Hi Simon,
How about this?
library(plotrix)
revlist<-grep("i",names(df),fixed=TRUE)
df[,revlist]<-sapply(df[,revlist],rescale,c(3,1))
Jim
On Fri, Apr 3, 2015 at 6:30 AM, Simon Kiss wrote:
> Hi there: I have a list of data frames with identical variable names.
> I’d like to reverse scale the s
Hi Collin,
Have a look at this:
http://stats.stackexchange.com/questions/70643/power-analysis-for-kruskal-wallis-or-mann-whitney-u-test-using-r
Although, thinking about it, this might have constituted your "perusal of
the literature".
Plus it always looks better when you spell the names properly
Hi Hurr,
Try this:
par(mar=c(7,4,4,2))
plot(x1,x2,type="l",xlab="")
require(plotrix)
fullaxis(1,at=pretty(x1),labels=round(1/pretty(x1),2),tick=TRUE,
col="red",col.axis="red",lwd=4,pos=-0.2)
mtext("x1",side=1,line=4.5)
Jim
On Fri, Apr 3, 2015 at 11:16 AM, Hurr wrote:
> I have been doing other
Hi Hurr,
The problem is with the quotes in xlab="". In your version the quotes are
not ASCII double quotes but some sort of "smart" quotes. Either your email
client has substituted them or more likely, you are editing your code in
something like MS Word and cutting and pasting the result into R. Tr
Hi Antonio,
Is it possible to use "add=TRUE" and display the plot in two passes?
Jim
On Sat, Apr 4, 2015 at 11:44 PM, Antonio Silva
wrote:
> Hello everybody
>
> The problem is that species names are shown in the ordination diagram when
> the data set has a maximum of 80 rows, and mine has 81 (
Hi John,
One way is to create an index variable that will divide your data into the
appropriate intervals. There are a number of ways to do this. Say you want
the "two month" version of bimonthly and you have a date variable
("raindate") for each observation like "1982-01-01".
date_order<-paste(re
Hi Catalin,
I'm not quite sure which values you want to use, but does this start to do
what you want?
Jq90<-quantile(spei$J,0.9)
Jq10<-quantile(spei$J,0.1)
require(plotrix)
plot(spei$year,spei$J,xlab="Year",ylab="Temerature anomaly",
main="Temperature anomalies by year (1901 to 2013)",
pch=ifels
hat on each list element using lapply or llply. I have
> about 4 data frames and a few other recodes to do so automating would be
> nice, rather than applying your code to each individual list element.
> simon
>
> On Apr 2, 2015, at 6:30 PM, Jim Lemon wrote:
>
> Hi Simon,
>
Hi Chistine,
The latticeExtra package should be included with your installation of R.
Enter:
library(latticeExtra)
in your R session to make it available.
Jim
On Tue, Apr 7, 2015 at 4:28 PM, David Winsemius
wrote:
>
> On Apr 6, 2015, at 9:38 PM, Christine Lee via R-help wrote:
>
> > Thank yo
Hi Sun,
No, I was thinking of something like hunspell, which seems to fit into the
sort of work that you are doing.
Jim
On Fri, Apr 10, 2015 at 11:42 PM, Sun Shine wrote:
> Thanks Jeff.
>
> I'll add that to the ever-growing list my current studies are generating
> daily. :-)
>
> Cheers
> S
>
>
Hi Alexandra,
The error probably comes from the first iteration of i in 0:23. As indexing
in R begins at 1, there is no element 0. Try using:
for(i in 1:24) {
...
and see what happens.
Jim
On Sat, Apr 11, 2015 at 7:06 AM, Alexandra Catena wrote:
> Update:
>
> I have this so far. * The first
> to the next loop, yet it doesn't. I will keep trying!
>
> Thanks,
> Alexandra
>
> On Fri, Apr 10, 2015 at 3:43 PM, Jim Lemon wrote:
> > Hi Alexandra,
> > The error probably comes from the first iteration of i in 0:23. As
> indexing
> > in R begins at 1, t
Hi Alejo,
The color.id function in plotrix will do this, one color at a time:
sapply(rainbow(6),color.id)
#FFFF #00FF #00FF00FF #00FF # #FF00
[1,] "red" "yellow" "green" "cyan""blue""magenta"
[2,] "red1""yellow1" "green1" "cyan1" "blue1" "magent
Hi jpm miao,
What sort of "Excel" graphs do you want to produce? There are a few
varieties, you know.
Jim
On Wed, Apr 15, 2015 at 1:54 PM, jpm miao wrote:
> Hi,
>
>I understand that there're many great graphic packages in R (e.g.,
> ggplot2) . Nevertheless, my office uses Excel extensively
Hi Dot,
Jeff's guess is probably correct, but perhaps you could describe the
crazy tick marks and the repeating labels a little more. I suspect
that if "newdate" was a character variable you wouldn't get a plot at
all, and if it is a factor, a few of the labels might identify what
went wrong.
Jim
Hi Hermann,
This isn't much more elegant, but
test.list<-sapply(test,function(x) { strsplit(x," ") },simplify=TRUE)
names(test.list)<-NULL
Jim
On 4/15/15, Hermann Norpois wrote:
> Hello,
>
> I try to open a text file test.txt with the content
>
> * a b d
> * z u i h hh
> * h bh kk
>
> so that
Hi merm,
In case Sergio's message is a little cryptic:
return_a_list<-function() {
a<-"First item of list"
b<-c(2,4,6,8)
c<-matrix(1:9,nrow=3)
return(list(a,b,c))
}
x<-return_a_list()
x
Jim
On Thu, Apr 16, 2015 at 7:21 PM, Sergio Fonda wrote:
> Collect in a vector or dataframe or list the
Hi Diego,
You may have to do some conversion as you have three fields in the
first line using the default space separator and five fields in
subsequent lines. If the first line doesn't contain any important data
you can just delete it or replace it with a meaningful header line
with five fields and
Hi Agathe,
You can start with the "maps" package:
# in an R session
install.packages("maps")
# assume you want a simple map containing France
map("world",xlim=c(-6.0,9.6),ylim=c(42,51.5))
then plot your data by the coordinates of the stations. You will
probably want to plot graphical elements to
ive,
>>>dadf$datetime <- as.POSIXct(paste(dadf$V1,dadf$V2))
>>> since the dates appear to be in the default format.
>>> (I generally prefer to work with datetimes in POSIXct class rather than
>>> POSIXlt class)
>>>
>>> -Don
>>>
>&
ot.
>>
>> Unfortunately, I get the following error:
>>
>>
>> st1_daily<-by(MyData$st1,MyData$date,mean)
>> Error in tapply(seq_len(0L), list(`MyData$date` = c(913L, 914L, 925L, :
>> arguments must have same length
>>
>>
>> This is particul
Hi Rod,
How about this?
scenarios <- expand.grid(A = c("pass", "fail"), B = c("pass", "fail"), C =
c("pass", "fail"), D = c("pass", "fail"), E = c("pass", "fail"))
scenarios$F<-ifelse(scenarios$E=="pass","fail","pass")
Jim
On Thu, Aug 2, 2018 at 11:20 AM, R Stafford wrote:
> I have 6 variable
Hi Petr,
I recently had to align the minima of deceleration events to form an
aggregate "braking profile" for different locations. It seems as
though you are looking for something like:
find_increase<-function(x,surround=10) {
inc_index<-which.max(diff(x))
indices<-(inc_index-surround):(inc_inde
Hi Kenneth,
My guess is that you have tried to send screenshots of your output and
these were blocked. Try to cut and paste the output into your message.
Jim
On Tue, Aug 7, 2018 at 6:38 PM, John wrote:
> On Mon, 6 Aug 2018 20:18:38 +0200
> kenneth Barnhoorn wrote:
>
> Your examples did not app
Hi malika,
You don't seem to have defined your functions correctly. For example:
H<-function(u,x1)
would define an empty function H if that command worked, but it doesn't
Jim
On Tue, Aug 7, 2018 at 3:51 PM, malika yassa via R-help
wrote:
> hellothis is my programmeyou can help me, i cann't fo
Hi Alex,
I don't use Ubuntu, but if I saw that error message I would upgrade my
C++ compiler and try again. With luck, this is what caused the cascade
of errors beneath it.
Jim
On Fri, Aug 10, 2018 at 1:36 AM, Alexandra Thorn
wrote:
> Hi all,
>
> Following some updates to R that I received via S
Hi Farid,
Whatever you attached has not gotten through.
Jim
On Sat, Aug 11, 2018 at 6:47 PM, Farid Ch wrote:
> Hi all,
>
> Please check the attached file.
>
> Thanks
> Farid
>
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>
Hi Stefano,
This was such a stinker of a problem that I just had to crack it:
# create some data the lazy man's way
year_dates<-c(paste(2000,rep("01",31),formatC(1:31,width=2,flag=0),sep="-"),
paste(2000,rep("02",29),formatC(1:29,width=2,flag=0),sep="-"),
paste(2000,rep("03",31),formatC(1:31,wid
Hi citc,
Try this:
geac<-matrix(c(9,9,8,8,8,23,23,23,23,22,27,27,27,25,24,
19,19,19,20,20,17,17,17,18,19,8,8,8,9,9,2,2,3,3,3),ncol=5,byrow=TRUE)
library(plotrix)
barp(geac,names.arg=2014:2018,main="A level grades chemistry",
xlab="Year",ylab="Percentage of each grade",ylim=c(0,30),
col=c("white
pink"))
and I thank you for alerting me to the fact that the legend arguments
in barp don't position the legend properly. I'll fix it.
Jim
On Sun, Aug 19, 2018 at 9:52 AM, Rolf Turner wrote:
>
> Jim:
>
> (a) There's no legend.
>
> (b) I am still curious as to
Hi Phillip,
Jose has the correct answer. You probably missed this sentence in the
"Note" section of the help page:
"If the date string does not specify the date completely, the returned
answer may be system-specific."
In your case, the function throws up its hands and returns NA as you
haven't sp
Hi David,
As you want the _values_ of Year from the initial data frame appended
to the _names_ of GW_Elevation, you can't do it the easy way:
dddf<-read.table(text="LocationDateYear GW_Elevation
127(I)5/14/2006 2006 752.46
119(I)5/14/2006 2006
Hi Richard,
This may be what you want:
data(mtcars)
m<-list()
for(i in 1:6) {
rhterms<-paste(paste0("I(hp^",1:i,")"),sep="+")
lmexp<-paste0("lm(mpg~",rhterms,",mtcars)")
cat(lmexp,"\n")
m[[i]]<-eval(parse(text=lmexp))
}
plot(mpg~hp,mtcars,type="n")
for(i in 1:6) abline(m[[i]],col=i)
Jim
On
Hi Pedro,
I have encountered similar situations in a number of areas. Great care
is taken to record significant events of low probability, but not the
non-occurrence of those events. Sometimes this is due to a problem
with the definition of non-occurrence. To use your example, how close
does an ani
Hi Tanya,
Have you looked at the return value of pheatmap?
ret<-pheatmap(counts_filtered_df,scale="row",cluster_col=FALSE,
cluster_row=TRUE,border_color=NA,show_rownames = TRUE)
str(ret)
names(ret$tree_row)
names(ret$tree_col)
Look at what is in "ret" to see if your numeric matrix is hidden ther
Hi Philip,
This may work:
library(dplyr)
`RefDate` <- as.Date(c("2010-11-1","2010-12-01","2011-01-01"))
`Number of vegetables` <- c(14,23,45)
`Number of people` <- c(20,30,40)
MyData <- data.frame(RefDate,`Number_of_vegetables`,
`Number_of_people`,check.names=FALSE)
MyVars <- c("Number of vegetab
Hi Akshay,
Try this:
table(cut(xht,breaks=seq(0,10,by=2)))
Jim
On Sat, Sep 8, 2018 at 8:26 PM akshay kulkarni wrote:
>
> dear members,
> I am facing difficulties in plotting histograms
> in R in Linux CLI.
>
> Is there a function in R which produces a table of freq
Hi David,
If you mean that you have two data frames named x and y and want the
correlations between the columns that would be on the diagonal of a
correlation matrix:
r<-list()
for(i in 1:n) r[[i]]<-cor(x[,i],y[,i])
If I'm wrong, let me know.
Jim
On Mon, Sep 10, 2018 at 3:06 PM David Disabato
Hi Kevin,
It might be just as easy to write R scripts that would do basic
analyses. Users could "source" these scripts in an R session or from
the command line. The scripts would be much more compact than the .exe
files that you describe.
Jim
On Tue, Sep 11, 2018 at 8:06 AM Kevin Kowitski via R-h
Hi Kristy,
Try this:
colname.mat<-combn(paste0("year",1:4),2)
samplenames<-apply(colname.mat,2,paste,collapse="")
k<-1
for(column in 1:ncol(colname.mat)) {
assign(samplenames[column],replicate(k,sample(unlist(daT[,colname.mat[,column]]),3,TRUE)))
}
Then use get(samplenames[1]) and so on to
l(colname.mat)) {
>
> assign(samplenames[column],replicate(k,sample(unlist(daT[,colname.mat[,
> column]]),3,TRUE)))
>
> }
>
> > get(samplenames[1])
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> year224 0.556 0.667 0.571 0.526 0.629 0.696 0.323
Hi Sonam,
You're right. Although the cex.axis argument is present, it doesn't
seem to be used. I will have to debug this, which may take a day or
two.
Jim
On Tue, Sep 11, 2018 at 6:37 PM Sonam Sandeep Dash
wrote:
>
> Respected Sir,
> I have created a Taylor plot using the plotrix package. Howeve
Hi Abou,
Surprisingly you can't omit the x axis in dotchart. This hack will work:
sink("dotchar_noax.R")
sink()
Edit the resulting file by joining the first two lines with the
assignment symbol (<-), delete the two lines at the bottom and comment
out the line "axis(1)".
source("dotchart.noax.R")
agnes (cluster)
Jim
On Wed, Sep 12, 2018 at 8:10 AM Bryan Mac wrote:
>
>
> Bryan Mac
> Data Scientist
> Research Analytics
> Ipsos Insight LLC
>
>
>
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSU
Hi Roslinazairimah,
You seem to be using the dotplot function from the lattice package. If so:
dotplot(cyl ~ mpg, data = mtcars, groups = cyl, cex=1.2,
scales=list(y=list(labels=sort(unique(mtcars$cyl,
main="Gas Milage for Car Models", xlab="Miles Per Gallon")
Jim
On Wed, Sep 12, 2018 at 4
Hi Rich,
Sometimes an empty image is due to not closing the image file. If you
forgot to put:
dev.off()
after the plotting commands, or there was an error in the plotting
commands, the file may be left open. Try manually entering "dev.off()"
after you have run the code. If you don't get an error:
Hi Fang,
Let's assume that you are using the "binom.confint" function in the
"binom" package and you have made a spelling mistake or two. This
function employs nine methods for estimating the binomial confidence
interval. Sadly, none of these is "lrt". The zero condition is
discussed in the help pa
Hi Luigi,
Maybe this:
testdf<-data.frame(A=1,B=2,C=3)
> testdf
A B C
1 1 2 3
toNull<-function(x) return(NULL)
testdf<-sapply(testdf,toNull)
Jim
On Thu, Sep 27, 2018 at 5:29 PM Luigi Marongiu wrote:
>
> Dear all,
> I would like to erase the content of a dataframe -- but not the
> dataframe itsel
ction, I get:
>
> > df
> $A
> NULL
>
> $B
> NULL
>
> $C
> NULL
>
> > str(df)
> List of 3
> $ A: NULL
> $ B: NULL
> $ C: NULL
>
> The dataframe has become a list. Would that affect downstream applications?
>
> Thank you,
> Lui
tion(x) return(NULL)
> > df<-as.data.frame(sapply(df,toNull))
> > df
> data frame with 0 columns and 0 rows
> > str(df)
> 'data.frame': 0 obs. of 0 variables
> On Thu, Sep 27, 2018 at 10:12 AM Jim Lemon wrote:
> >
> > Ah, yes, try 'as.
Hi Knut,
As Bert said, you can start with diff and work from there. I can
easily get the text for the subset, but despite fooling around with
"parse", "eval" and "expression", I couldn't get it to work:
# use a bigger subset to test whether multiple runs can be extracted
kkdf<-subset(airquality,Te
Bugger! It's
eval(parse(text=paste0("kkdf[c(",paste(starts,ends,sep=":",collapse=","),"),]")))
What a mess!
Jim
On Fri, Sep 28, 2018 at 8:35 AM Jim Lemon wrote:
>
> Hi Knut,
> As Bert said, you can start with diff and work from there. I
Hi Luigi,
An easy way is to use "points" to overplot the outliers:
grbxp<-boxplot(dfA$Y ~ dfA$X,
ylim=c(0, 200),
col="green",
ylab="Y-values",
xlab="X-values"
)
points(grbxp$group,grbxp$out,col="green")
On Fri, Sep 28, 2018 at 7:51 PM Luigi Marongiu wrote:
>
Hi David,
I think you want this:
SampledWells <- MyData[!( MyData$Location %in% c("MW-09", "MW-10")), ]
Jim
On Thu, Oct 4, 2018 at 9:02 AM David Doyle wrote:
>
> I'm sure this is a simple question but I'm not sure where to find the
> answer.
>
> I want to remove some of the data. For example w
Hi Subhamitra,
Where I think the error arises is in the line:
N<-nrow(mat)
Since we don't know what "mat" is, we don't know what nrow(mat) will
return. If "mat" is not a matrix or data frame, it is likely to be
NULL. Try this:
print(N)
after defining it and see what it is.
Jim
On Sat, Oct 6,
ibrary(pracma)
>
> N<-nrow(ts)
> r<-matrix(0, nrow = N, ncol = 1)
> for (i in 1:N){
> r[i]<-approx_entropy(ts[i,], edim = 2, r = 0.2*sd(ts[i,]), elag = 1)
> }
>
> After calculating for 1 series, I need to calculate the same things for the
> multiple s
ritten a code
>>
>> library(pracma)
>>
>> N<-nrow(ts)
>> r<-matrix(0, nrow = N, ncol = 1)for (i in 1:N){
>> r[i]<-approx_entropy(ts[i,], edim = 2, r = 0.2*sd(ts[i,]), elag = 1)}
>>
>> * After calculating for 1 series, I need to c
Hi Marc,
I do this quite often with Tcl-Tk. All you have to do is make sure
that the PATH contains the correct location for the R executable if
you don't specify it within your program.
Jim
On Wed, Oct 10, 2018 at 1:58 AM Marc Capavanni via R-help
wrote:
>
> Hi there,
>
> I'm currently using R on
Hi Leslie,
Keeping track of any sort of text (or music or pictures) can be
challenging when the number of files becomes large. One way to
organize a large collection is to categorize it in some way that makes
sense to you. Say you create a directory structure:
R_code___
Hi Tranh,
I'm not sure why you are converting your variables to factors, and I
think the model you want is:
lm(KIC~temperature+AC+AV+Thickness+temperature:AC+
temperature:AV+temperature:thickness+AC:AV+
AC:thickness+AV:thickness,SR)
Note the colons (:) rather than asterisks (*) for the interact
Hi again,
Two things, I named the data frame SR as shown in the model.
The other is for those who may wish to answer the OP. The mediafire
website is loaded with intrusive ads and perhaps malware.
Jim
On Thu, Oct 11, 2018 at 9:02 AM Jim Lemon wrote:
>
> Hi Tranh,
> I'm not su
Yes, I thought that as well and had worked out this but didn't send it:
add_Pscores<-function(x) {
return(sum(unlist(x),na.rm=TRUE))
}
by(rzdf[,c("PO1M", "PO1T", "PO2M", "PO2T")],rzdf$STUDENT_ID,FUN=add_Pscores)
rzdf$STUDENT_ID: AA15285
[1] 724.8
---
Hi Roberto,
Here is a snippet of code that translates the text responses of the
BIS-11 into numeric values. Note the reversal of the order in the
second item:
BIS$Q1<-as.numeric(factor(BIS$Q1,
levels=c("Almost","Often","Occasionally","Rarely/Never")))
BIS$Q2<-as.numeric(factor(BIS$Q2,
levels=c("
Hi Myriam,
This may not be the ideal way to do this, but I think it works:
mcdf<-read.table(text="41540 41540 41442 41599 41709 41823 41806 41837
41898 41848
41442 0.001
41599 0.002 0.001
41709 0.004 0.003 0.003
41823 0.002 0.001 0.002 0.001
41806 0.004 0.004 0.005 0.006 0.005
41837 0.004 0.004 0.
low but could not find the right answer
> Best Roberto
>
>
> Op do 1 nov. 2018 om 00:25 schreef Jim Lemon :
>>
>> Hi Roberto,
>> Here is a snippet of code that translates the text responses of the
>> BIS-11 into numeric values. Note the reversal of the order in the
&g
> dataname$"very long name".
> So, is ther a way to do this procedure for all columns?
>
> Roberto
>
>
> Op do 1 nov. 2018 om 10:50 schreef Jim Lemon :
>>
>> Hi Roberto,
>> What I suggested is a brute force method of translating response
>&g
ginal column lables which
> I need in my barplot. Isn't it?
>
> Op do 1 nov. 2018 om 11:03 schreef Jim Lemon :
>>
>> I would use the "names" or "colnames" functions to change them to Q1,
>> Q2, ... as I did.
>>
>> Jim
>>
>>
Hi Ferri,
One way is to snip out a Google Maps image of the area you want, then
using the "maps" package, start a plot bounded by the corner
coordinates of your Google Maps image. You can get those by clicking
on the corners of the area that you selected. Then use the
"readbitmap" package to create
Hi Medic,
Perhaps this:
medic_df<-read.table(text="name number
ds6277
lk 24375
ax46049
dd70656
az216544
df 220620
gh641827",
header=TRUE)
library(plotrix)
options(scipen=10)
barp(medic_df$number,names.arg=medic_df$name,width=0.5)
As others have noted, this is really a
Hi lily,
Something like this should work:
DF1<-read.table(text=
"latitude longitude Precip
45.5 110.5 3.2
45.5 1115.0
45.5 111.5 1.8
45.5 1122.0
46 110.5 6.1
46 1114.5
Hi Subhamitra,
As Bert noted, you are mixing base and grid graphics. Here is a simple
way to get a plot like what you described. It will probably take more
work to find what you actually do want and discover how to get it.
for(i in 1:38) assign(paste0("veh",i),rep(sample(10:35,1),10)+runif(10,-4,4
gn=signaturevirality5&;>
> Sender
> notified by
> Mailtrack
> <https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&;>
> 11/21/18,
> 7:02:18 AM
>
>
> On Wed, Nov 21, 2018 at 4:38 AM Jim Lemon wrote:
>
>>
teful to
> you.
>
> Thank you very much for your kind help.
>
>
>
> [image: Mailtrack]
> <https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&;>
> Sender
> notified by
> Mailtrack
> <https://mailtrack.
edium=signature&utm_campaign=signaturevirality5&;>
> Sender
> notified by
> Mailtrack
> <https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&;>
> 11/21/18,
> 9:12:14 AM
>
> On Wed, Nov 21, 2018 at 8
<https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&;>
> Sender
> notified by
> Mailtrack
> <https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&;>
> 11/21/18,
> 10:35:03 AM
1. xaxt="n" means "Don't display the X axis". See the help for "par" in the
graphics package
2. axis(1,at=1:nrows,labels=names(MPG3))
This means, "Display the bottom axis (1) with ticks at 1 to the number of
rows in the data frame"
"Use the values of MPG$Year as labels for the ticks". see the help
Hi Vicci,
It's very clunky, but I think it will do what you want.
rrdf<-read.csv(text="No,date,chamber,d13C,ppm_CO2,ppm_13CO2
1,10.14.2018 10:43 PM,IN,-0.192,439.6908,4.9382
2,10.14.2018 10:47 PM,101,-0.058,440.7646,4.9509
3,10.14.2018 10:50 PM,103,-1.368,535.6602,5.9967
4,10.14.2018 10:53 PM,1
Hi Ogbos,
If we assume that you have a 3 column data frame named oodf, how about:
oodf[,4]<-floor((cumsum(oodf[,1])-1)/28)
col2means<-by(oodf[,2],oodf[,4],mean)
col3means<-by(oodf[,3],oodf[,4],mean)
Jim
On Wed, Nov 28, 2018 at 2:06 PM Ogbos Okike wrote:
>
> Dear List,
> I have three data-column
Hi david,
The formatting of the data frame looks like the Province and Year
columns have gotten stuck together. This probably has something to do
with your Excel spreadsheet or the function that you are using to read
it in. If there is are fewer column names than columns, this error is
likely to ha
Hi Dagmar,
This will probably involve creating a variable to differentiate the
two days in each data.frame:
myframe$day<-as.Date(as.character(myframe$Timestamp),"%d.%m.%Y %H:%M:%S")
days<-unique(myframe$day)
Then just sample the two subsets and concatenate them:
myframe[c(sample(which(myframe$da
Hi Ogbos,
Here is a slight modification of a method I use to display trip
density on a map:
oolt<-read.table(text="Lat Lon
30.1426 104.7854
30.5622 105.0837
30.0966 104.6213
29.9795 104.8430
39.2802 147.7295
30.2469 104.6543
26.4428 157.7293
29.4782 104.5590
32.3839 105.3293
26.4746 157.8411
25.10
ke to add horizontal color bar legend that could be used to
> > explain the number of lightning counts at different points on the
> > latitude band as plotted.
> >
> > Thank you
> > Warm regards
> > Ogbos
> >
> > On Sun, Dec 9, 2018 at 9:46 PM Jim Lemon
Hi Ogbos,
Back on the air after a few days off. I don't have your data ("QUERY
2"), but I think this will fix your problem.
library(maps)
map("world")
box()
library(plotrix)
color.legend(-180,-150,100,-130,legend=c(0,25000,5,75000,10),
rect.col=color.scale(1:5,extremes=c("blue","red")),gr
track.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&;>
>>> Sender
>>> notified by
>>> Mailtrack
>>> <https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&;>
>&g
Hi Subhamitra,
Thanks. Now I can provide some assistance instead of just complaining. Your
first problem is the temporal extent of the data. There are 8613 days and
6512 weekdays between the two dates you list, but only 5655 observations in
your data. Therefore it is unlikely that you have a comple
Hi Ek,
I thought there would be a simple fix for this, but had to write a
little function:
fillList<-function(x) {
maxrows<-max(unlist(lapply(x,length)))
return(lapply(x,"[",1:maxrows))
}
that fills up the rows of each list with NAs. I got the expected result with:
testlist<-list(a=1:8,b=1:9,c
uld like to discuss another 2 queries.
>>
>> Thank you very much Sir for educating a new R learner.
>>
>> [image: Mailtrack]
>> <https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&;>
>> Sender
>> notified by
Hi Subhamitra,
As for the error that you mention, it was probably:
Error in axis(1, at = year_mids, labels = 3 - 1 - 1994:3 - 8 - 2017) :
'at' and 'labels' lengths differ, 24 != 1992
Anything more than a passing glance reveals that you didn't read the
explanation I sent about the arguments pass
Hi Subhamitra,
My apologies, I caught a mistake. To have the first tick in the middle of
the first year, you want half of the _observations_ in a year, not half of
the days. As I now have your data at my fingertips:
3567/15.58385
[1] 228.8908
Almost exactly what was calculated for the first serie
Hi Tasha,
I may be right off the track, but you could plot RGB proportions on a
3D plot. The easiest way I can think if would be to convert your 0-255
values to proportions:
rgb_prop<-read.table(text="Red Green Blue pct
249 158 37 56.311
249 158 68 4.319
249 158 98 0.058
249 128 7 13.965
249 128 3
Gunter
>
> "The trouble with having an open mind is that people keep coming along and
> sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Tue, Dec 18, 2018 at 3:10 PM Jim Lemon wrote:
>>
>> H
Hi Ek,
Look at unlist and the argument "recursive". You can step down through
the levels or a nested list to convert it to a single level list.
Jim
On Thu, Dec 20, 2018 at 1:33 PM Ek Esawi wrote:
>
> Thank you Bert. I don't see how unlist will help. I want to combine
> them but keep the "rectang
file = "Temp.txt",append = TRUE,quote = TRUE)}
> Error in (function (..., row.names = NULL, check.rows = FALSE,
> check.names = TRUE, :
> arguments imply differing number of rows: 3, 55, 56, 53, 54, 16, 21,
> 23, 50, 24
>
>
> On Wed, Dec 19, 2018 at 9:36 PM
n. Sometimes a bit tricky but quite
>> handy, especially when you consider some grouping.
>>
>> Cheers
>> Petr
>>
>> >
>> > -- Bert
>> >
>> >
>> > Bert Gunter
>> >
>> > "The trouble with having an open mi
Hi Rachel,
I'll take a guess and assume that you are monitoring the mobile phones
of 36 people, adding an observation every time some specified change
of state is sensed on each device. I'll also assume that you are only
recording four types of measurement. It seems that you want to
aggregate these
a new data frame etc, but I still feel a
> bit lost. Do I need to make one per subject or per Probe etc..
>
>
> Thanks for your help. I hope that you can help me resolve this issue.
>
>
> Best,
>
>
> Rachel
>
>
>
>
>
>
> On Sat, Jan 5, 2019
Hi Meriam,
I don't have the packages loaded that you use, but a first guess would
be to start a wider device. For example, the default x11 device is
7x7, so:
x11(width=10)
would give you a rectangular output device that might move the columns
of labels outward. The same applies for any other devi
Hi Halllie,
As Jeff noted, a data frame is not a matrix (it is a variety of list),
so that looks like your problem.
hkdf<-data.frame(sample(3:5,4,TRUE),sample(1:3,4,TRUE),sample(2:4,4,TRUE),
sample(3:5,4,TRUE),sample(1:3,4,TRUE),sample(2:4,4,TRUE))
library(irr)
kripp.alpha(hkdf)
kripp.alpha(as.ma
Hi Hallie,
If I understand your email correctly, you have four repeated
observations by the three raters of the same six variables. This is a
tougher problem and I can't solve it at the moment. I'll return to
this later and see if I can offer a solution.
Jim
On Wed, Jan 30, 2019 at 3:56 AM Halli
),occasion=rep(rep(1:4,each=6),3),
var=rep(rep(1:6,4),3),obs=runif(72))
library(cccrm)
cccUst(hkdf,"obs","rater","occasion")
cccvc(hkdf,"obs","rater","occasion")
Jim
On Wed, Jan 30, 2019 at 1:01 PM Jim Lemon wrote:
>
> Hi Hallie,
501 - 600 of 3432 matches
Mail list logo
