the receiver to deal with any text file newline decoding
> choice/task after the file transfer is completed.
>
> On December 6, 2018 7:03:48 AM PST, Duncan Murdoch <
> murdoch.dun...@gmail.com> wrote:
> >On 06/12/2018 7:45 AM, Kate Stone wrote:
> >> Hello r-help,
&
t sure whether this is an R problem or something to do with my
OS specifically, or even with the file itself. Any ideas?? I've
attached a small script demonstrating the issue.
Many thanks,
Kate
--
Kate Stone
PhD candidate
Vasishth Lab | Department of Linguistics
Potsdam University, 14467
certainly not wisdom."
>-- Clifford Stoll
>
>
> On Fri, Jun 26, 2015 at 10:58 AM, Kate Ignatius
> wrote:
>> "reading in a tab delimited file using args"
>>
>> What I mean by that is that I'm using a bash script to call in an R
>> script and using the
When reading in a tab delimited file using args I keep getting the error:
Error: unexpected symbol in "Name index"
Execution halted
The code is this:
a <- read.table(args[1],sep="\t",header=T, stringsAsFactors=F)
When inputting the file directly, as follows, this produces no errors:
a <- read
> for (j in 2:4) {
> data[[j]] <- match(data[[j]], uniqStrings, nomatch = 0L)
> }
> data
> }
>
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
> On Fri, May 29, 2015 at 9:58 AM, Kate Ignatius
> wrote:
>>
>> I have a pe
I have a pedigree file as so:
X0001 BYX859 0 0 2 1 BYX859
X0001 BYX894 0 0 1 1 BYX894
X0001 BYX862 BYX894 BYX859 2 2 BYX862
X0001 BYX863 BYX894 BYX859 2 2 BYX863
X0001 BYX864 BYX894 BYX859 2 2 BYX864
X0001 BYX865 BYX894 BYX859 2 2 BYX865
And I was hoping to change
.OO#. .OO#. rocks...1k
> ---
> Sent from my phone. Please excuse my brevity.
>
> On May 9, 2015 7:59:31 AM PDT, Kate Ignatius wrote:
>>I have some data that I've trouble importing..
blank fields are implicitly added. See 'Details'.
>
>
> --
> Don MacQueen
>
> Lawrence Livermore National Laboratory
> 7000 East Ave., L-627
> Livermore, CA 94550
> 925-423-1062
>
>
>
>
>
> On 5/9/15, 7:59 AM, "Kate Ignatius" wr
I have some data that I've trouble importing...
A B C D E
A 1232 0.565
B 2323 0.5656 0.5656 0.5656
C 2323 0.5656
D 2323 0.5656
E 2323 0.5656
F 2323 0.5656
G 2323 0.5656
G 2323 0.5656 0.5656 0.5656
When I input the data it seems to go like this:
SampleID ItemB ItemC ItemD ItemE
A 1232 0.565
B 232
colnames(af), value=FIXED)]
aps <- af[,grep(as.character(list(ap),colnames(af))]
and also aps <- unique (grep(ap, colnames(af))
Is there another way I can do this - maybe without using grep?
Thanks!
Kate.
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Hi,
Supposed I had a data frame like so:
A B C D
0 1 0 7
0 2 0 7
0 3 0 7
0 4 0 7
0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 1 5
0 5 1 5
0 4 1 5
0 8 4 7
0 0 3 0
0 0 3 4
0 0 3 4
0 0 0 5
0 2 0 6
0 0 4 0
0 0 4 0
0 0 4 0
For each row, I want to count how many max column values appear to
adventurely get the
ot;", columns[sa_ind])
> days <- paste0(days,"$")
> selected <- lapply(days, function(x) grep(x,columns))
> selected <- sort(unique(unlist(all_ind)))
>
> columns[selected]
> [1] "SA_TUES" "SA_MON" "CH_TUES" "CH_MON" &q
Hi,
I've got a complicated grep problem (or not)... I currently have a
file with the headings as follows:
DAY
MONTH
YEAR
SA_TUES
SA_MON
SU_WED
CH_TUES
CH_WED
CH_MON
AR_TUES
AR_WED
AR_MON
SA_THUR
SU_FRI
CH_THUR
CH_FRI
AR_THUR
AR_FRI
I want to grep out all columns that have SA at the beginning of
I have genetic data as follows (simple example, actual data is much larger):
comb =
ID1 A A T G C T G C G T C G T A
ID2 G C T G C C T G C T G T T T
And I wish to get an output like this:
ID1 AA TG CT GC GT CG TA
ID2 GC TG CC TG CT GT TT
That is, paste every two columns together.
I have this
9: AA AA RA C 3 300
>>> 10: AA RRRR C 3 302
>>> 11: AA RRRA C 2 200
>>> 12: AA RRRR C 2 213
>>> 13: AA AAAA C
, David Winsemius wrote:
>
>> On Jan 1, 2015, at 5:07 PM, Kate Ignatius wrote:
>>
>> Apologies - mix up of syntax all over the place, a habit of mine. The
>> last line was in there because of code beforehand so it really doesn't
>> need to be there. Here is t
eriesO.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> ---
> Sent from my phone. Please excuse my brevity.
>
> On January 1, 2015 4:16:52 AM PST, Kate
ld
> 1: AA RRRA A 2 20
> 2: AA RRRR A 2 21
> 3: AA AAAA B 4 55
> 4: AA AAAA B 4 55
> 5: RA AARR B 0 55
> 6: RR AARR B
correct code:
childseg<-0
x:=sumchild <-0
span<-rle(x)$lengths[rle(x)$values==TRUE
childseg[x]<-rep(seq_along(span), times = span)
childseg[childseg == 0]<-''
On Thu, Jan 1, 2015 at 1:56 AM, Kate Ignatius wrote:
> Is it possible to add the following code or similar
RA A 2 20
> 2: AA RRRR A 2 21
> 3: AA AAAA B 4 55
> 4: AA AAAA B 4 55
> 5: RA AARR B 0 55
> 6: RR AARR B 4 55
>
I'm trying to use both these packages and wondering whether they are possible...
To make this simple, my ultimate goal is determine long stretches of
1s, but I want to do this within groups (hence using the data.table as
I use the "set key" option. However, I'm I'm not having much luck
making thi
ay pdf files is what
>> belongs there.
>>
>> On Macintosh we can avoid knowing by using 'open', which means use the
>> system standard.
>> I don't know what the linux equivalent is, either the exact program or
>> the instruction to use the standard
ting the options.
> the latex function doesn't do pdflatex (by default it does regular
> latex) unless you set the options
> as I indicated.
>
> On Tue, Dec 9, 2014 at 3:11 PM, Kate Ignatius wrote:
>> Ah yes, you're right.
>>
>> The log has this error:
>>
.OO#. .OO#. rocks...1k
> ---
> Sent from my phone. Please excuse my brevity.
>
> On December 9, 2014 8:43:02 AM PST, Kate Ignatius
> wrote:
>>Thanks! I do get several errors though when running on Linux.
>>
>>Running your code, I get this:
>>
>
:\\progra~2\\Adobe\\Reader~1.0\\Reader\\AcroRd32.exe')
> ## 64 bit windows
>
> ## Linux
> ## I don't know the xdvicmd value
>
>
> ## this works on all R systems
> library(Hmisc)
> tmp <- matrix(1:9,3,3)
> tmp.dvi <- dvi(latex(tmp))
> print.default(tmp
Hi,
I have a simple question. I know there are plenty of packages out
there that can provide code to generate a table in latex. But I was
wondering whether there was one out there where I can generate a table
from my data (which ever way I please) then allow me to save it as a
pdf?
Thanks
K.
I have genetic information for several thousand individuals:
A/T
T/G
C/G etc
For some individuals there are some genotypes that are like this: A/,
C/, T/, G/ or even just / which represents missing and I want to
change these to the following:
A/ A/.
C/ C/.
G/ G/.
T/ T/.
/ ./.
/A ./A
/C ./C
/G
neer (Solar/BatteriesO.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> ---
> Sent from my phone. Please excuse my brevity.
>
> On October 14, 2014 7:23:55 AM PDT, Kate Ignatius
> wrote:
>>I'm having
T RA
A/A AA
T/G RA
it will have a problem grepping out this single column.
On Tue, Oct 14, 2014 at 10:38 AM, John McKown
wrote:
> On Tue, Oct 14, 2014 at 9:23 AM, Kate Ignatius
> wrote:
>> I'm having an issue with grep:
>>
>> I have numerous columns that end with .a
I'm having an issue with grep:
I have numerous columns that end with .at... when I use grep like so:
df[,grep(".at",colnames(df))]
it works fine. When I have one column that ends with .at, it does not
work. Why is that? As this is loop with varying number of columns
ending in .at I would like
something working but I'm having trouble with the gt part...
I'm getting the error: object of type 'closure' is not subsettable.
The vcf is my original file that I want to match with so not sure
whether this a problem.
On Mon, Oct 13, 2014 at 4:46 PM, Kate Ignatius wrote:
> Hi
Hi all,
I need help with a function. I'm trying to write a function to apply
to varying number of columns in a lot of files - hence the function...
but I'm getting stuck. Here it is:
gt<- function(x) {
alleles <- sapply(x, function(.) strsplit(as.character(.), "/"))
gt <- apply(x, funct
, 2014 at 6:54 PM, Peter Alspach
wrote:
> Tena koe Kate
>
> If kateDF is a data.frame with your data, then
>
> apply(kateDF, 1, function(x) isTRUE(all.equal(x[2], x[1], check.attributes =
> FALSE, tolerance=0.1)))
>
> comes close to (what I think) you want (but not to what yo
Is there an easy way to check whether a variable is within +/- 10%
range of another variable in R?
Say, if I have a variable 'A', whether its in +/- 10% range of
variable 'B' and if so, create another variable 'C' to say whether it
is or not?
Is there a function that is able to do that?
eventua
is column X$IID1new != '' does not exist in X
>
> Here you clearly ask for nonexistent column, and why the heck you want to
> select column by number of rows?
>
>> as.character(as.matrix(X[,(2*nrow(X)+1)]))
> Error in `[.data.frame`(X, , (2 * nrow(X) + 1)) :
>
I have two data frames
For simplicity:
X=
V1 V2 V3 V4 V5 V6
samas4 samas5 samas6 samas4_father samas5_mother samas6_sibling
samas4 samas5 samas6 samas4_father samas5_mother samas6_sibling
samas4 samas5 samas6 samas4_father samas5_mother samas6_sibling
Y=
FID IID
FAM01 samas4
FAM01 samas5
FAM0
ledge
> is certainly not wisdom."
> Clifford Stoll
>
>
>
>
> On Sun, Sep 28, 2014 at 6:38 AM, Kate Ignatius
> wrote:
>> Strange that,
>>
>> I did put everything with as.character but all I got was the same...
>>
>> class of dbpmn[,2]) = factor
&g
its all about factors and data frames and
characters...
K.
On Sun, Sep 28, 2014 at 1:15 AM, Jim Lemon wrote:
> On Sun, 28 Sep 2014 12:49:41 AM Kate Ignatius wrote:
>> Quick question:
>>
>> I am running the following code on some variables that are factors:
>>
>> dbp
Quick question:
I am running the following code on some variables that are factors:
dbpmn$IID1new <- ifelse(as.character(dbpmn[,2]) ==
as.character(dbpmn[,(21)]), dbpmn[,20], '')
Instead of returning some value it gives me this:
c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1))
Playing around wi
Hi,
I hope I can explain my problem clearly
I have a plink output file that I want to graph a heat map of the
PI_HAT estimates. I have the following code that I has worked in the
past but this time I'm getting the error:
In `levels<-`(`*tmp*`, value = if (nl == nL) as.character(labels) else
3456 sibling 349 0
> #2702.3 2702 9980 sibling 349 0
> #3064.4 3064 3 father 0 0
> #3064.5 3064 4 mother 0 0
> #3064.6 3064 5 sibling 4 3
> #3064.7 306486 sibling 4 3
> #
sibling 0 842
3064 3 father 0 0
3064 4 mother 0 0
3064 5sibling 879 880
3064 86 sibling 879 880
3064 87 sibling 879 880
On Sat, Aug 16, 2014 at 9:31 PM, Jorge I Velez wrote:
> Dear Kate,
>
> Try this:
>
> res <- do.call(rbind, lapply(xs, function(l){
&
ccount for this?
On Sat, Aug 16, 2014 at 8:02 PM, Kate Ignatius wrote:
> Thanks!
>
> I think I know what is being done here but not sure how to fix the
> following error:
>
> Error in l$PID[l$\Relationship == "sibling"] <- l$Sample.ID[father] :
> replacement has le
Thanks!
I think I know what is being done here but not sure how to fix the
following error:
Error in l$PID[l$\Relationship == "sibling"] <- l$Sample.ID[father] :
replacement has length zero
On Sat, Aug 16, 2014 at 6:48 PM, Jorge I Velez wrote:
> Dear Kate,
>
> Ass
Hi,
I have a data.table question (as well as if else statement query).
I have a large list of families (file has 935 individuals that are
sorted by famiy of varying sizes). At the moment the file has the
columns:
SampleID FamilyID Relationship
To prevent from having to make a pedigree file by
I'm trying to have a layout of two graphs on a page... this has worked
before... but I changed up the way I do my venn diagrams so now
instead of the Venn Diagram being at the bottom of the page below the
bar/line graph it takes up the whole page and its overlays the
bar/line graph placed on the to
Thanks!
On Sat, Jun 21, 2014 at 11:05 AM, Jorge I Velez
wrote:
> Hi Kate,
>
> You could try
>
> sum(X[, 1] == 1 & X[, 2] == 1)
>
> where X is your data set.
>
> HTH,
> Jorge.-
>
>
>
> On Sun, Jun 22, 2014 at 12:57 AM, Kate Ignatius
> wrote:
>
I have 4 columns, and about 300K plus rows with 0s and 1s.
I'm trying to count how many rows satisfy a certain criteria... for
instance, how many rows are there that have the first column == 1 as
well as the second column == 1.
I've tried using rowSums and colSums but it keeps giving me this type
Hi All,
I'm trying to merge two files together using:
combinedfiles <- merge(comb1,comb2,by=c("Place","Stall","Menu"))
comb1 is about 2 million + rows (158MB) and comb2 is about 600K+ rows (52MB).
When I try to merge using the above syntax I get the error:
Error in merge.data.frame(comb1, comb
I have a list of files that I have called like so:
main_dir <- '/path/to/files/'
directories <- list.files(main_dir, pattern = '[[:alnum:]]', full.names=T)
filenames <- list.files(file.path(directories,"/tmpdir/"), pattern =
'[[:alnum:][:punct:]]_eat.txt+$', recursive = TRUE, full.names=T)
This
I'm trying to plot a GWAS (in you will) with lined segments
representing an overall p-value for each gene. Here is my code:
skatg <- ggplot(comm, aes(x = position,y = p, colour = grey)) +
geom_point(size = 0.75) +
geom_segment(data=rare, aes(x = txStart,
n't provide reproducible example so who knows.
>
> R> set.seed(1)
> R> x <- data.frame(matrix(runif(150), ncol=10))
> R> # col is a function, so not a good name
> R> col <- colMeans(x)
> R> mean(col)
> [1] 0.5119
>
> It's polite to include th
Hi All,
I've successfully gotten out the colMeans for 60 columns using:
col <- colMeans(x, na.rm = TRUE, dims = 1)
My next question is: is there a way of getting a mean of all the
column means (ie a mean of a mean)?
Thanks!
__
R-help@r-project.org ma
My code that I've used is:
mcgc <- ggplot(sam, aes(x = person,y = m, colour = X)) +
geom_point(size = 0.75) +
scale_colour_gradient2(high="red", mid="green",
limits=c(0,1), guide = "colourbar") +
geom_hline(aes(yintercept = mad, linetype =
I've used geom_point and geom_hline in ggplot2 and have gotten
satisfactory legends for both. However, I have one black line and one
blue line in the figure but in the legend they are both black - how
can I correct this in the legend to be the right colors?
mcgc <- ggplot(sam, aes(x = m,y = a
I'm trying to work out the average of a certain value by chromosome.
I've done the following, but it doesn't seem to work:
Say, I want to find average AD for chromosome 1 only and paste the
value next to all the positions on chromosome 1:
sam$mmad[sam$chrom == '1'] <-
(sam$ad)/(colSums(sam[c(1:nr
I'm not doing a Manhattan plot, but plotting AD (coloured by DP) along
the genome:
points <- ggplot(sam,aes(x = midpoint,y = ad, colour = dp, size = 3)) +
geom_point() +
scale_y_continuous(breaks=c(0,20,30,40)) +
labs(x = "chr",y = "ad") +
scale_colour_gradient2(high="red", mid="green")
address 0x4, cause 'memory not mapped'
*My question: is it possible to remove the all columns from above file
to *achieve* the desired result?*
* *
* **Thank you for help*
Kate Dresh
[[alternative HTML version deleted]]
_
:
"Error in procGPA(data, tangentresiduals = FALSE) : object 'out' not found"
Any help is appreciated.
Thanks,
Kate
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PLEASE do read the posting guide ht
Thanks for all of your help. It works to me.
Kate
On Mon, Nov 15, 2010 at 10:06 AM, David Winsemius wrote:
>
> On Nov 15, 2010, at 10:42 AM, Kate Hsu wrote:
>
> Hi r users,
>>
>> I have two data sets (X1, X2). For example,
>> time1<-c( 0, 8, 15, 22, 43,
0. Anyone know
how to use merge command to deal with this?
Thanks,
Kate
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PLEASE do read the posting guide http://www.R-p
1
I want to replace by 0, when I tried the following command, I get som
error message.
data[is.na(data)]<-0
Warning message:
In `[<-.factor`(`*tmp*`, thisvar, value = 0) :
invalid factor level, NAs generated
Anyone knows how to deal with this?
Thanks,
Kate
[[alternative HTM
)(0.8)(0.1)
YYNY (0.6)(0.5)(0.2)(0.9)
..
..
..
Any efficient way to do this?
Thanks,
Kate
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can't simply input the raw numbers for successes/failures since the ORs are all
adjusted for various factors.]
Thanks!
Kate Snedeker
--
Kate Snedeker, PhD
Postdoctoral Fellow
Centre for Public Health and Zoonoses & Department of Population Medicine
MacNabb House
Ontario Veterinary
,ylab="y coordinate movement (meters)",xlab="x
coordinate movement (meters)")
abline(h=0, v=0, col = "grey", lty=2)
#legend??
Any help would be appreciated!
Kate
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nction(mu)
{
log(prod(dcauchy(x,mu,s)))
}
maxNR(loglik,start=median(x))$estimate
Does anyone know how to solve this problem?
Thanks,
Kate
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erence
between the two sets of cases?
Any advice would be greatly appreciated!
Kate
. You need to check the @hole
slot of the @polygons[[1]] object if you care whether it's a hole
or an island!
Barry
__
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ht
ot; within this file and despite my best efforts (and
much reading!), I'm at a lost.
Many thanks,
Kate
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PLEASE do read the posting guide http://www.R-project.org/
individually for each tree ensemble? Is it possible to calculate these
predictors for the new random forest object after calling the combine
function? Any help would be greatly apprecaited. Thanks, Kate
[[alternative HTML version deleted]]
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wever they generated regressions of data
I don't care about (i.e., when "model1"==0).
Regards -- Kate
Kate Rohrbaugh
Independent Project Analysis, Inc.
44426 Atwater Drive
Ashburn, VA 20147
office - 703.726.5465
fax - 703.729.8301
email - [EMAIL PROTECTED]
website - www.
n either.
I'll just go back into Stata and create separate new datasets if I have to, but
there HAS to be a more elegant way.
Thank you for ANY feedback!
Kate
Kate Rohrbaugh
Independent Project Analysis, Inc.
44426 Atwater Drive
Ashburn, VA 20147
office - 703.726.5465
fax
n either.
I'll just go back into Stata and create separate new datasets if I have to, but
there HAS to be a more elegant way.
Thank you for ANY feedback!
Kate
Kate Rohrbaugh
Independent Project Analysis, Inc.
44426 Atwater Drive
Ashburn, VA 20147
office - 703.726.5465
fax
In "lm" command, we can use "vcov" option to get variance-covariance matrix.
Does anyone know how to get variance-covariance matrix in nlrq?
Thanks,
Kate
[[alternative HTML version deleted]]
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Has anybody used fuzzy time series to forecast enrollments? I have some code
that does not work so well. Thanks.
Kate
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PLEASE
oves)
plot(scoves,col='red', type='l')
lines(ses, col='blue',type='l')
lines(st,col='black',type='l')
The shapes are what I want, but I want x-axis consistent to the correspond to
n. Does anyone know the solution?
Thanks,
Kate
) : generates NaNs
2: In dt(x, df, log) : generates NaNs
3: In dt(x, df, log) :generates NaNs
4: In log(s) : generates NaNs
5: In dt(x, df, log) : generates NaNs
6: In dt(x, df, log) : generates NaNs
Thanks a lot,
Kate
- Original Message -
From: "Prof Brian Ripley" <[EMAIL PR
nvalid class "mle" object: invalid object for slot "fullcoef" in class
"mle": got class "list", should be or extend class "numeric"
When I typed warnings(), I get
In dt(x, ncp = ncp, df = df, log = TRUE) :
full precision was not achieved in
istribution?
Thanks a lot!!
Kate
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide co
Thank you very much for your help. This should take care of my problem.
@Romain: it would be greatly appreciated if you could release A2R on CRAN.
On Wed, May 7, 2008 at 5:31 AM, Romain Francois <
[EMAIL PROTECTED]> wrote:
> Hello Kate,
>
> Sorry I've only seen that thread.
it working, but we don't even know your OS and R
> version.
>
>
>
> On Tue, 6 May 2008, Kate wrote:
>
> Hi there,
> >
> > I've tried to install the A2R package using the files from
> > http://addictedtor.free.fr/packages/A2R/lastVersion/
>
anyone please help? Thanks.
Kate
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commen
need to use the observed y
and x run regression first and then assign the value to the missing y later.
Thanks,
Kate
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PLEASE do
th(V),length(sigmaV))
ps: TSE[i,j] corresponds to V[i] and sigmaV[j].
Thanks,
Kate
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Hi,
My example is as the following,
W<-matrix(c(2,4,2,8,1,3),ncol=3)
which(W==min(W))
I would like to find the index for min of the matrix W, i.e. (1, 3) instead of
5 as the output. What command could I use?
Thanks,
Kate
[[alternative HTML version dele
#x27;l')
I have the discontinuous point at ratio=0.7. I do not want to have the line at
0.7, and would like a solid dot at payoff=100 and a hollow dot at payoff=40 at
the jump point 0.7. How would I do this?
Thanks,
Kate
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t")),col=c("brown","red","blue","black"),lty=1:4,cex=1)
I would like to change "Chi-square" to the greek letter chi^2, how could I do
it in R?
Thanks,
Kate
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