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
Hi,
I have a list of 150 strings, say, ap,:
aajkss
dfghjk
sdfghk
...
xxcvvn
And I would l like to grep out these strings from column names in
another file, af,. I've tried the following but none seem to work:
aps <- af[,grep(ap, colnames(af), value=TRUE)]
aps <- af[,grep(ap, colnames(af), va
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
u have illustrated
> in your 'eventual outcome'). Anyhow, it may be enough to allow you to get
> there.
>
> HTH
>
> Peter Alspach
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of K
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$IID1new <- ifelse(as.character(dbpmn[,2]) ==
as.character(dbpmn[,(21)]), dbpmn[,20], '')
Instead of returning some value it gives me this:
c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1))
Playing around wi
Hi,
I hope I can explain my problem clearly
I have a plink output file that I want to graph a heat map of the
PI_HAT estimates. I have the following code that I has worked in the
past but this time I'm getting the error:
In `levels<-`(`*tmp*`, value = if (nl == nL) as.character(labels) else
3456 sibling 349 0
> #2702.3 2702 9980 sibling 349 0
> #3064.4 3064 3 father 0 0
> #3064.5 3064 4 mother 0 0
> #3064.6 3064 5 sibling 4 3
> #3064.7 306486 sibling 4 3
> #
Sample.ID[father]
> if(sum(mother) == 0)
> l$MID[l$Relationship == 'sibling'] <- 0
> else l$MID[l$Relationship == 'sibling'] <- l$Sample.ID[mother]
> l
> }))
>
> It is assumed that when either parent is not available the M/PID is 0.
>
> Best,
>
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$Sample.ID[father] :
> replacement has le
with(l, Relationship == 'father')
> mother <- with(l, Relationship == 'mother')
> l$PID[l$Relationship == 'sibling'] <- l$Sample.ID[father]
> l$MID[l$Relationship == 'sibling'] <- l$Sample.ID[mother]
> l
> }))
> res
>
> HTH
Hi,
I have a data.table question (as well as if else statement query).
I have a large list of families (file has 935 individuals that are
sorted by famiy of varying sizes). At the moment the file has the
columns:
SampleID FamilyID Relationship
To prevent from having to make a pedigree file by
I'm trying to have a layout of two graphs on a page... this has worked
before... but I changed up the way I do my venn diagrams so now
instead of the Venn Diagram being at the bottom of the page below the
bar/line graph it takes up the whole page and its overlays the
bar/line graph placed on the to
Thanks!
On Sat, Jun 21, 2014 at 11:05 AM, Jorge I Velez
wrote:
> Hi Kate,
>
> You could try
>
> sum(X[, 1] == 1 & X[, 2] == 1)
>
> where X is your data set.
>
> HTH,
> Jorge.-
>
>
>
> On Sun, Jun 22, 2014 at 12:57 AM, Kate Ignatius
> wrote:
>
I have 4 columns, and about 300K plus rows with 0s and 1s.
I'm trying to count how many rows satisfy a certain criteria... for
instance, how many rows are there that have the first column == 1 as
well as the second column == 1.
I've tried using rowSums and colSums but it keeps giving me this type
Hi All,
I'm trying to merge two files together using:
combinedfiles <- merge(comb1,comb2,by=c("Place","Stall","Menu"))
comb1 is about 2 million + rows (158MB) and comb2 is about 600K+ rows (52MB).
When I try to merge using the above syntax I get the error:
Error in merge.data.frame(comb1, comb
I have a list of files that I have called like so:
main_dir <- '/path/to/files/'
directories <- list.files(main_dir, pattern = '[[:alnum:]]', full.names=T)
filenames <- list.files(file.path(directories,"/tmpdir/"), pattern =
'[[:alnum:][:punct:]]_eat.txt+$', 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")
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