharts.
The data is a summary table (data frame) of likert-scale-evaluations
(strongly agree, agree...strongly disagree) to different issues
constructed as follows (L1=precentage of strongly agree evaluations,
Hi again,
Deepayan Sarkar wrote (21.4.2008):
Write your own panel function (which may or may not be a simple
exercise depending on your level of expertise in R). You could use
panel.barchart as a starting point. Basically, you need to insert
some calls to panel.text() (or something
Hi,
My problem is simple: since having updated the lattice package, I cannot
load lattice anymore. If I type in the command 'library(lattice)' the
loading fails with the following message:
--- cut here ---
Error in library.dynam(lib, package, package.lib) :
shared library 'lattice' not
Hi,
thanks for the quick reply :)
Prof Brian Ripley kirjoitti viestissään (06.05.2008):
Try installing again by
install.packages(lattice, .Library)
(from an account with suitable privileges).
Tried (as root) - not working :(
If that still fails, we need to see the output produced during
Prof Brian Ripley kirjoitti viestissään (06.05.2008):
Does starting R --vanilla help?
I am wondering if you have another corrupt copy of lattice somewhere.
The latter was the problem, many thanks for this! I use Rkward as GUI
and obviously some packages have been installed into the user
Hi,
Monica Pisica wrote:
- There is no perfect “beginner” book.
How about
- Crawley, Michael (2007). The R book, Wiley Sons.
- Maindonald, John John Braun (2007): Data Analysis and Graphics Using
R (2nd edition), Cambridge University Press.
As a political scientist (with programming
Hi!
mathallan wrote:
How can I from the summary function, decide which glm (fit1, fit2 or fit3)
fits to data best? I don't know what to look after, so I would please
explain the important output.
Start with the AIC value (Akaike Information Criterion). The model
having the lowest AIC is the
Hi!
Wacek Kusnierczyk wrote:
m = matrix(1:4, 2)
apply(m, 1, cat, '\n')
# 1 2
# 3 4
# NULL
why the null?
Could it be the return value of 'cat'. See ?cat, where:
---snip ---
Value
None (invisible NULL).
---snip ---
Kind regrads,
Kimmo
Hi,
I am quite new to R (but like it very much!), so please apologize if
this is a too simple question.
I have a large data frame consisting of data from a survey. There is,
for example, information about age and education (a numeric value from
1-9). Now I would like to extract the total
Hi,
https://stat.ethz.ch/mailman/listinfo/r-help
and there You'll find the section:
To unsubscribe from R-help, get a password reminder, or change your
subscription options enter your subscription email address:
Hope this helps,
Kimmo
__
Hi,
minben wrote:
I am a new R-language user. I have set up a data frame mydata,one of
the colume of which is skill. Now I want to select the observations
whose skill value is equal to 1,by what command can I get it?
Try this:
mydata1-mydatasubset(mydata, skill==1)
Maybe You should also read
Hi,
Crosby, Jacy R wrote:
i.e. I'd like to have aov(Phen1~L1) use only Pat1-Pat4,and Pat 10.
Similarly, aov(Phen1~L2) should use Pat1, 6, and 10.
Etc.
Is this something I can do in the aov function, or do I need to modify my
dataset before running aov? In either case, I need
Hi Yannick,
yannick misteli wrote:
I have a vector x with certain ID numbers in it and want to create a
subset from my dataset Y with this vector i.e extract only the data with
the given IDs from vector x.
example:
x
[1] 10066924 10207314 10257322 10334594 10348247
and now I want to
ivo welch kirjoitti:
dear R experts:
I am playing with boxplots for the first time. most of it is
intuitive, although there was less info on the web than I had hoped.
alas, for some odd reason, my R boxplots have some fat black dots, not
just the hollow outlier plots. Is there a
Hi Ivo,
ivo welch wrote:
alas, for some odd reason, my R boxplots have some fat black dots, not
just the hollow outlier plots. Is there a description of when R draws
hollow vs. fat dots somewhere?
[and what is the parameter to change just the size of these dots?]
Have you tried the command
Hi,
kayj wrote:
Hi,
I do have a data set with some missing values that appear as blanks. I want
to fill these blanks with an NA. How can this be done? Thanks for your help
Something like this?
my.data-data.frame(var=c(1,2,5,,66,4,3,,67,5,3,2,1,4,32,56,23),
stringsAsFactors=F)
Hi Jens!
23.03.2010 17:18, koj wrote:
The problem is: I want to group the data. I want to have ten groups. The
first two bars should be [1,1] and [2,1] together in one bar and in the
second bar of the first obervation should be [1,2] and [2,2] (stacked with
beside =TRUE). Therefore the first
Hi Jens!
24.03.2010 14:48, koj wrote
Hi Kimmo, thank you for your answer, but this is not the thing I am searching
for. Unfortunately, I have not described the problem very good. But just in
this moment I have a good idea: I use add=TRUE and paint two plots. And so I
am sure that I can
Dear Jim,
17.03.2011 20:54, Jim Silverton wrote:
I have a matrix say:
23 1
12 12
00
0 1
0 1
0 2
23 2
I want to count of number of distinct rows and the number of disinct element
in the second column and put these counts in a column. SO at the end of the
day I should have:
c(1,
Hi!
I have used the following command:
densityplot(~PV1CIV, groups=SGENDER, data=ISGFINC2,
lwd=2, col=1, lty=c(1,2), pch=c(+,o),
key=list(text=list(lab=levels(ISGFINC2$SGENDER), col=1),
space=bottom, columns=2, border=T, lines=T, lwd=2,
lty=c(1,2), col=1), ref=T, plot.points=F)
to
Hi,
thank you, Pascal, for your quick reply. Unfortunately your suggestion
is not working. Please have a look on the attachment, I have added
manually the mean lines I am trying to plot. The problem with 'abilne'
seems to be that the argument 'v' is relative to the graph area, not the
x-axis
HI again,
thanks for the replies. Unfortunately, due to my deadlines, I had no
time to test Petr's suggestion. I will test it later. I had to figure
out an alternate solution, so I decided to use normal plotting functions:
with(subset(ISGFINC2, as.numeric(SGENDER)==1), plot(density(PV1CIV),
Hi,
some sample data would be *very* helpful...
Kind regards,
Kimmo
16.03.2012 15:44, statquant2 wrote:
Hello I am looking for a special plot.
Let's suppose I have *100 days and
*each day I have a (1D)
distribution of the same variable.
I would
Hi!
Maybe not the most elegant solution, but works:
for(i in seq(1,length(data)-(length(data) %% 3), 3)) {
ifelse((length(data)-i)3, { print(sort(data)[ c(i:(i+2)) ]);
print(mean(sort(data)[ c(i:(i+2)) ])) }, { print(sort(data)[
c(i:length(data)) ]); print(mean(sort(data)[ c(i:length(data))
Dear Romezo,
a solution maybe not that elegant and effective, but seems to work:
test_calculate-function() {
tarrow-1
TARGET-data.frame(Thesis=0, Day=0,A=0,B=0,C=0)
for (i in c(unique(my.data$Thesis))) {
for (j in c(unique(my.data$Day[ my.data$Thesis==i ]))) {
Hi,
19.12.2011 11:12, Ville Iiskola wrote:
Error in if (abs(x - oldx) ftol) { : missing value where TRUE/FALSE
needed
The reason for this error is in the row 563. There the choice has
value 1 and Ie has missing value. If the choice has value 0 and Ie
has missing values, then there is no
Hello,
three commands might do the job (NOTE: df=your data frame,
obser=Observation, var1=Variable 1 [TYPE: string], var1flag=Variable 1
flag [TYPE: string])
1. df$var1flag-NA
2. df$var1flag[ is.na(as.numeric(df$var1)) ]-df$var1[
is.na(as.numeric(df$var1)) ]
3. df$var1-as.numeric(df$var1)
Hi!
I recently posted a similar question (entitled Adding mean line to a
lattice density plot). Have not got any usable solution forcing my to
fall back to the use of the normal 'plot' function. The problem was the
same as yours: using panel.abline simply did not work, the position of
the
Hi!
28.09.2012 08:41, Atte Tenkanen wrote:
Sorry. I should have mentioned that the order of the components is important.
So c(1,4,6) is accepted as a subvector of c(2,1,1,4,6,3), but not of
c(2,1,1,6,4,3).
How to test this?
How about this:
--- code ---
g1- c(2,1,1,4,6,3)
g2-
Hi!
28.09.2012 09:13, Bhupendrasinh Thakre wrote:
Statement I tried :
b - unclass(Sys.time())
b = 1348812597
c_b - rnorm(1,2,1)
Do you mean this:
--- code ---
df-data.frame(x=0,y=0)
colnames(df)
[1] x y
colnames(df)[2]-paste(b,unclass(Sys.time()),sep=_)
colnames(df)
[1] x
Hi!
I want to use R for network analysis and have tried to install the
'igraph' package. Unfortunately, the installation is aborted by an error:
--snip--
gcc -std=gnu99 -I/usr/lib64/R/include -DNDEBUG -I/usr/local/include
-DUSING_R -I. -Ics -Iglpk -Iglpk/amd -Iglpk/colamd
Hi!
On Thu, 2 Aug 2012, Uwe Ligges wrote:
R.h should be part if your R installation, given the output above
probably in /usr/lib64/R/include ?
But there is no such file R.h in my system???
~$ locate \/R.h
~$
Or with 'find':
# find / -name R.h
#
Any ideas?
Kind regards,
Kimmo
Hi!
On Thu, 2 Aug 2012, Prof Brian Ripley wrote:
You may bave a micro-packaged distribution (some form of SuSE as I recall):
is there a separate R-devel RPM? (Fedora had one an one time.)
Thanks, this made the trick. After having R-base-devel installed I
succeeded in intalling the 'igraph'
Hi!
How about trying this:
data[ data$col1!=data$col2 !is.na(data$col3), ]
col1 col2 col3
2a1 ST001
3b2 ST002
HTH, Kimmo
28.05.2014 15:35, jeff6868 wrote:
Hi everybody,
I have a little problem in my R-code which seems be easy to solve, but I
wasn't able to find the
Dear Katharina,
08.02.2011 11:21, Katharina Lochner wrote:
The following appeared on my console:
install.packages(psych)
Warnung in install.packages(psych) :
'lib = C:/PROGRA~1/R/R-212~1.1/library ist nicht schreibbar
Apparently You do not have permissions to write to the target
Hi!
21.02.2011 08:50, Schmidt, Lindsey C (MU-Student) wrote:
What is plot.new? How can I fix this data or add plot.new so it works?
?plot.new
?plot
plot(u[h],v[h],type=l,asp=1) seems to work for me...
HTH,
Kimmo
__
R-help@r-project.org mailing list
Hi!
21.02.2011 08:50, Schmidt, Lindsey C (MU-Student) wrote:
What is plot.new? How can I fix this data or add plot.new so it works?
?plot.new
?plot
plot(u[h],v[h],type=l,asp=1) seems to work for me...
HTH,
Kimmo
__
R-help@r-project.org mailing list
Hi!
10.06.2015, 13:20, khatri wrote:
My date column is in following format : %m/%d H:M:S
There is not mention of year in the data.So how can I read this using
strptime function.
I have tried strptime(dates,%m/%d H:M:S) but this is returning NA.
Thanks
How about:
strptime(dates,%m/%d
Hi!
25.08.2015, 18:17, Sam Albers wrote:
Hi all,
This is a process question. How do folks efficiently identify column
numbers in a dataframe without manually counting them. For example, if I
want to choose columns from the iris dataframe I know of two options. I can
do this:
Hi!
You can download the example file with this link:
https://www.dropbox.com/s/tlf1gkym6d83log/example.json?dl=0
BTW, I have used a JSON validator and the problem seems to related to
wrong/missing EOF.
--- snip ---
Error: Parse error on line 1:
...:"1436705823768"} {"created_at":"Sun J
Hi!
I have collected 500.000+ tweets with a Python script using 'tweepy',
which stored the data in JSON format. I would like to use R for data
analysis, but have encountered problems when trying to import the data
file with 'jsonlite'.
Here what I have tried:
>
Hi,
thanks to Duncan and Jeroen to quick replies. I was actually my thinking
error :) I suppoed 'fromJSON' to cope with a multi-line file or a list,
but this seems not to be the case. So I first read the file with
'readLines' into a list and processed all items with 'fromJSON' within a
Hi!
Many thanks to Duncan and Jim for their quick replies.
27.05.2016, 01:08, Jim Lemon wrote:
Hi Kimmo,
par(mar=c(5,7,4,2))
dotchart(kedf$x)
mtext(kedf$Group.2,side=2,at=1:6,line=0.5,
las=2,cex=log(abs(kedf$Freq))+1)
Jim
This 'dotchart' solution worked fine and I got what I wanted :)
Dear R-helpers!
I have a data frame storing data for word co-occurrences, average
distances and co-occurence frequency:
Group.1Group.2 x Freq
1 deutschland achtziger 2.001
2 deutschlandalt 1.254
3 deutschland anfang -2.001
4 deutschlandansehen
Hi Juho!
01.06.2016, 14:40, Juho Kiuru wrote:
Hi all, I am new to R and TwitteR and would love to get some advice from
you.
I managed to get list of tweets containing word 'innovation' tweeted in
Helsinki with following script:
searchTwitter('innovation', n=1,
Hi again!
21.06.2016, 21:33, chalabi.el...@yahoo.de wrote:
Hi Kimmo, Thanks for your reply. I think now my problem is that I
don't understand what does factor(df.classes[training]) do?
Sorry, my mistake, should habe been 'df$speed'. Please try the following:
--- snip ---
set.seed(7)
Hi!
Some sample data could help us to help you...
But have you read '?xyf' in order to ensure that your 'Y' is what 'xyf'
expects it to be?
What kind of error messages do you get?
Regards,
Kimmo
16.06.2016, 15:13, ch.elahe via R-help wrote:
Is there any answer?
Hi all, I have a df and I
my data based on speed, to see the coming costumer's
protocols fall into which speed group and I think I need to bring this speed
column in Y element of xyf
On Thursday, June 16, 2016 2:29 PM, K. Elo <mailli...@pp.inet.fi> wrote:
Hi!
Some sample data could help us to help you...
But have
Hi!
22.06.2016, 22:00, chalabi.el...@yahoo.de wrote:
Dear Kimmo,
I already used df$speed[training] in df.xyf but I get this error:
Error in xyf(Xtraining,factor(df$speed[training]),grid=somgrid(5, :
NA/NaN/Inf in foreign function call (arg 1)
Please check for zeros (0) and NAs in
Hi,
what is the exact problem? I tried you code and it works fine...
Best,
Kimmo
24.08.2016, 10:07, Serpil ŞEN wrote:
Dear Authorized Sir / Madam,
I need your help on clustering with R.
I have symmetric distance matrix which i created usign ClustalOmega program.
and used this R codes for
Hi!
18.10.2016, 14:38, Abhinaba Roy wrote:
Hi R helpers,
I have json inputs from an app which I want to convert to dataframes. Below
are the two inputs. Can someone help me in converting these to dataframes
[...]
IMHO, the best way is to use the package 'jsonlite', see:
*
Dear all,
I am currently working a research project on social media interaction.
As a part of this project, mostly for teaching purposes, I should
develop a R-based approach for real-time visualisation of streamed data
(from Twitter).
My idea is simple (and working :) ): A Python-script
Hi!
Maybe this helps:
http://r.789695.n4.nabble.com/Error-in-normalizePath-path-with-McAfee-td2532324.html
Best,
Kimmo
15.12.2016, 08:18, Amelia Marsh via R-help wrote:
Hi
I had installed R studio Desktop 1.0.44. However whenever I wanted to write any
command, before I could complete, I
Hi!
2017-09-18 07:13 -0500, Therneau, Terry M., Ph.D. wrote:
> This question likely has a 1 line answer, I'm just not seeing
> it. (2, 3, or 10 lines is
> fine too.)
>
> For a vector I can do group <- match(x, unqiue(x)) to get a vector
> that labels each
> element of x.
Actually, you get a
Hi!
Seems to be an encoding problem. This worked for me (have not full-
checked the output, though):
fromJSON(encodeString(zWebObj))
HTH,
Kimmo
2018-05-08 12:49 -0700, David Winsemius wrote:
> >
> > On May 8, 2018, at 10:08 AM, Evans, Richard K. (GRC-H000) > k.ev...@nasa.gov> wrote:
> >
> >
Hi!
How about this:
--- snip --
for (i in 1:(length(split_str)-1)) {
assign(paste("DF",i,sep=""),DF[
c((which(DF$name==split_str[i])+1):(which(DF$name==split_str[i+1])-1)),
])
}
--- snip ---
'assign' creates for each subset a new data.frame DFn, where n ist a
count (1,2,...).
But note:
Hi!
Maybe not the most elegant solution, but a workaround is to have a
function:
> save2<-function(y, ...) { save(y,...)}
> save2(x1,x2,file="test.RData")
The point is to include the variables to be "renamed" as parameters (in
my example: y). The function will use the parameter variable names
Hi!
An alternative with 'assign':
for ( i in 71:75) {
setwd(paste("C:/Awork/geneAssociation/removed8samples/neuhausen", i,
sep=""))
temp.df<-read.csv("seg.pr3.csv", head=T)
temp.df$id<-paste0("sn",i,sep="")
assign(paste0("seg",i,sep=""),temp.df)
}
rm(temp.df,i) # Clean up
HTH,
Kimmo
Hi!
Not just an gmail issue. After my last reply I have gotten tons of
spams from "Samantha Smith". Keep hitting my "rank as spam"-button in
the hope that my MUA could learn :)
Best,
Kimmo
ti, 2018-04-17 kello 19:34 +, Ding, Yuan Chun kirjoitti:
> No, I do not use gmail, still got dirty
Hi!
Have you tried to use 'fromJSON' with the parameter 'simplifyDataFrame'
set to TRUE?
See:
https://cran.r-project.org/web/packages/jsonlite/vignettes/json-aaquickstart.html
-> Section "Data Frames" explains how this affects the data frame
structure. IMHO this should solve your problem...
Hi!
Maybe this would do the trick:
--- snip ---
library(reshape2) # Use 'reshape2'
library(dplyr)# Use 'dplyr'
datatransfer<-data %>% mutate(letter2=letter) %>%
dcast(id+letter~letter2, value.var="weight")
--- snip ---
Or did I misunderstood something?
Best,
Kimmo
2019-01-06, 13:16
Hi!
Not having a data chunk prevents me from testing abit, but maybe you
should take a look on:
?table
?xtabs
to start with.
But as already suggested by other users, a small data set would be of
great help :)
HTH,
Kimmo
su, 2019-01-06 kello 13:49 -0500, Rachel Thompson kirjoitti:
> Hi Rich,
Hi!
2019-03-25 kello 09:30 +0800, Steven Yen wrote:
> The second command is ugly. How can I print the 25 numbers into 2
> rows
> of ten plus a helf row of 5? Thanks.
Something like this?
x<-1:25; for (i in seq(1,length(x),10))
print(x[i:ifelse((i+9)>length(x),length(x),i+9)])
HTH,
Kimmo
Hi!
2019-02-27 22:51 -0500, Aimin Yan wrote:
> I have a question about assigning color based on the value of a
> matrix
>
> The following is my matrix.
>
> d
> lateRT earlyRT NAD ciLAD
> lateRT 1.0 0.00 0.006224017 0.001260241
> earlyRT
Hi Rajesh,
2019-05-05 10:23 +0530, Rajesh Ahir_GJ wrote:
> Hello R users,
>
> I am getting an error while running following code.
>
> library(ggplot2)
> ggplot(hourly_data1,aes(hour, power))+
> geom_boxplot(aes(fill=monthname),outlier.shape=NA) +
> facet_wrap(~monthname)
>
Hi Drake,
2019-05-04, 17:34 -0700, Drake Gossi wrote:
> Hello everyone,
>
> I'm trying to learn how to put together a citation network, and, in
> doing so, I'm playing around with a data set of my own making. I'm
> going back and forth between two .csv files. One has two columns and
> is simply
nha kirjoitti:
> Thanks for providing the code but I also needed the output sheet in
> .csv format with all the four columns corresponding to the value
> (Chrom,
> Start_pos, End_pos & Value ranging from what I specified earlier).
>
> Puja
>
> On Fri, Jan 31, 2020 at 10:23
Hi!
Let's assume your data is stored in a data frame called 'df'. So this
code should do the job:
df$Value[ (df$Value>=0.2 & df$Values<=0.4) | df$Value>=0.7 ]
Best,
Kimmo
pe, 2020-01-31 kello 09:21 -0500, pooja sinha kirjoitti:
> Hi All,
>
> I have a .csv file with four columns (Chrom,
Hi!
Oh, sorry, one "s" too much in my code. Here the correct one:
df$Value[ (df$Value>=0.2 & df$Value<=0.4) | df$Value>=0.7 ]
Best,
Kimmo
pe, 2020-01-31 kello 17:12 +0200, K. Elo kirjoitti:
> Hi!
>
> Let's assume your data is stored in a data frame calle
Hi,
cannot reproduce, either, on my Linuxmint 19.3 + R 3.6.2.
Here the outputs:
--- snip ---
> test(mean, 1:10)
[1] 5.5
> test(NULL, 1:10)
NULL
Error in FUN(args) : could not find function "FUN"
> test(mean, list(x=1:10, na.rm=TRUE))
[1] NA
Warning message:
In mean.default(args) : argument is
Hi!
Have you already read this:
https://cran.r-project.org/web/packages/rtweet/vignettes/auth.html
I think they explain rather well how to use Twitter tokens with
rtweet...
HTH,
Kimmo
ke, 2020-04-08 kello 17:19 +1200, Patrick Connolly kirjoitti:
> I'm using the rtweet package which makes use
llo Kimmo,
>
> Yes. I did that and it worked fine -- as far as it goes. But it
> didn't cover what to do when using the same twitter account on a
> computer with a different user name -- which is what my question was
> about.
>
>
> On Wed, 08-Apr-2020 at 08:55AM +0300
Hi!
With 'dplyr':
dt_count %>% mutate(STATUS=ifelse(STATUS %in%
c("Resolved","Closed"),"Resolved/Closed",STATUS)) %>% group_by(STATUS)
%>% summarise(n=sum(N))
Output:
1 Assigned 135
2 Cancelled 20
3 In Progress56
4 Pending75
5 Resolved/Closed 1180
HTH,
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