Hi Rolf,
Following the docs back to draw.key, It looks like the ellipsis
argument is ignored. I was hoping for a brilliant solution along the
lines of:
adj=0
that could be passed down the functions like a hot potato, but was disappointed.
Jim
On Tue, Dec 24, 2019 at 9:26 AM Rolf Turner wrote:
Hi Kimberley,
Given the number of posts that read "I have a problem, please advise",
your concern for our mental welfare is a great Xmas present.
Jim
On Tue, Dec 24, 2019 at 10:38 AM Steinmann, Kimberly@CDPR
wrote:
>
> I am not sure how to close the thread - I hate to waste anyone's time on a
Hi Paul,
For your first question:
max(file.info(".")$mtime)
[1] "2019-12-21 21:04:19 AEDT"
As for the second, I didn't know what an AIS file was, so I googled
it. I still don't know, so I don't have a clue how to turn a string
like that into a number.
Jim
On Tue, Dec 24, 2019 at 5:14 AM Paul
Hi Kimberley,
Since you are using a loop and therefore testing one value of
v_trends_lbs at a time, the "&" in the "if" statement should be "&&".
Pinching Bert's example but using a for loop instead of ifelse:
x <- seq(-2,2,.25)
v_lbs<-rep("",length(x))
for(i in 1:length(x)) {
if(is.na(x[i]))
Hi Neha,
Well, that's a clue to why you are getting NAs:
log10(0)
[1] -Inf
Another possibility is that the values used in the initial calculation
have been read in as factors.
Jim
On Mon, Dec 23, 2019 at 10:55 AM Neha gupta wrote:
>
> Hi Jim
>
> The objective function is passed to san_res
Hi Neha,
The error message looks suspicious, as it refers to "all the MAEs"
while there is only one NA value in the summary. I would carefully
check the object that you are passing to san_res.
Jim
On Mon, Dec 23, 2019 at 4:17 AM Neha gupta wrote:
>
> I am using the following code to tune the 4
Hi Thomas,
Perhaps this is what you are seeking:
my_read_excel<-function(filename) {
serials<-read_excel(filename,sheet="Flow Data",range=("c6"))
flow.data<-read_excel(filename,sheet="Flow Data",range=("c22:c70"))
dates<-read_excel(filename,sheet="Flow Data",range=("h14"))
I'm probably misunderstanding what you want. I get this from the code I sent:
VC
[[1]]
Prob.of.exceedance_1 Prob.of.exceedance_2 Prob.of.exceedance_3
100 0.005343027
Prob.of.exceedance_4
1 0.01947477
[[2]]
Prob.of.exceedance_1
Hi Ioanna,
We're getting somewhere, but there are four unique combinations of
Taxonomy and IM.type:
ER+ETR_H1,PGA
ER+ETR_H2,PGA
ER+ETR_H1,Sa
ER+ETR_H2,Sa
Perhaps you mean that ER+ETR_H1 only occurs with PGA and ER+ETR_H2
only occurs with Sa. I handled that by checking that there were any
rows
Hi Ioanna,
For simplicity assume that the new data frame will be named E:
E<-D[,c("Taxonomy","IM.type",paste("VC,1:4,sep="_"))]
While I haven't tested this, I'm pretty sure I have it correct. Just
extract the columns you want from D and assign that to E.
Jim
On Fri, Dec 20, 2019 at 9:02 PM
Hi Ioanna,
I looked at the problem this morning and tried to work out what you
wanted. With a problem like this, it is often easy when you have
someone point to the data and say "I want this added to that and this
multiplied by that". I have probably made the wrong guesses, but I
hope that you can
Hi Medic,
mydata$Temperature[sample(1:N,N)
should do the trick. You will just get a pseudo-randomly shuffled set
of the same values.
Jim
On Thu, Dec 19, 2019 at 7:34 AM Medic wrote:
>
> Variable temperature:
> mydata$temperature
> has N values.
> With what code to сhoice (without return) n
0435,0.000405)
> )
>
> Basically I am using the total probability theorem to calculate a best
> estimate. I am stuck how to do it for many cases. Many thanks for your
> patience.
>
> -Original Message-
> From: Jim Lemon [mailto:drjimle...@gmail
Hi Ana,
Is this what you want?
a<-read.table(text="GENErs BETA
1 ENSG0154803 rs2605134 0.0360182
2 ENSG0154803 rs7405677 0.0525463
3 ENSG0154803 rs7211573 0.0525531
4 ENSG0154803 rs2746026 0.0466392
5 ENSG0141030 rs2605134 0.0806140
6 ENSG0141030
Hi Philip,
This may be a starter:
attach(airquality)
heights <- tapply(Temp,Month,mean)
temp_sd<-tapply(Temp,Month,sd)
lower <- tapply(Temp,Month,function(v) t.test(v)$conf.int[1])
upper <- tapply(Temp,Month,function(v) t.test(v)$conf.int[2])
library(plotrix)
Hi Francesca,
Do you want something like this?
Jim
On Thu, Dec 5, 2019 at 6:58 PM Francesca wrote:
>
> Hi, sorry for bothering again.
> I was wondering how I can reshape the data, if in your code,
> I would like to have only two panels, where in the panel with Participation
> =0, I represent
Hi vod,
Now that I have your data I can do a bit better:
# save data as "vv.csv"
vv<-read.csv("vv.csv")
# order the points in order of theta
vv<-vv[order(vv$theta),]
oldpar<-polar.plot(vv$r,vv$theta,main="Polar plat of vv.csv",lwd=3,line.col=4,
Hi val,
You had a "conditional leak" in your ifelse statements:
dat2 <-read.table(text="ID d1 d2 d3
A 0 25 35
B 12 22 0
C 0 0 31
E 10 20 30
F 0 0 0",
header=TRUE,stringsAsFactors=FALSE)
dat2$d4<-
ifelse(dat2$d1,dat2$d1,ifelse(dat2$d2,dat2$d2,ifelse(dat2$d3,dat2$d3,0)))
Even though it works, it
Hi Chelsea,
A brute force method, but I think it does what you want:
# create a sequence of integers to make checking easy
null.cz<-1:68
separate_interdigitated_vectors<-function(x,nv=2,vlen=17) {
xlen<-length(x)
starts<-seq(1,xlen-vlen*nv+1,by=vlen*nv)
cat(xlen,starts,"\n")
for(start in
Hi Josh,
I couldn't work out how to do this in ggplot, but here is a possible solution:
tagSummary<-read.csv(text="speciesSci,recvDeployName,nDet
Arenaria interpres,Bucktoe Preserve,96
Arenaria interpres,CHDE,132
Arenaria interpres,Fortescue,22133
Arenaria interpres,Mispillion,2031
Arenaria
While the remedy for your dissatisfaction with my previous solution
should be obvious, I will make it explicit.
# that is rows containing at most one value > 0.8
# ignoring the diagonal
keeprows<-apply(ro246,1,function(x) return(sum(x>0.8)<2))
ro246.lt.8<-ro246[keeprows,keeprows]
Jim
I thought you were going to trick us. What I think you are asking now
is how to get the variable names in the columns that have at most one
_absolute_ value greater than 0.8. OK:
# I'm not going to try to recreate your correlation matrix
calc.jim<-matrix(runif(100,min=-1,max=1),nrow=10)
for(i in
Hi Ana,
Rather than addressing the question of why you want to do this, Let's
get make the question easier to answer:
calc.rho<-matrix(c(0.903,0.268,0.327,0.327,0.327,0.582,
0.928,0.276,0.336,0.336,0.336,0.598,
0.975,0.309,0.371,0.371,0.371,0.638,
0.975,0.309,0.371,0.371,0.371,0.638,
so far that this kind of QQ plot is an indication that
> data has non zero mean:
> https://stats.stackexchange.com/questions/280634/how-to-interpret-qq-plot-not-on-the-line
>
> but is that an indication that something is wrong with the analysis?
>
> Thanks
> Ana
>
>
I thought about this and did a little study of GWAS and the use of
p-values to assess significant associations. As Ana's plot begins at
values of about 0.001, this seems to imply that almost everything in
the genome is associated to some degree. One expects that most SNPs
will not be associated
Hi Baojun,
You probably need the right condition:
x<-c(1,2,3.1415926535,4)
> nonint<-floor(x)!=x
> nonint
[1] FALSE FALSE TRUE FALSE
> x[nonint]<-round(x[nonint],3)
> x
[1] 1.000 2.000 3.142 4.000
Jim
On Fri, Nov 8, 2019 at 12:54 AM Baojun Sun wrote:
>
> I have a Excel spreadsheet with two
ually everything is good, but I want to 'kill' my curiosity.
>
> year<-c(1981:2015)
> month<-c(3, 1, 12, 11, 2, 1, 12, 1, 2, 1, 12, 3, 2, 12, 2, 7, 2, 6, 2, 1, 1,
> 12, 3, 12, 3, 12, 2, 2, 9, 2, 1, 4, 12, 3, 4)
> plot(month~year,xaxt="n", type="b", yla
Hi Ani,
There are a number of ways to modify this sort of plot. Here is one:
x11(width=7,height=5)
par(cex.axis=.8)
plot(month~year,xaxt="n", type="l", ylab="Month", xlab="Year",
main="Month of occurrence in year")
axis(1,at=seq(1981,2014,3))
library(plotrix)
Hi Greg,
I tried this:
cp<-read.table(text="Birey Grup Time y
11 Cp1 0.7916386
11 Cp3 1.7463777
11 Cp7 1.2008390
11 Cp14 0.6311380
11 Cp21 2.1563557
11 Cp28 1.2008390",
header=TRUE)
library(prettyR)
xtab(Grup~Time,cp)
Crosstabulation of Grup by Time
Time
Homework Chandeep, homework.
Jim
On Tue, Nov 5, 2019 at 9:40 PM Chandeep Kaur wrote:
>
> Dear Team,
>
> Could you please help me with the below question? How can I get the desired
> output?
>
> Produce the following sequence using only rep(), seq() and potentially
> other functions/operators.
Hi Christofer,
This is a guess, but have you tried:
save(AAA31,file="Save.RData")
Jim
On Thu, Oct 31, 2019 at 8:10 PM Christofer Bogaso
wrote:
>
> Hi,
>
> I wanted to save a few R objects in RData file for some future use.
> The names of such R objects are actually dynamic so I used below code
Hi Thomas,
Perhaps you should be doing something like writeLines(txt[1],...) or just:
sink("10619.txt")
cat(txt[1])
sink()
Jim
On Thu, Oct 31, 2019 at 4:48 PM Thomas Subia wrote:
>
> Colleagues,
>
> I'm trying to convert a pdf to a text file with the following code.
>
> # pdf to excel
>
Hi Ana,
Seems to work without error for me:
# installed qvalue_1.26.0 from CRAN archive
library(qvalue)
pvals<-c(6.919239e-02,1.073784e-01,1.218613e-01,1.586202e-01,
1.370340e-01,3.452574e-02,2.545619e-01,1.676715e-02,8.571197e-01,
8.649025e-01,1.777414e-02,6.801867e-01,6.873085e-01,
in via SQL or whatever language is available.
Jim
On Thu, Oct 24, 2019 at 10:17 AM Ana Marija wrote:
>
> no can you please send me an example how the command would look like in my
> case?
>
> On Wed, Oct 23, 2019 at 6:16 PM Jim Lemon wrote:
> >
> > Yes. Have you tri
t; so my example would not reproduce the error
>
> On Wed, Oct 23, 2019 at 6:05 PM Jim Lemon wrote:
> >
> > Hi Ana,
> > When I run this example taken from your email:
> >
> > l4<-read.table(text="X1 X2 X3 X4 X5 variant_id pval_nominal gene_id.LCL
> > ch
Ah, it looks like a memory allocation problem.
Jim
On Thu, Oct 24, 2019 at 10:05 AM Ana Marija wrote:
>
> I also tried left_join but I got: Error: std::bad_alloc
>
> > df3 <- left_join(l4, asign, by = c("chr","pos"))
> Error: std::bad_alloc
> > dim(l4)
> [1] 166941635 8
> > dim(asign)
>
Hi Ana,
When I run this example taken from your email:
l4<-read.table(text="X1 X2 X3 X4 X5 variant_id pval_nominal gene_id.LCL
chr1 13550 G A b38 1:13550:G:A 0.375614 ENSG0227232
chr1 14671 G C b38 1:14671:G:C 0.474708 ENSG0227232
chr1 14677 G A b38 1:14677:G:A 0.699887
w more once I'm done.
>
> Can you suggest any other sources?
>
> Thanks.
>
> -Original Message-
> From: Jim Lemon
> Sent: Tuesday, October 22, 2019 3:26 PM
> To: Phillip Heinrich
> Cc: r-help
> Subject: Re: [R] If Loop I Think
>
> Hi Philip,
> Try this:
>
Hi Philip,
Try this:
phdf<-read.table(
text="Row Outs RunnerFirst RunnerSecond RunnerThird R1 R2 R3
1 0
2 1
3 1
4 1 arenn001
5 2 arenn001
6 0
7 0 perad001
8 0 polla001 perad001
9 0 goldp001 polla001 perad001
10 0 lambj001 goldp001
11 1 lambj001 goldp001
12 2 lambj001
13 0
14 1 ",
Hi Yeasmin,
I suspect that you didn't intend to have conditions like:
a<0 && b>0 && b 0 && abs(b) < abs(a)
If this is the case, the following function seems to return the values
of phase that you want:
assign_phase<-function(x,y) {
phase<-c(1,2,7,8,3,4,6,5)
phase_index<-4 * (x > 0) + 2 * (y >
Hi Subhamitra,
This is not the only way to do this, but if you only want the monthly
averages, it is simple:
# I had to change the "soft" tabs in your email to commas
# in order to read the data in
spdf<-read.table(text="PERMNO,DATE,Spread
111,19940103,0.025464308
111,19940104,0.064424296
I have thought about this one myself, and just reading the posts and
links has afforded me a more informed viewpoint. My guess is that it
boils down to a contest between mathematics and prosody. To speakers
of English, "square" in the mathematical sense implies the active
form such as "I square
Hi ani,
Sorry, a typo in the function - should be:
makeNA(x)<-function(x,varname,value) {
x[,varname][x[,varname]==value]<-NA
return(x)
}
Jim
On Fri, Oct 18, 2019 at 2:01 PM Jim Lemon wrote:
>
> Hi ani,
> You say you want to replace with NA, so:
>
> # it will be ea
Hi ani,
You say you want to replace with NA, so:
# it will be easier if you don't use numbers for the names of the data frames
names(test) <- paste0("Y",1986:2015)
makeNA(x)<-function(x,varname,value) {
x[,varname][x[,varname]<-value]<-NA
return(x)
}
lapply(test,makeNA,list("RR",))
gt; It`s like this?
>
> barplot(CONTBR_RESULT[order(CONTBR_RESULT)][16:30])
>
> Regards
> Jorge
>
> On Wed, 16 Oct 2019 at 02:51, Jim Lemon wrote:
>>
>> Hi Jeff,
>> Let's say you have the following data:
>>
>> set.seed(12345)
>> CONTBR_RESULT<-s
Hi Jeff,
Let's say you have the following data:
set.seed(12345)
CONTBR_RESULT<-sample(20:200,30)
If you don't mind ordering the results, you can do this:
barplot(rev(sort(CONTBR_RESULT))[1:15],...)
If you want the values in the original order:
barplot(CONTBR_RESULT[order(CONTBR_RESULT) >
Hi Nancy,
The chickwts dataset contains one sort-of continuous variable (weight)
and a categorical variable (feed). Two things that will help you to
understand what you are trying to do is to "eyeball" the "weight"
data:
# this shows you the rough distribution of chick weights
Hi Jui-Kun,
Are you using the plm package and talking about a pvcm object that is
returned by the function of the same name? If so, it is not a "file"
but a list of values returned by the function. Maybe this will move
your question in the direction of intelligibility.
Jim
On Sat, Oct 5, 2019 at
Hi April,
Try this:
# this could be done from a file
textlines<-read.table(text="color=green
shape=circle
age=17
name=Jim",
stringsAsFactors=FALSE)
for(i in 1:length(textlines)) {
nextline<-unlist(strsplit(textlines[i,1],"="))
assign(nextline[1],nextline[2])
}
color
[1] "green"
shape
[1]
uot;1","2","4": 1 4 2 1 1 2 2 2 3 1
> $ v3 : Factor w/ 4 levels "0","1","2","3": 1 1 1 2 2 1 3 3 4 1
> $ v4 : Factor w/ 4 levels "0","2","3","4": 1 1 1 1 1 1 1 3 4 2
> $ v5 : Factor w/
Hi Phillip,
The following seems to do what you want:
phdf<-read.table(text="v1 v2 v3 v4 v5 code
0 0 0 0 01
1 4 0 0 01
1 1 0 0 01
1 0 1 0 01
2 0 1 0 01
0 1 0 0 01
0 1 2 0 01
0 1 2 3 01
0 2 3 4 41
0 0 0 2 31",
Hi Ana,
You seem to have a p-value at the top of the second plot. Do you just
want that p-value in a different place?
My first guess would be the "annotate" argument. Say you wanted your
p-value in the middle of the plot.
# your ggplot line
p+annotate("text",x=1.5,y=0.05,label="p = 1.6x10-16")
p
Hi Leo,
The easiest way to do this is:
par(las=1)
# twoord.plot command
par(las=0)
Jim
On Sat, Sep 28, 2019 at 8:56 PM leo wu wrote:
>
> Dear Jim Lemon:
> I am currently using the twoord.plot() function in plotrix package. I
> want to inquire, how am I supposed to rotate my
On Sat, Sep 28, 2019 at 2:04 AM Martin Maechler
wrote:
>
> For back compatibility reasons, the old command line option will
> continue to work so the many shell and other scripts that use
> it, will not stop working.
>
That's a relief. I was getting worried that we would become:
The knights who
Hi Phillip,
Try this:
as.Date(c("20180329","20180330","20180331"),"%Y%m%d")
[1] "2018-03-29" "2018-03-30" "2018-03-31"
Note that the format argument has to match the date format exactly.
Jim
On Wed, Sep 25, 2019 at 9:54 AM Phillip Heinrich wrote:
>
> The date is imbedded in the GameID
Hi Ana,
Your minimum value in pvalR is very small and may be causing trouble.
As the qvalue package seems to be in Bioconductor, perhaps posting to
that help list would get an answer.
Jim
On Wed, Sep 25, 2019 at 1:48 AM Ana Marija wrote:
>
> Hello,
>
> I tried using qvalue function:
>
>
Hi Phillip,
While I really like Ana's solution, this might also help:
phdf<-read.table(text="Date count
2018-03-29 1
2018-03-29 1
2018-03-29 1
2018-03-30 1
2018-03-30 1
2018-03-30 1
2018-03-31 1
2018-03-31 1
2018-03-31 1",
Hi rhotuser,
Your question is really not about R, but about understanding IR
spectroscopy methods for soil composition. That's not my field, and
you will be lucky to find someone on this help list who is:
1) an expert in the field
2) willing to explain the methods used in the prospectr package
Hi Christophe,
Your call to pyramid.plot is okay. I would make a couple of suggestions:
par(mar=pyramid.plot(males.overweight,females.overweight,
top.labels=c("Males","Labels","Females"),laxlab=seq(0,60,by=10),
raxlab=seq(0,30,by=10),labels=agelabels,lxcol="#74c476",rxcol="#9e9ac8",
Hi Edward,
Say your "data frame" is named "epdat". This may do it:
epmat<-matrix(epdat[10:289],nrow=28)
colnames(epmat)<-sub("1","",names(epdat[10:289])[seq(1,270,by=28)])
This one looks like the Sorceror's Apprentice tangled with one of
those experimental schedule scripting programs.
Jim
On
Hi Cristofer,
If you just ask for a number of breaks, you will get what "hist"
thinks you should. Try this or something similar:
hist(x,breaks=seq(min(x),max(x),length.out=21))
Jim
On Wed, Sep 18, 2019 at 8:55 PM Christofer Bogaso
wrote:
>
> Hi,
>
> I have a numerical vector as below
>
> x =
Hi Jennifer,
This is one way:
library(plotrix)
pyramid.plot(my.dta[,1],my.dta[,2],
labels=c("Statement 1","Statement 2","Statement 3",
"Statement 4","Statement 5","Statement 6",
"Statement 7","Statement 8","Statement 9",
"Statement 10","Statement 11","Statement 12","Statement 13"),
Sorry, forgot to include the list.
On Sat, Sep 14, 2019 at 10:27 AM Jim Lemon wrote:
>
> See inline
>
> On Fri, Sep 13, 2019 at 11:20 PM Subhamitra Patra
> wrote:
>>
>> Dear Sir,
>>
>> Yes, I understood the logic. But, still, I have a few queries
Hi Subhamitra,
I'll try to write my answers adjacent to your questions below.
On Fri, Sep 13, 2019 at 6:08 PM Subhamitra Patra
wrote:
> Dear Sir,
>
> Thank you very much for your suggestion.
>
> Yes, your suggested code worked. But, actually, I have data from 3rd
> January 1994 to 3rd August
Hi Subhamitra,
Your data didn't make it through, so I guess the first thing is to
guess what it looks like. Here's a try at just January and February of
1994 so that we can see the result on the screen. The logic will work
just as well for the whole seven years.
# create fake data for the first
Hi Lubo,
Have a look at the "irr" package. I transferred any unique functions
from concord to it to reduce duplication of functions in different
packages.
Jim
On Thu, Sep 12, 2019 at 7:03 PM Lubo Larsson wrote:
>
> Hello,
>
> I would like to know if there is an R utility for computing some
>
Hi bickis,
Putting on my dark glasses and flailing about with a big white stick*,
I would suggest that you look at what "atv" actually produces from
those three objects. I wouldn't be surprised to find quite different
things.
Jim
* blind guess
On Thu, Sep 12, 2019 at 7:04 PM wrote:
>
> Here is
Hi Zuhri,
Try:
mydf<-mydf[,c(2,1)]
You might be surprised.
Jim
On Tue, Sep 10, 2019 at 12:20 PM smart hendsome via R-help
wrote:
>
> Hi R-user,
> I have a problem regarding R. How can I move my 2nd column into 1st column.
> For example I have data as below:
> mydf <-
Hi Alex,
At a guess you may want something like this:
data.frame(item="SKU",price=10,qty=2,ID="1")
This produces a data frame with one row. You will probably want many more rows.
Jim
On Wed, Aug 28, 2019 at 2:21 PM Alex Naverniak wrote:
>
> Hi,
> I am trying to create Inventory structure
Hi Arnaud,
The reason I wrote the following function is that it always takes me
half a dozen tries with "reshape" before I get the syntax right:
amdf<-read.table(text="A 10
B 5
C 9
A 5
B 15
C 20")
library(prettyR)
stretch_df(amdf,"V1","V2")
V1 V2_1 V2_2
1 A 105
2 B5 15
Hi Ogbos,
Hope things are going well for you. Perhaps this is what you want:
date_x1<-seq(as.Date("1953-01-02"),as.Date("2006-11-15"),length.out=8)
value_x1<-sample(1000:5000,8)
date_x2<-seq(as.Date("1957-07-26"),as.Date("1994-07-17"),length.out=6)
value_x2<-sample(0:1000,6)
Hi Greg,
I replied because I thought the name of the "expand.grid" function can
be puzzling. While "expand.grid" is a very elegant and useful
function, it is much easier to see what is happening with explicit
loops rather than loops buried deep inside "expand.grid". Also note
Bill's comment about
Hi Greg,
One problem is that you have misplaced the closing brace in the third
loop. It should follow the assignment statement. Because you used
loops rather than Bert's suggestion, perhaps you are trying to order
the values assigned. In your example, the ordering will be ssn, then
month of birth,
Hi Ana Marija,
It would help if we had the data, but I suspect that the data that we
don't have is not uniformly distributed:
library(plotrix)
ams_norm<-rnorm(100)
png("ams_norm.png")
qqunif(rescale(ams_norm,c(0,1)))
dev.off()
ams_unif<-runif(100)
png("ams_unif.png")
qqunif(ams_unif)
dev.off()
Hi Ana,
Or just for a bit of fun:
pt<-read.table(text="eidQ phenoQ phenoH
117 -9 -9
125 -9 -9
138 -9 1
142 -9 -9
156 -9 -9
174 -9 -9
138 -9 1
1000127 2 1
1000690 2 -9
1000711 2
Hi Tolulope,
The "in" operator steps through each element of the vector on the
right. You only have one element. Therefore you probably want:
for(x in 1:5)
...
Jim
Jim
On Tue, Aug 6, 2019 at 6:54 PM Tolulope Adeagbo
wrote:
>
> Hey guys,
>
> I'm trying to write a loop that will repeat an
Hi Tolulope,
Get the Monte Carlo package:
https://cran.r-project.org/web/packages/MonteCarlo/vignettes/MonteCarlo-Vignette.html
and look at an online tutorial:
https://www.youtube.com/watch?v=T_igE6bb6hU
Jim
On Mon, Jul 29, 2019 at 5:29 PM Tolulope Adeagbo
wrote:
>
> Hello Everyone,
>
>
Hi Faradj,
There is a problem with your structure statement in that the hyphen
(-) following the left angle bracket (<) has been transformed into a
fancy hyphen somewhere in the process. I replaced it with an ordinary
hyphen and it worked okay. Also, your coding for "B" seems to include
the first
Hi Faradj,
Rui's advice is correct, here's a way to do it. Note that I have
replaced the comma decimal points with full stops for my convenience:
fkdf<-read.csv(text="Year,Country,X1,X2
1990,United States,0,0.22
1991,United States,0,0.22
1992,United States,0,0.22
1993,United States,0,0.22
Hi Spencer,
While Sarah has already given you very good advice, there is a rough
method of checking whether one data set can be substituted for
another:
str(anno)
str(Dilution)
If you're lucky and the objects are not too complicated, this will
give you a start on whether one can be substituted
>
> library(plotly)
> plot_ly(sydf,
> x = ~year,
> y = ~rate,
> type = 'scatter', mode = 'lines') %>%
> layout(
> xaxis = list(
> ticktext = sydf$monthday[seq(1, length(sydf$monthday), 3)],
> tickvals = sydf$year,
> tickmode = &quo
ot;2002-07-13"
So making the labels these mid-year dates as character strings might
do what you want:
yrlabels<-as.character(yrticks)
Jim
On Sat, Jul 13, 2019 at 8:33 AM Jim Lemon wrote:
>
> Hi Steven,
> year1 is a number (e.g. 1993), monthday (e.g. 05-01) is not.
>
>
nthday,
> tickvals = sydf$yrticks,
> tickmode = "array",
> tickangle = 270
> ))
>
> But the chart didn't show any tick labels.
> I guess I need to sample sydf$monthday, right? Because that's what I want to
> show as tick labels. But the probl
axis lables, and
> use that for the "ticktext" parameter.
> I thought it must be some variation of the seq(from, to, by= ). Can I use
> that with a list of strings?
>
> Thanks,
> Steven
>
> -Original Message-
> From: Jim Lemon
> Sent: Thursday,
axis, but still
> have the plot showing the chart as rate based on year?
> I tried this:
> plot(sydf$year,sydf$rate,type="b",
> xlab="month-day",ylab="Rate")
>
> but this only changes the title of the x axis to "month-day". I want
Hi Ana,
The first error message may be the critical one. "weight//NA" is
almost certainly not the name of the file you want, and two
consecutive slashes in a path is also incorrect. You may not have
placed the data files in the correct file structure, often a problem
with programs that are written
Hi Steven,
A basic plot can be displayed like this:
sydf<-read.table(text="year rate
1993 0.608
1994 0.622
1996 0.623
1998 0.647
2000 0.646
2002 0.625
2004 0.628
2006 0.685
2008 0.679
2010 0.595
2012 0.567
2014 0.599
2016 0.642
2018 0.685",
header=TRUE)
Hi Sarah,
The size_n_color function in the plotrix package does something like
this. To get the colors that you want using "color.scale" in the same
package, look at the help page for that function. The values in
A1.matrix[,3] probably don't range from 0 to 2, so getting exactly
what you want can
Yep, you're right.
Jim
On Tue, Jul 2, 2019 at 7:52 AM William Dunlap wrote:
>
> Should that encoding="UTF-8" be encoding="UTF-16"?
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
>
> On Mon, Jul 1, 2019 at 2:45 PM Jim Lemon wrote
Hi Marvin,
One way to get around the problem with "sd" is to only process those
columns of a data frame for which the variance is defined. It is an
opportunity to show students how to write wrapper functions as well:
sd_num<-function(x) return(ifelse(is.numeric(x),sd(x),NA))
Hi Javad,
Unicode characters do have embedded nulls. Try this:
d4<-read.table("./4.csv",sep=";",header=TRUE,encoding="UTF-8")
Jim
On Tue, Jul 2, 2019 at 3:47 AM javad bayat wrote:
>
> Dear R users;
> I am trying to read an excel CSV file (1.csv). When I read it as csv file
> in R, the R shows
Hi Eleftheria,
If a non-parametric test doesn't sink your preferred hypothesis, it
can offer some peace of mind with regard to scrupulous reviewers. If
it does, it is wise to ponder the reliability of your results.
Jim
On Fri, Jun 28, 2019 at 10:34 PM Eleftheria Dalmaris
wrote:
>
> Dear all,
>
Hi Janet,
This might help:
jcdf<-read.table(text="scen trans evap flow
1 1.10.10.09
1 1.20.20.10
1 1.30.30.20
2 2.10.10.09
2 2.20.20.10
2 2.30.30.20
3 3.10.10.09
3 3.20.20.10
3
Hi Jeff,
Well, the output of "format" is character strings, so that is what you
are missing. I assume that you want to get two numeric values for each
date.
date1<-"06/20/2019 09:07"
> as.POSIXct(date1,format="%m/%d/%Y %H:%M")
[1] "2019-06-20 09:07:00 AEST"
Hi Sam,
If I have the structure of your data right, this might help:
scdf<-read.csv(text="DD Pack0.00
FTA English News0.00
FTA Complimentary0.00
WB1.18
WION1.18
Al Jazeera0.00
Animal Planet2.36
Asianet Movies17.70
Calcutta News0.00
Comedy Central5.90",
header=FALSE,
stringsAsFactors=FALSE)
Hi Nevil,
In case you are still having trouble with this, I wrote something in R
that should do what you want:
mystrings<-c("ABC","A(B)C","AB[C]","BC","{AB}C")
get_enclosed<-function(x,left=c("(","[","<","{"),right=c(")","]",">","}")) {
newx<-rep("",length(x))
for(li in 1:length(left)) {
ess
>> distance between each data point, but the points corresponding to the
>> labels can be clearly accessible in the zoom mode.
>>
>> Thank you very much Sir.
>>
>>
>>
>>
>>
>> [image: Mailtrack]
>> <https://mail
label
> (mentioned in column 1, and 3) for the different data points of column 2,
> and 4.
>
> Hope I successfully answered your question.
>
> Thank you.
>
>
>
> [image: Mailtrack]
> <https://mailtrack.io?utm_source=gmail_medium=signature_campaign=signatureviralit
Hi Subhamitra,
It is time to admit that I had the wrong idea about what you wanted to
do, due to the combination of trying to solve two problems at once
while I was very tired. I appreciate your patience.
>From your last email, you have a data frame with four columns. The
first and third are
Hi Subhamitra,
I don't have the factoextra package, but this may give you what you want:
ts<-read.table(text="Name DMs Name EMs
A 2.071 a 2.038
B 2.0548 b 2.017
C 2.0544 c 2.007
D 2.047 d 1.963
E 2.033 f 1.947
F 2.0327 g 1.942
G 2.0321 h 1.932
H 2.031 i 1.924
I 2.0293 j 1.913
J 2.0291 k 1.906
K
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