Hi,
According to daisy function from cluster documentation, it can compute
dissimilarity when NA (missing) value(s) is present.
http://stat.ethz.ch/R-manual/R-devel/library/cluster/html/daisy.html
But why when I tried this code
library(cluster)
x <- c(1.115,NA,NA,0.971,NA)
y <- c(NA,1.006,NA,N
I have a data which contain some NA value in their elements.
What I want to do is to **perform clustering without removing rows**
where the NA is present.
I understand that `gower` distance measure in `daisy` allow such situation.
But why my code below doesn't work?
__BEGIN__
# plot heat map
I have the following code that perform hiearchical clustering and plot
them in heatmap.
__
library(gplots)
set.seed(538)
# generate data
y <- matrix(rnorm(50), 10, 5, dimnames=list(paste("g", 1:10, sep=""),
paste("t", 1:5, sep="")))
# the actual data is much larger that the above
# perform hiear
I have the following list of strings:
name <- c("hsa-miR-555p","hsa-miR-519b-3p","hsa-let-7a")
What I want to do is for each of the above strings
replace the text after second delimiter with "zzz".
Yielding:
hsa-miR-zzz
hsa-miR-zzz
hsa-let-zzz
What's the way to do it?
[[alternative HTM
I have a data which I plot using this code.
Attached is the plot
_BEGIN_
library(ggplot2)
dat.m <- read.delim("http://dpaste.com/1269939/plain/",sep="";)
colnames(dat.m) <- c("ensg","mirna_hgc","variable","value")
dat.m.y <- subset(dat.m,dat.m$variable=="y")
qplot(value,data=dat.m.y, geom="bar",
I have the following result of expand grid:
> d <- expand.grid(c("x","y","z"),c("x","y","z"))
What I want is to create a combination of strings
but only the half of the all combinations:
Var1 Var2
1xx
2yx
3 yy
4 zy
5 xz
6zz
What's the way to do it?
I have the following data:
v <- rnorm(13)
w <- rnorm(13)
x <- rnorm(13)
y <- rnorm(13)
z <- rnorm(13)
Using GGPLOT facet, what I want to do is to create a 5*5 matrix,
where each cells plot the correlation between
each pair of the above data. E.g. v-v,v-w; v-x,...,z-z
What's the way to do it?
A
I have the following list of data each has 10 samples.
The values indicate binding strength of a particular molecule.
What I want so show is that 'x' is statistically different from
'y', 'z' and 'w'. Which it does if you look at X it has
more values greater than zero (2.8,1.00,5.4, etc) than othe
Is there a native way to produce SE of correlation in R's cor() functions
and p-value from T-test?
As explained in this web
http://www.sjsu.edu/faculty/gerstman/StatPrimer/correlation.pdf
(page 14.6)
The standard error is sqrt((1-r^2)/(n-2)), where n- is the number of sample.
- G.V.
[[
I have the following list of strings:
x <- c("foo, foo2, foo3", "bar", "qux, qux1")
what I want to do is to obtain
foo, bar qux
Namely for each element in the vector obtain only string
before the first comma.
What's the way to do it?
- G.V.
[[alternative HTML version deleted]]
_
I have the following data frame:
> foo
w x y z
n 1.51550092 1.4337572 1.2791624 1.1771230
q 0.09977303 0.8173761 1.6123402 0.1510737
r 1.17083866 1.2469347 0.8712135 0.8488029
What I want to do is to change it into :
> newdf
1 nw 1.51550092
2 q w
I have a data that looks like this:
mRNA Value
---
mRNA1 30
mRNA2 199
...... ...
mRNA1000 13
Then I'll normalize the value based on the s
I have only *two* datasets from normal and cancer samples.
CancerNormal
--
mRNA1 3049
mRNA2 199200
... ... ...
mRNA1000 1340
Each samples contain several thousan mRNA microarray express
Given the attached plot,
how can I locate the center text with "Mean and SD" so that it can be
placed exactly under "---emp".?
The current code I have is this:
L = list(bquote(Em.Mean ==.(new_avg)),bquote(Em.SD==.(new_std)),
bquote(Th.Mean ==.(theor_avg)),
bquote(Th.SD==.(theor_sd)))
leg
I have the following matrix:
> dat
[,1] [,2] [,3][,4]
foo 0.7574657 0.2104075 0.02922241 0.002705617
foo 0.000 0.000 0. 0.0
foo 0.000 0.000 0. 0.0
foo 0.000 0.000 0. 0.0
foo 0.000 0
The colors generated by heat.color(x)
is too saturated.
Where is there alternative command similar to that but "non-heated'
which also I can input the value 'x'.
- G.V.
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Dear expert
How can we plot diagonal across (from bottom-left-hand corner to top
right-hand corner),
at any given coordinate range
For example
> plot(c(-2,3), c(-1,5), type = "n", xlab="x", ylab="y", asp = 1)
or
> plot(c(0,1000), c(0,334), type = "n", xlab="x", ylab="y", asp = 1)
I tried abline
Dear expert,
Given such data:
#Cutpoint SN (1-PPV)
5 0.560.01
7 0.780.19
9 0.910.58
How can I plot ROC curve with R that produce similar result like the
attached file?
I know ROCR package but it doesn't take such input.
- GV
_
, main="");
> plot(bar,lwd=2,lty=3,col="blue");
> legend(0.6,0.6,c('Default','Probabilistic'), col=c('red','blue'),lwd=3);
>
> A second option would be setting up a suitable layout. For this, please take
> a look at ?layout.
> B
I have a R script that contain these lines for plotting:
plot(foo,lwd=2,lty=3,col="red", main="");
plot(bar,lwd=2,lty=3,col="blue");
legend(0.6,0.6,c('Default','Probabilistic'), col=c('red','blue'),lwd=3);
But it generate 1 file (Rplot.pdf) with two pages. Each page for 1 plot.
Is there a way I
Dear sirs,
I have a data that is generated like this:
> dat1 <- data.frame(V1 = rep(1, 5), V2 = sample(c(40:45), 5))
> dat2 <- data.frame(V1 = sample(c(0,1), 5, replace = TRUE), V2 =
> sample(c(40:45), 5, replace = TRUE))
What I want to do is to obtain a data frame that contain list of list.
>
1 43
4 1 43
5 1 24
6 0 24
- G.V.
>
> Ivan
>
>
> Le 1/31/2011 10:50, Gundala Viswanath a écrit :
>>
>> Dear Experts,
>>
>> I have a data that looks like this.
>>
>>> file1="dat1.tab"
>&
Dear Experts,
I have a data that looks like this.
> file1="dat1.tab"
> file2="dat2.tab"
> dat1<-read.table(file1)
> print(dat1)
V1 V2
1 1 43
2 1 43
3 1 43
>
> dat2<-read.table(file2)
> print(dat2)
V1 V2
1 1 43
2 1 21
3 1 43
4 1 43
5 1 24
6 0 24
>
The column V1 refer to labels and V2
Is there a way to do it? At best what I can achieve
is non integer:
> runif(10, min=1, max=100)
[1] 51.959151 56.654146 63.630251 3.172794 4.073018 11.977437 86.601869
[8] 75.788618 11.734361 6.770962
-G.V.
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ht
I have a data frame that looks like this:
> print(df)
V2 V3 V4 V5 V6 V7 V8 V9V10V11V12
1 FN 8.637 28.890 31.430 31.052 29.878 33.215 32.728 32.187 29.305 31.462
2 FP 19.936 30.284 33.001 35.100 30.238 34.452 35.849 34.185 31.242 35.635
3 TN 0.000 17.190
I have the following structure:
> str(propn)
table [1:2(1d)] 0.674 0.326
- attr(*, "dimnames")=List of 1
..$ label: chr [1:2] "o" "x"
> print(propn)
label
o x
0.6738347 0.3261653
How can I access the value of "o" and "x"
I tried this but fail
> print(propn$label[,"o"];
-
I have a variable that looks like this:
> print(pred$posterior)
ox
1 2.356964e-03 9.976430e-01
2 8.988153e-01 1.011847e-01
3 9.466137e-01 5.338627e-02
4 2.731429e-11 1.00e+00
Now what I want to do is to access "o" and "x"
How come this approach fa
I have a vector that looks like this:
> foo
[1] "o" "o" "o" "x" "o" "o" "o" "o" "o" "x" "x" "o" "x
How can we find the percentage of "o" and "x" in
that vector in R?
- G.V
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I am refering to a function call like this:
>data(iris)
>x <- svmlight(Species ~ ., data = iris)
I tried to see the content of it by typing:
> Species ~ .
but it gives nothing. How can I see it's content ?
- P.Dubois
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Dear expert,
I have a series of number that looks like this
x <- c(-0.005282,
0.000314,
0.002851,
-2.5059217162,
-0.007545,
-1.0317758496,
0.001598,
-1.2981735068,
0.072411)
How can I normalize it in R so that the new numbers
is ranging from 0 to 1 ?
- G.V.
___
Suppose I have two vectors of same dimensions:
x <-c(0.49534,0.80796,0.93970,0.8)
count <-c(0,33,0,4)
How can I group the vectors 'x' into two vectors:
1. Vector `grzero` that contain value in x with `count` value greater
than 0 and
2. Vector `eqzero` with value in x with `cou
Dear Expert,
I have a data that looks like this:
for_y_axis <-c(0.49534,0.80796,0.93970,0.8)
for_x_axis <-c(1,2,3,4)
count <-c(0,33,0,4)
What I want to do is to plot the graph using "for_x_axis" and
"for_y_axis" but will mark
each point with "o" if the value is equal to 0(zero) and with
Dear Experts,
I have a input file that looks like this
-0.438185,svm,1
-0.766791,svm,1
0.695282,svm,-1
0.759100,svm,-1
0.034400,svm,1
0.524807,svm,1
-0.27647800,nn,1
-0.16120810,nn,-1
0.63911350,nn,1
0.400554110,nn,1
0.429192240,nn,-1
0.454239140,nn,1
How can I create a data structure in R so th
Hi all,
I have a frequency data that looks like this.
3
2
1
5
What I want to get is the "decreasing" cumulative of this data
yielding
11
8
6
5
0
Is there any?
I am aware of cumsum(), which will yield
3
5
6
11.
But it is not what I want.
- G.V.
__
Dear masters,
I have data that looks like this:
#val Freq1 Freq2
0.000 178 202
0.001 4611 5300
0.002 99 112
0.003 26 30
0.004 17 20
0.005 15 20
0.006 11 14
0.007 11 13
0.008 13 13
...many more lines..
Full data can be found here:
http://dpaste.com/173536/plain/
What I intend to do is to have a
I have a data file that looks like this.
__DATA__
D7KAR5Z02F447V 176 G 0.22
D7KAR5Z02J3WLG 94 A 1.0529
D7KAR5Z02F4K6L 198 a 0.13
D7KAR5Z02J4SYO 67 C 0.9528
D7KAR5Z02J4SYO 83 C 1.0129
D7KAR5Z02J4SYO 97 T 0.13
D7KAR5Z02J4SYO 166 A 0.9427
I want the rows whe
I have a data frame (dat). What I want to do is for each row,
divide each row with the sum of its row.
The number of row can be large > 1million.
Is there a faster way than doing it this way?
datnorm;
for (rw in 1:length(dat)) {
tmp <- dat[rw,]/sum(dat[rw,])
datnorm <- rbind(datnorm, tmp
How can I truncate the scientific value keeping two digits decimal.
For example from:
6.95428812397439e-35
into
6.95e-35
-E.W.
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PLEASE do read the posting guide http://w
Is there any?
- G.V.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
How do people usually use the result of density function (e.g. dnorm)?
Especially when its value can be greater than 1.
What do they do with such density >1?
> dnorm(2.02,2,.24)
[1] 1.656498
- G.V.
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https://stat.et
AFAIK, R only has "pnorm" which computes the probability of getting a
value smaller or equal to "x" from
a normal distribution N[mean,stdev]. For example:
R> pnorm(0, 4, 10)
[1] 0.3446
means there is 34.46% chance of getting a value equal to or smaller
than 0 from a N(4, 10) distribution.
What
I have a data that looks like this:
http://dpaste.com/88988/plain/
How can I extract/subset the data frame
based on selected uniq ID.
Let's say I want the first K uniq ID.
I want to be able to specify the parameter "K" here,
(i.e. given K=3, we hope to extract dat$V2 = 0,1,2).
I'm stuck with thi
Currently, I am doing it this way.
x <- mtcars$mpg
h<-hist(x, breaks=10, col="red", xlab="Miles Per Gallon",
main="Histogram with Normal Curve")
xfit<-seq(min(x),max(x),length=40)
yfit<-dnorm(xfit,mean=mean(x),sd=sd(x))
yfit <- yfit*diff(h$mids[1:2])*length(x)
lines(xfit, yfit, col="blue", lwd=
of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
> -
t; The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
> -Oorspronkelijk bericht-
> Van: Gundala Viswanath [mailto:gunda...@gmail.com]
> Verzonden: woensdag 2 se
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
> -Oorspronkelijk bericht-
> Van: Gundala Viswanath [mailto:gunda...@gmail.com]
> Verzonden: woensd
ble to say what
> the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body
for an answer does not
> ensure that a reasonable answer can be extracted from a given body of
> data.
> ~ John Tukey
>
> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> Namens Gundala Viswanath
> Verzonden: woensd
I have a data that looks like this:
http://dpaste.com/88561/plain/
And I intend to create multiple density curve into one plot, where each curve
correspond to the unique ID.
I tried to use "sm" package, with this code, but without success.
__BEGIN__
library(sm)
dat <- read.table("mydat.txt");
pl
I have the following data which I tried to draw
the probability density plot.
Here is the code I have:
x <- read.table("mydat.txt");
d <- rep(x$V2,times=x$V3);
hist(d,probability=T, xlab="FlowSignal");
But why the y-axis range from 0 to 6,
instead of 0 to 1? What's the correct way to plot it?
ata frame where
the entry in V1 has (x.000) as its decimal.
yielding
> wanted_dat
V1 V2 V3
1 0.000 2 554889
2 123.000 03209
What's the way to do it in R?
- Gundala Viswanath
Jakarta - Indonesia
_
Dear all,
I have no problem coloring the axis and plot,
following these procedures:
http://www.nabble.com/Coloring-X-and-Y-axis-tt22989739r0.html#a22989739
However the X,Y and Main Label stays in black.
How can we change their colors?
- Gundala Viswanath
Jakarta - Indonesia
Dear all,
I have a plot with 2 x 2 figures matrix in it.
pdf("~/Desktop/myplot.pdf",width=13,height=7)
par(mfrow=c(2,2))
# follow by some code
Now the distance between figures of row 1 and row 2
is too wide. How can I modify?
- Gundala Viswanath
Jakarta -
Dear all,
Is there a way to do it?
The following code:
> pdf("test.pdf")
> plot(1,1,col="red")
> dev.off()
Only colors the plot into red, but not
x and y axis (inclusive the tick marks).
- Gundala Viswanath
Jakarta - Indonesia
__
Is there a way to do it in R?
Especially generating plot in EPS/PDF format.
By transparent I mean clear (not white) background.
I want to attached it to dark PPT slides.
- Gundala Viswanath
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33.2
16 6 44.5
17 7 5.00
Is there a way to do it in R?
- Gundala Viswanath
Jakarta - Indonesia
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PLEASE do read the posting guide http://www.R-project.org/p
marks.
- Gundala Viswanath
Jakarta - Indonesia
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained
- Gundala Viswanath
Jakarta - Indonesia
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible
27;s wrong with my approach.
So yes, I do want the V1 information, hence I am aware that
I dont' want:
> plot(hist(dat[dat >=-500 & dat <=500]))
- Gundala Viswanath
Jakarta - Indonesia
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__
library(grid)
library(gridBase)
opar <- par(no.readonly=TRUE)
par(opar)
grid.newpage()
pushViewport(viewport(width=0.5, height=0.5))
grid.rect(gp=gpar(col="grey", lty="dashed"))
par(omi=gridOMI())
par(mfrow=c(2, 2), mfg=c(1, 1), mar=c(3, 3, 1, 0))
for (i in 1:4) {
plot(1:10
;, j, "\n", sep="")
test.plot(i,j);
}
}
dev.off()
__END__
- Gundala Viswanath
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0.015241
6 0.0141398 0.0142373
7 0.0101515 0.0102948
8 0.0308843 0.031294
9 0.0095504 0.00960626
10 0.00729676 0.0073713
- Gundala Viswanath
Jakarta - Indonesia
__
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7;s wrong with the data type so that it gives such error?
- Gundala Viswanath
Jakarta - Indonesia
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Hi,
I have the following data that looks like this:
> names(dat)
[1] "(-2329,-2319]" "(-1399,-1389]" "(-669.4,-659.4]"
How can I modify those names into just this?
[1] -2329 -1399 -669.4
- Gund
V11V12 V13V14
V15 V16
0 0.00380959 0.00380959 0.00380959 0 .00380083 0.00380083 0.00380083
Is there a quick way to do it in R?
- Gundala Viswanath
Jakarta - Indonesia
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ht
Hi Jim,
> or sorting the columns of the above table if that is what you are using to
> plot.
How do you do that? Yes I am using that data exactly for the plotting.
- GV.
>
> Jim
>
>
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the second bar (value = 76) id placed on the right to the first.
(8.048,18.05] (-21.95,-11.95]
81 76
This is the command I use to plot:
barplot(x,ylab="Number of Unique Tags", xlab="Expected - Observed")
- Gundala Viswanath
Hi all,
Suppose I hve this vector:
> x
[1] 3 4 7 17 22 12 15 12 3 3 1 1
How can I remove the top-3 element.
Yielding only:
[1] 17 22 12 15 12 3 3 1 1
- Gundala Viswanath
Jakarta - Indonesia
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ht
RL
http://docs.google.com/Doc?id=dcvdrfrh_5cm5qkchw
How can I fix the code so that it gives a exact plot
to "dat" above, with same number of bins and its
respective frequency?
- Gundala Viswanath
Jakarta - Indonesia
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h
Dear all,
Thanks so much, the latest version by Gustav work just fine.
In the first version, the outcome is more digital
than contiguous.
- Gundala Viswanath
Jakarta - Indonesia
On Wed, Feb 11, 2009 at 10:35 PM, Gustaf Rydevik
wrote:
> On Wed, Feb 11, 2009 at 2:15 PM, Ben Bolker wr
Dear all,
Is there a way to generate K numbers of integer (K = 10^6).
The maximum value of the integer is 200,000 and minimum is 1.
And the occurrences of this integer follows
a lognormal distribution.
- Gundala Viswanath
Jakarta - Indonesia
__
R
0.200
[4,] 0.200
[5,] 0.200
My questions are:
1. How does %*% work in the above example?
2. Is there a more understandable (naive) way to implement such
product in this context?
- Gundala Viswanath
Jakarta - Indonesia
__
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es = TRUE, qmethod = c("escape", "double"))
- Gundala Viswanath
Jakarta - Indonesia
On Sun, Jan 18, 2009 at 12:09 AM, Henrik Bengtsson
wrote:
> Do it in chunks of rows. /H
>
> On Sat, Jan 17, 2009 at 6:55 AM, Gundala Viswanath wrote:
>> Hi Hadley,
>>
>
tried your suggestion already, I have memory problem
> x <- cbind(dat$V1, as.character(dat$V2))
Error: cannot allocate vector of size 4.2 Gb
Execution halted
- Gundala Viswanath
Jakarta - Indonesia
On Sat, Jan 17, 2009 at 11:39 PM, hadley wickham wrote:
> On Sat, Jan 17, 2009 at 7:59
Hi,
> Unless you specify an in-memory database the database is stored on disk.
Thanks for your explanation.
I just downloaded 'sqldf'.
Where can I find the option for that? In sqldf I can't see the command.
I looked at:
envir = parent.frame()
doesn't appear to be the on
ssue?
Why is that?
Sorry for my naive question.
- Gundala Viswanath
Jakarta - Indonesia
On Fri, Jan 16, 2009 at 9:09 PM, Gabor Grothendieck
wrote:
> On Fri, Jan 16, 2009 at 5:52 AM, r...@quantide.com wrote:
>> I agree on the database solution.
>> Database are the rigth too
ding value for each query above,
yielding:
0.02
0.7
However, I want to avoid slurping whole repo.txt into an object (e.g. hash).
Is there any ways to do that?
The reason I want to do that because repo.txt is very2 large size
(milions of lines,
with tag length > 30 bp), and my PC memory i
Sorry for my late reply.
Thank you so much Jim. This script of yours
is very2 useful. I have used it.
- Gundala Viswanath
Jakarta - Indonesia
On Wed, Jan 14, 2009 at 12:17 AM, jim holtman wrote:
> Here is a function I use to see how big the objects in my workspace are:
>
&g
ain all.
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 9 18 8 17 18
[26] 13 9 5 6 7 9 10 11 13 14 3 17 2 1 14 3 3 13 4 4 16
Is there a quick way to do it?
I tried "grep("[0-9]", vect)" but fail.
- Gundala Viswan
22 22 22 22
Slot "dimension":
[1] 100 1
__ END__
How can I acces "Slot 'ra'" only?
I tried
print(x$ra)
but fail.
- Gundala Viswanath
Jakarta - Indonesia
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oesn't give the exact byte size.
- Gundala Viswanath
Jakarta - Indonesia
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and provide comm
Thanks for your most reasonable reply, Henrique.
- Gundala Viswanath
Jakarta - Indonesia
On Tue, Jan 13, 2009 at 8:01 PM, Henrique Dallazuanna wrote:
> Try this:
>
> unlist(ifelse(q %in% names(x), x[q], NA))
>
> On Tue, Jan 13, 2009 at 8:49 AM, Gundala Viswanath
> wro
But why this failed?
> extracted <- x[[q]]
Error in x[[q]] : subscript out of bounds
we expect the output of 'extracted' to be
a vector as well. When the key is not present
to give "NA" in vector
- Gundala Viswanath
Jakarta - Indonesia
___
t; "AAC" "AAG" "ATA" "ATT"...
>
>> print(new_repo)
> [1] "AAA" "AAT" "AAC" "AAG" "ATA" "ATT"...
- Gundala Viswanath
Jakarta - Indonesia
On Tue, Jan 13, 2009 at 2:40 PM, wrote:
hr [1:100] "AAA" "AAT" "AAC" "AAG" "ATA" "ATT"...
> print(new_repo)
[1] "AAA" "AAT" "AAC" "AAG" "ATA" "ATT"...
I tried as.vector(), but it remains the same factor.
- Gundala V
"ABDCA" "ACACC" "DADAA"
> "ABCAD" ...
>> str(z)
> chr [1:200] "EAEDC" "DACCD" "BEAAD" "CDDDA" "ABDCA" "ACACC"
> "DADAA" "ABCAD" ...
>> system.time(y <- matc
Yes Jim, exactly.
BTW, I found from ?match
" Matching for lists is potentially very slow and best avoided
except in simple cases."
Since I am doing this for million of tags. Is there a faster alternatives?
- Gundala Viswanath
Jakarta - Indonesia
On Tue, Jan 13, 2009 at 12:
hat the query contain repeating elements, yet
the output of which only returns unique.
How can I make it returning
[1] 3 6 6
instead?
- Gundala Viswanath
Jakarta - Indonesia
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Hi Jorge and all,
How can I modified your code when
query size can be bigger than repository,
meaning that it can contain repeats.
e.g. qr <- c("AAC", "ATT", "ATT","AAC", "ATT", "ATT", "AAT", "ATT", &qu
index in repository in a fast way.
Giving:
[1] 3 6
Typically the size of repo is around ~12million element, and
query around ~1 million element.
- Gundala Viswanath
Jakarta - Indonesia
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AAAT" ...
I want to destroy the attribute yielding only this:
> str(mat)
Named chr [1:32268] "yQAAA" "jQAAQ" "UQAAg" "FQAAw" "1QABA" ...
But why this command fail to do it?
> dimnames(mat) <- NULL
Btw, the object size for
of Column in matrix = length of string (= 3)
2. Number of Row in matrix = length of vector ( = 4).
3. Character "a" encode as "0",
"c" -> "1",
"g" -> "2",
"t" -> "3"
Length of strings are
'b*', $string;
print "COMP: $compressed\n";
printf "%d bytes\n", length $compressed;
my @data;
# Store the compressed bit into array
push @data, $compressed;
# process the array
foreach my $dat ( @data ) {
my $decompressed = unpack 'b*', $dat;
$decompressed
Dear all,
The basename() function returns the extension also:
> myfile <- "path1/path2/myoutput.txt"
> basename(myfile)
[1] "myoutput.txt"
Is there any other function where it just returns
plain base:
"myoutput"
i.e. without 't
favorite questions; "tell me what you
> want to do, not how you want to do it".
>
> On Thu, Jan 8, 2009 at 6:12 AM, Gundala Viswanath wrote:
>> Dear all,
>>
>> I found that printing with 'cat' is very slow.
>>
>> For example in my mach
s already very large (>2Gb)
for ( s in 1:length(x) ) {
cat(as.character(foo[s]),"\t",bar[s],"\t", qux[s],"\n")
}
__END__
for "x" of size ~1.5million, takes more than 10 hours to print.
On my Linux 1994.MHz AMD processor.
Is there any faster al
Dear all,
Does R has any function that measures how much
memory hold by any particular object?
- Gundala Viswanath
Jakarta - Indonesia
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PLEASE do read the posting guide
0 0 0 0 0 0 0 0 ...
- Gundala Viswanath
Jakarta - Indonesia
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contain
0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 1
[3,] 0 0 0 0 0 0 0 0 02
I tried:
> as.matrix(x)
But failed.
- Gundala Viswanath
Jakarta - Indonesia
__
3 0
In principle:
1. Number of Column in matrix = length of string (= 3)
2. Number of Row in matrix = length of vector ( =4).
3. Character "a" encode as "0",
"c" -> "1",
"g" -> "2",
"t" -> "3"
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