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
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 -
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 HTML
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, origin=-0.05,
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?
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?
G.V.
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
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
I have only *two* datasets from normal and cancer samples.
CancerNormal
--
mRNA1 3049
mRNA2 199200
... ... ...
mRNA1000 1340
Each samples contain several thousan mRNA microarray
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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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
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)))
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 with the
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
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 to
:50, Gundala Viswanath a écrit :
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
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.
hiv
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 can
=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.
Best,
Jorge
On Tue, Feb 1, 2011 at 12:20 AM, Gundala Viswanath wrote:
I have a R script that contain these lines
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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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$label[,o];
- G.V.
I am refering to a function call like this:
data(iris)
x - svmlight(Species ~ ., data = iris)
I tried to see the content of it by typing:
Species ~ .
but it gives nothing. How can I see it's content ?
- P.Dubois
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I have a vector that looks like this:
foo
[1] o o o x o o o o o x x o x
How can we find the percentage of o and x in
that vector in R?
- G.V
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PLEASE do read the posting
I have a variable that looks like this:
print(pred$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 fail?
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
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
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 x if
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
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
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
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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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.
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
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 this
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);
plotfn
-project.org]
Namens Gundala Viswanath
Verzonden: woensdag 2 september 2009 12:10
Aan: r-h...@stat.math.ethz.ch
Onderwerp: [R] Howto Superimpose Multiple Density Curves Into One Plot
I have a data that looks like this:
http://dpaste.com/88561/plain/
And I intend to create multiple density
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 september 2009 14:33
Aan: ONKELINX
a given body of data.
~ John Tukey
-Oorspronkelijk bericht-
Van: Gundala Viswanath [mailto:gunda...@gmail.com]
Verzonden: woensdag 2 september 2009 14:47
Aan: ONKELINX, Thierry
CC: r-h...@stat.math.ethz.ch
Onderwerp: Re: [R] Howto Superimpose Multiple Density Curves Into One Plot
Hi
-
Van: Gundala Viswanath [mailto:gunda...@gmail.com]
Verzonden: woensdag 2 september 2009 14:54
Aan: ONKELINX, Thierry
CC: r-h...@stat.math.ethz.ch
Onderwerp: Re: [R] Howto Superimpose Multiple Density Curves Into One Plot
str(dat)
'data.frame': 200 obs. of 2 variables:
$ V1: num 0.98
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
-Oorspronkelijk bericht-
Van: Gundala Viswanath [mailto:gunda...@gmail.com]
Verzonden
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=2)
But
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?
#id
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
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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 - Indonesia
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
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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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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/posting
- 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
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
information, hence I am aware that
I dont' want:
plot(hist(dat[dat =-500 dat =500]))
- Gundala Viswanath
Jakarta - Indonesia
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PLEASE do read the posting guide http
(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)
}
__ END__
- Gundala Viswanath
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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https
);
}
}
dev.off()
__END__
- Gundala Viswanath
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained
that it gives such error?
- 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
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
- Gundala Viswanath
Jakarta - Indonesia
__
R
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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https
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
Jakarta - Indonesia
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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V12 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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https
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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PLEASE do read
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
gustaf.ryde...@gmail.com wrote:
On Wed, Feb 11, 2009 at 2:15 PM
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
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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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 h.wick...@gmail.com wrote:
On Sat, Jan 17, 2009 at 7:59 AM
How do you do that Henrik? Write.table doesn't have that option.
Usage:
write.table(x, file = , append = FALSE, quote = TRUE, sep = ,
eol = \n, na = NA, dec = ., row.names = TRUE,
col.names = TRUE, qmethod = c(escape, double))
- Gundala Viswanath
Jakarta
,
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 is too small to keep it.
- Gundala Viswanath
for my naive question.
- Gundala Viswanath
Jakarta - Indonesia
On Fri, Jan 16, 2009 at 9:09 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Fri, Jan 16, 2009 at 5:52 AM, r...@quantide.com r...@quantide.com wrote:
I agree on the database solution.
Database are the rigth tool to solve
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 one.
- Gundala Viswanath
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 jholt...@gmail.com wrote:
Here is a function I use to see how big the objects in my workspace
How can I acces Slot 'ra' only?
I tried
print(x$ra)
but fail.
- Gundala Viswanath
Jakarta - Indonesia
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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 Viswanath
Jakarta - Indonesia
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
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Thanks for your most reasonable reply, Henrique.
- Gundala Viswanath
Jakarta - Indonesia
On Tue, Jan 13, 2009 at 8:01 PM, Henrique Dallazuanna www...@gmail.com wrote:
Try this:
unlist(ifelse(q %in% names(x), x[q], NA))
On Tue, Jan 13, 2009 at 8:49 AM, Gundala Viswanath gunda...@gmail.com
the exact byte size.
- Gundala Viswanath
Jakarta - Indonesia
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1QABA ...
But why this command fail to do it?
dimnames(mat) - NULL
Btw, the object size for keeping matrix with dim names is greater
than without, right?
- Gundala Viswanath
Jakarta - Indonesia
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https
, and
query around ~1 million element.
- 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, ATT, )
Sorry, I should have mentioned this earlier.
- Gundala Viswanath
Jakarta - Indonesia
On Tue, Jan
instead?
- Gundala Viswanath
Jakarta - Indonesia
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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 12:14 PM, jim
DACCD BEAAD CDDDA ABDCA ACACC
DADAA ABCAD ...
system.time(y - match(x,z))
user system elapsed
0.2 0.0 0.2
str(y)
int [1:2000] 1 2 3 4 5 6 7 8 9 10 ...
On Mon, Jan 12, 2009 at 10:17 PM, Gundala Viswanath gunda...@gmail.com
wrote:
Yes Jim, exactly.
BTW, I found from
(new_repo)
[1] AAA AAT AAC AAG ATA ATT...
I tried as.vector(), but it remains the same factor.
- Gundala Viswanath
Jakarta - Indonesia
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...
- Gundala Viswanath
Jakarta - Indonesia
On Tue, Jan 13, 2009 at 2:40 PM, bill.venab...@csiro.au wrote:
as.character()
Bill Venables
http://www.cmis.csiro.au/bill.venables/
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
of Row in matrix = length of vector ( = 4).
3. Character a encode as 0,
c - 1,
g - 2,
t - 3
Length of strings are assumed to be uniform within the vector,
and it can be greater than 3 (up to 40 characters).
- Gundala Viswanath
Jakarta - Indonesia
my $dat ( @data ) {
my $decompressed = unpack 'b*', $dat;
$decompressed =~ s/(..)/$digmap{$1}/ge;
print $decompressed\n;
# or do further processing on $dat
}
__END__
- Gundala Viswanath
Jakarta - Indonesia
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(2Gb)
for ( s in 1:length(x) ) {
cat(as.character(foo[s]),\t,bar[s],\t, qux[s],\n)
}
__END__
for x of size ~1.5million, takes more than 10 hours to print.
On my Linux 1994.MHz AMD processor.
Is there any faster alternatives to cat ?
- Gundala Viswanath
Jakarta - Indonesia
want to do, not how you want to do it.
On Thu, Jan 8, 2009 at 6:12 AM, Gundala Viswanath gunda...@gmail.com wrote:
Dear all,
I found that printing with 'cat' is very slow.
For example in my machine this snippet
__BEGIN__
# I need to resolve to use this type of loop.
# because using write
Dear all,
The basename() function returns the extension also:
myfile - path1/path2/myoutput.txt
basename(myfile)
[1] myoutput.txt
Is there any other function where it just returns
plain base:
myoutput
i.e. without 'txt'
- Gundala Viswanath
Jakarta - Indonesia
0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 1
[3,] 0 0 0 0 0 0 0 0 02
I tried:
as.matrix(x)
But failed.
- Gundala Viswanath
Jakarta - Indonesia
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R
Dear all,
I have the following structure of a matrix
str(mdat)
num [1:32268, 1:10] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : NULL
How can I destroy the attributes such that it simply
gives:
str(mdat)
num [1:32268, 1:10] 0 0 0 0 0 0 0 0 0 0 ...
- Gundala
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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TTCTCCGGCGACCACCGAGATCTACACTCTTTCC 18.0 7 9.00 18.0 12.0 18.0 15.0
16 1 TTCTCCGGCGACCACCGCGATCTACACTCTTTCC 18.0 7 9.00 18.0 12.0 18.0 15.0
My question is how can I extract the column V3 up to V9 into another
new data frame.
I tried this but failed:
str - paste(V, 3:9, sep=)
print(dat$str)
- Gundala
:
__BEGIN__
open INFILE, '' , 'filename.txt' or die $!;
while (INFILE) {
my $line = $_;
# then process line by line
}
__END__
the reason I want to do that is because the data
I am processing are large (~ 5GB), my PC may not
be able to handle that.
- Gundala Viswanath
Jakarta - Indonesia
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