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 2.
States
> underneath my data points so that I can see them over the location to
> which they correspond.
>
> Nicola
>
> On Mon, May 13, 2019 at 9:09 PM Jim Lemon wrote:
> >
> > Hi Nicola,
> > Getting the blank rows will be a bit more difficult and I don't see
&g
Hi Nicola,
Getting the blank rows will be a bit more difficult and I don't see
why they should be in the final data frame, so:
townzip<-read.table(text="waltham, Massachusetts 02451
Columbia, SC 29209
Wheat Ridge , Colorado 80033
Charlottesville, Virginia 22902
Fairbanks, AK 99709
Montpelier, VT
Hi Andrew,
First, a little mind reading. My crystal ball says that "cw" can be
interpreted as "carapace width". It didn't tell me the parameters of
the distribution, so:
set.seed(1234)
mf<-list(rnorm(400,145,15),rnorm(400,160,15))
library(plotrix)
multhist(mf, xlab="CW", ylab="Frequency", ylim=c(0
Hi Nicola,
There seems to be a big database of this here:
http://download.geonames.org/export/dump/
Jim
On Mon, May 6, 2019 at 2:47 PM Nicola Ruggiero
wrote:
>
> Hi there,
>
> I'm looking for a R package that would enable me to add a column to a
> data.frame that is itself a translation of zip
Hi Senaka,
Your attachments didn't make it through. If the datasets are not
large, perhaps use "dput" to create a text version that can be pasted
into the email. Large data sets may have to be uploaded to a public
URL>
Jim
On Sat, May 4, 2019 at 3:10 PM Senaka Amarakeerthi wrote:
>
> Hi All,
>
atediffs<-aggregate(dAT$date,list(dAT$Id),diffs,3)
initdates<-as.character(as.Date(as.vector(by(dAT$date,dAT$Id,min)),
origin=as.Date("1970-01-01")))
Result<-data.frame(Id=datediffs$Group.1,Release=initdates,
'T1-T2'=datediffs$x[,1],'T1-T3'=datediffs$x[,1]+dat
Hi Marna,
You can get the information you need with this:
dAT$date<-as.Date(dAT$date,"%d-%b-%y")
diffs<-function(x,maxn) return(diff(x)[1:maxn])
initdate<-function(x) return(min(x))
datediffs<-aggregate(dAT$date,list(dAT$Id),diffs,3)
I can't do the manipulation of the resulting values at the mome
tsplit),nrow=length(allhits)))
# change the names of the list
names(tmmdf)<-mmdf$Regulator
for(column in 1:length(hitsplit)) {
hitmatches<-match(hitsplit[[column]],allhits)
hitmatches<-hitmatches[!is.na(hitmatches)]
tmmdf[hitmatches,column]<-allhits[hitmatches]
}
Jim
On Fri, May 3, 20
Hi Matthew,
I'm not sure whether you want something like your initial request or
David's solution. The result of this can be transformed into the
latter:
mmdf<-read.table(text="Regulator hits
AT1G69490
AT4G31950,AT5G24110,AT1G26380,AT1G05675,AT3G12910,AT5G64905,AT1G22810,AT1G79680,AT3G02840,AT5G2
Hi Matthew,
Is this what you are trying to do?
mmdf<-read.table(text="Regulatorhits
AT1G69490AT4G31950,AT5G24110,AT1G26380,AT1G05675
AT2G55980AT2G85403,AT4G89223",header=TRUE,
stringsAsFactors=FALSE)
# split the second column at the commas
hitsplit<-strsplit(mmdf$hits,",")
# define a f
Hi Spencer,
Just download it to your R working directory and:
load("GBM_data.Rdata")
Worked okay for me (all 53.9 Mb)
Jim
On Mon, Apr 15, 2019 at 8:39 AM Spencer Brackett
wrote:
>
> I am also looking to be able to read this file on an appropriate
> application. As of now, it’s too large to v
Hi Val,
For this particular problem, you can just replace NAs with zeros.
vdat[is.na(vdat)]<-0
vdat$xy <- 2*(vdat$x1) + 5*(vdat$x2) + 3*(vdat$x3)
vdat
obs Year x1 x2 x3 xy
1 1 2001 25 10 10 130
2 2 2001 0 15 25 150
3 3 2001 50 10 0 150
4 4 2001 20 0 60 220
Note that this is not a gen
Hi Matthew,
How about this?
library(lattice)
xyplot(mpg ~ wt | cyl,
data=mtcars,
col = mtcars$gear,
pch = mtcars$carb
)
library(plotrix)
grange<-range(mtcars$gear)
xyplot(mpg ~ wt | cyl,
data=mtcars,
col = color.scale(mtcars$gear,extremes=c("blue","red"),xrange=g
Hi Bienvenue,
Perhaps you should ask whether you really want to "sort it out". The
warning is telling you that you are converting the NA values to NA in
the returned numeric vector. I can't think of anything more sensible
to do with NA values. You may also have character strings that cannot
be conv
Hi Eric,
When I run your code (using the MASS library) I find that
rstudent(fit2) also returns NaN in the seventh position. Perhaps the
problem is occurring there and not in the "influence" function.
Jim
On Wed, Apr 3, 2019 at 9:12 AM Eric Bridgeford wrote:
>
> I agree the influence documentatio
Hi Cyrus,
Try this:
pcr<-data.frame(Ct=runif(66,10,20),Gene=rep(LETTERS[1:22],3),
Type=rep(c("Std","Unkn"),33),Rep=rep(1:3,each=22))
testagg<-aggregate(pcr$Ct,c(pcr["Gene"],pcr["Type"],pcr["Rep"]),
FUN=function(x){c(mean(x), sd(x), sd(x)/sqrt(sd(x)))})
nxcol<-dim(testagg$x)[2]
newxs<-paste("x",1
quot;AT2G02010" "AT2G18690" "AT2G30750" "AT2G39200"
> "AT2G43620"
> [22] "AT3G01830" "AT3G54150" "AT3G55840" "AT4G03460" "AT4G11470" "AT4G11890"
> "AT4G14370"
> [29] "AT4G15417"
Hi Matthew,
First thing, don't put:
mydf3 <- data.frame(myenter)
inside your loop, otherwise you will reset the value of mydf3 each
time and end up with only "myenter" and the final list. Without some
idea of the contents of comatgs, it is difficult to suggest a way to
get what you want.
Jim
On
Hi Javed,
Easy.
A<-c(2000,2100,2300,2400,6900,7000,7040,7050,7060)
median(A)
[1] 6900
B<-c(3300,3350,3400,3450,3500,7000,7100,7200,7300)
median(B)
[1] 3500
wilcox.test(A,B,paired=FALSE)
Wilcoxon rank sum test with continuity correction
data: A and B
W = 26.5, p-value = 0.233
alternative
Hi Bert,
Good reference and David Urbina's example showed that a simple swap
was position dependent. The reason I pursued this is that it seems
more efficient to sequentially apply the precedence rules to the
arbitrarily sorted elements of the vector than to go through the
directed graph approach.
Hi Pedro,
This looks too simple to me, but it seems to work:
swap<-function(x,i1,i2) {
tmp<-x[i1]
x[i1]<-x[i2]
x[i2]<-tmp
return(x)
}
mpo<-function(x) {
L<-unique(as.vector(x))
for(i in 1:nrow(x)) {
i1<-which(L==x[i,1])
i2<-which(L==x[i,2])
if(i2 wrote:
>
> Dear All,
>
> This should be
Hi Luigi,
Upon careful reading of the help page, you can do it with scatter3D:
scatter3D(X, Y, Z, col.var = Z, pch = 16, cex = 2,clim=c(0.5,3))
scatter3D(X, Y, K, col.var = K, pch = 16, cex = 2,clim=c(0.5,3))
Jim
On Thu, Mar 14, 2019 at 9:32 PM Luigi Marongiu wrote:
>
> Dear all,
> I am trying
Hi Smruti,
In the example in question, you are probably doing something like this:
# I didn't see any attachment for the data
X<-rnorm(99,mean=59.96753)
t.test(X,mu=50,alternative="less")
One Sample t-test
data: X
t = 101.29, df = 98, p-value = 1
alternative hypothesis: true mean is less
Hi Bowie,
As David suggested, you can substitute the R missing value (NA) for
999 (probably an SPSS missing value). If you don't want to change it,
you could probably just subset your data like this:
V<-create_infotables(data=Test[Test[n] != 999,-1],y="class",bins=10)
where "n" is the column num
Hi John,
You seem to have 1569 days of data, so perhaps you can get around your
axis problem like this:
plot(Nile,xaxt="n",xlab="Week")
...
axis(1,at=seq(0,200,50),labels=seq(0,200,50)*7)
(untested)
Jim
On Thu, Mar 7, 2019 at 10:46 AM Sparks, John wrote:
>
> Hi R Helpers,
>
> I am doing some wo
Hi Darren,
You're probably looking for the %% (remainder) operator:
x<-1:10
# get odd numbers
x[as.logical(x%%2)]
# get even numbers
x[!(x%%2)]
Jim
On Sun, Mar 3, 2019 at 4:10 PM Darren Danyluk wrote:
>
> Hello,
>
> I found this email when looking for some help with R Studio. It's actually
>
Hi Aimin,
This example uses a log transformation to spread the colors out:
d<-read.table(text=" lateRT earlyRT NAD
ciLAD LAD
1.0 0.00 0.006224017 0.001260241 0.0069699285
0.0 1.00 0.001425649 0.007418436 0.0007096344
0.006224017 0.00
.
>
> > 1-(10-c(0.1, 1))* (10-c(0.1,1))/(10^2)
> [1] 0.0199 0.1900
>
> Cheers.
> Petr
>
> > -Original Message-
> > From: Jim Lemon
> > Sent: Thursday, February 21, 2019 12:24 AM
> > To: Rolf Turner
> > Cc: PIKAL Petr ; r-help@r-project.or
Okay, suppose the viewing field is circular and we consider two
particles as in the attached image.
Probability of being within the field:
R0 > sqrt((x1+R1-x0)^2 + (y1+R1-y0)^2)
Probability of being outside the field:
R0 < sqrt((x2-R1-x0)^2 + (y2-R1-y0)^2)
Since these are the limiting cases, it l
Hi Petr,
This is off the top of my head, but I assume that the shape of the
particle is not considered in counting. Assume particles are uniformly
distributed in the viewing field. If all particles entirely within the
field are counted, large particles will be undercounted. If all
particles within
Hi rain1290,
I have recently been working on a similar project, building a grid of
event densities for geographic coordinates. If you are stuck, I may be
able to provide some assistance.
Jim
On Sun, Feb 17, 2019 at 10:49 PM rain1290--- via R-help
wrote:
>
> Hello there,
>
> I am trying to overla
d
> check if I can still go the way u have suggested or if the correction would
> call for another approach.
> Best regards
> Ogbos
>
>
> On Sun, Feb 17, 2019, 02:22 Jim Lemon wrote:
>>
>> Hi Ogbos,
>> It may be easier to use strptime:
>>
>> dta<
Hi Ogbos,
It may be easier to use strptime:
dta<-data.frame(year=rep(2005,5),month=rep("05",5),
day=c("01","06","11","16","21"),
hour=c(2,4,6,8,10),minute=rep(0,5),second=rep(0,5),value=1:5)
dta$Ptime<-strptime(paste(paste(dta$year,dta$month,dta$day,sep="-"),
paste(dta$hour,dta$minute,dta$secon
Hi Keith,
Perhaps you do not want to go with calendar weeks:
365/7 = 52.14
as there are not an even number of weeks in a year, you may want to
plot by week from your initial observation. As you are using as.Date,
you could simply calculate weeks as:
data$Week<-1+as.numeric(data$Date - data$Date[
Hi Sasha,
I'll take a wild guess that your column names have periods (.)
replacing the spaces in the names you use:
species occurrence -> species.occurrence
The error message means that R can't find the variable name you have
used in the "by" argument. The second wild guess is that your column
na
),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,
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
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 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
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 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
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
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
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
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 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
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 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
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 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
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
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 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
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,
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
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$da
Hi david,
The formatting of the data frame looks like the Province and Year
columns have gotten stuck together. This probably has something to do
with your Excel spreadsheet or the function that you are using to read
it in. If there is are fewer column names than columns, this error is
likely to ha
Hi Ogbos,
If we assume that you have a 3 column data frame named oodf, how about:
oodf[,4]<-floor((cumsum(oodf[,1])-1)/28)
col2means<-by(oodf[,2],oodf[,4],mean)
col3means<-by(oodf[,3],oodf[,4],mean)
Jim
On Wed, Nov 28, 2018 at 2:06 PM Ogbos Okike wrote:
>
> Dear List,
> I have three data-column
Hi Vicci,
It's very clunky, but I think it will do what you want.
rrdf<-read.csv(text="No,date,chamber,d13C,ppm_CO2,ppm_13CO2
1,10.14.2018 10:43 PM,IN,-0.192,439.6908,4.9382
2,10.14.2018 10:47 PM,101,-0.058,440.7646,4.9509
3,10.14.2018 10:50 PM,103,-1.368,535.6602,5.9967
4,10.14.2018 10:53 PM,1
1. xaxt="n" means "Don't display the X axis". See the help for "par" in the
graphics package
2. axis(1,at=1:nrows,labels=names(MPG3))
This means, "Display the bottom axis (1) with ticks at 1 to the number of
rows in the data frame"
"Use the values of MPG$Year as labels for the ticks". see the help
<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
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
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.
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:
>
>>
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
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 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 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
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
>>
>>
> dataname$"very long name".
> So, is ther a way to do this procedure for all columns?
>
> Roberto
>
>
> Op do 1 nov. 2018 om 10:50 schreef Jim Lemon :
>>
>> Hi Roberto,
>> What I suggested is a brute force method of translating response
>&g
low but could not find the right answer
> Best Roberto
>
>
> Op do 1 nov. 2018 om 00:25 schreef Jim Lemon :
>>
>> Hi Roberto,
>> Here is a snippet of code that translates the text responses of the
>> BIS-11 into numeric values. Note the reversal of the order in the
&g
Hi Myriam,
This may not be the ideal way to do this, but I think it works:
mcdf<-read.table(text="41540 41540 41442 41599 41709 41823 41806 41837
41898 41848
41442 0.001
41599 0.002 0.001
41709 0.004 0.003 0.003
41823 0.002 0.001 0.002 0.001
41806 0.004 0.004 0.005 0.006 0.005
41837 0.004 0.004 0.
Hi Roberto,
Here is a snippet of code that translates the text responses of the
BIS-11 into numeric values. Note the reversal of the order in the
second item:
BIS$Q1<-as.numeric(factor(BIS$Q1,
levels=c("Almost","Often","Occasionally","Rarely/Never")))
BIS$Q2<-as.numeric(factor(BIS$Q2,
levels=c("
Yes, I thought that as well and had worked out this but didn't send it:
add_Pscores<-function(x) {
return(sum(unlist(x),na.rm=TRUE))
}
by(rzdf[,c("PO1M", "PO1T", "PO2M", "PO2T")],rzdf$STUDENT_ID,FUN=add_Pscores)
rzdf$STUDENT_ID: AA15285
[1] 724.8
---
Hi 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
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 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 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
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
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
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,
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 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:
>
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 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
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.
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
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
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 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 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
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 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")
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
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 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
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 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 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
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