Hi
>
> hello, this is my script:
>
> #1) read in data:
>
daten<-read.table('K:/Analysen/STRUCTURE/input_STRUCTURE_tab_excl_5_282_559.txt',
> header=TRUE, sep="\t")
> daten<-as.matrix(daten)
If there is any column with nonnumeric values it will transfer all numeric
values from daten data.frame
Hi
I do not understand much about your equations. I think you shall look to
Practical Regression and Anova Using R from J.Faraway.
Having data frame DF with columns - users, groups, results you could do
fit <- lm(results~groups, data = DF)
Regards
Petr
>
> Hi,
>
> I'm a newbie to R. My
Hi
r-help-boun...@r-project.org napsal dne 05.12.2010 00:19:17:
>
> year1 is the time series data set below
> 11.64
> 11.50
> 11.49
> 11.16
> 11.15
> 11.37
> 11.37
> 11.57
> 11.83
> 11.87
> 11.85
> 11.92
> 11.77
> 11.71
> 11.57
> 11.24
> 11.27
> 11.33
> 11.17
> .
>
> With a tota
Please,when i export the output from R to excel.I am not getting all the
15,000 observations but only 2000.Thank you
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__
Hi
write.table(tab, "clipboard", sep = "\t", row.names = FALSE)
or
write.table(tab, "somefile.xls", sep = "\t", row.names = FALSE)
Regards
Petr
r-help-boun...@r-project.org napsal dne 13.08.2010 01:47:40:
>
> I need a code to export my output to excel 2007.I am dealing with
> observations o
Hi
r-help-boun...@r-project.org napsal dne 24.07.2010 11:48:11:
>
> Thanks Murphy and pikal,
>
> I need another help,for fitting first fourier transformation ,i used
> following thing .Please advise on this
>
>
> beer_monthl has 400+ records
>
> EXample:
> > head(beer_monthly)
> beer
> 1 9
Thanks Murphy and pikal,
I need another help,for fitting first fourier transformation ,i used
following thing .Please advise on this
beer_monthl has 400+ records
EXample:
> head(beer_monthly)
beer
1 93.2
2 96.0
3 95.2
4 77.1
5 70.9
6 64.8
time<-seq(1956,1995.2,length=length(beer_monthly)
Hi
r-help-boun...@r-project.org napsal dne 21.07.2010 17:46:02:
>
> Hi R-community,
> I have the code as follows,i Fitted model as follows
> lbeer<-log(beer_monthly)
> t<-seq(1956,1995.2,length=length(beer_monthly)) #beer_monthly contains
400+
> entries
not needed
> t2=t^2
> beer_fit_parabola
Hi
if you do not insist on switch you can use factor to get desired result
set.seed(666)
x<-sample(1:5,20, replace=T)
x
[1] 4 1 5 2 2 4 5 3 1 2 4 1 1 1 2 5 1 5 3 3
factor(x, labels=letters[1:5])
[1] d a e b b d e c a b d a a a b e a e c c
Levels: a b c d e
r-help-boun...@r-project.org napsal
Hi
r-help-boun...@r-project.org napsal dne 01.06.2010 13:01:38:
> Dear All,
>
> I am newbie to R, and I wanted to plot a barplots with R and in such a
way
> that It will also show me position which I can plot on the bar line.
>
> Here is my code that I am using to plot,
>
> > chromosome <- c(
Hi
again a work for cut.
See ?cut and notice a labels option.
Regards
Petr
r-help-boun...@r-project.org napsal dne 26.03.2010 11:37:20:
> Hi All
>
> I have a column/variable called time difference. It has a whole list of
> numbers from 0 through to the hundreds eg 236. I want to assign a
>
Hi
r-help-boun...@r-project.org napsal dne 30.10.2009 14:40:06:
> Hi All:
>
> I have a data frame called all_corn. This has 31 columns. The first
column is
> a character key. The next 15 columns (stat1,stat2,...,stat15) are the
> statistical forecast. The last 15 columns (sls1,sls2,...,sls5)
Hi
r-help-boun...@r-project.org napsal dne 26.10.2009 10:48:51:
> Dear list,
>
> I have been searching for a week to fit a simple linear model to my
data. I
> have looked into the previous posts but I haven't found anything
relevant to
> my problem. I guess it is something simple...I just can
Hi
r-help-boun...@r-project.org napsal dne 26.05.2009 15:34:25:
>
> Hi
>
> I have a vector of data lets call zz (40 values from 4 samples)
Are you sure it is a vector? Your indexing suggest data frame or matrix.
> the data is already in groups, i can even split up the samples using
>
> SampA
Hi
r-help-boun...@r-project.org napsal dne 28.04.2009 15:04:01:
>
> My data contains a variable "observation_date" and it contains values
as:
>
> 1985-09-02
> 1985-09-15
> 1985-07-31
> 1985-09-02
>
>
> I need to process data annually rather than daily, therefore I'm trying
to
> 1) either e
Petr Pikal
petr.pi...@precheza.cz
724008364, 581252140, 581252257
r-help-boun...@r-project.org napsal dne 16.04.2009 16:45:15:
>
> I have csv files imported in r each with 2 columns and many many rows. I
have
> sorted the data in them but want to extract some values.
>
> The first column is
Hi
r-help-boun...@r-project.org napsal dne 24.02.2009 06:23:23:
> Hi R users,
>
> I have a question. How can I use for loop to do pair comparisons. For
> example,
>
> > x<-c(1,2,3)
>
> > result<-matrix(data=NA, nrow=choose(3,2), ncol=1)
>
> > for(i in 1: length(x))
>
> +{ result[i,]<-ife
Hi
r-help-boun...@r-project.org napsal dne 14.01.2009 14:34:14:
> Dear ALL
> suppose "x=7.5",and i need of only integer part of variable "x" that is
"7"
> only then what command i can use in R.
If you looked at help page for integer you would quickly find links to
trunc, round, floor, ceiling
1/(x-axis value) = frequency in time
(the x-axis is in cycles per time)
2008/9/12 Petr PIKAL <[EMAIL PROTECTED]>:
> Hi
>
> [EMAIL PROTECTED] napsal dne 09.09.2008 18:44:34:
>
>> For the command 'spectrum' I read:
>>
>> The spectrum here is defined with scaling 1/frequency(x), following
> S-PLUS.
>
Hi
[EMAIL PROTECTED] napsal dne 09.09.2008 18:44:34:
> For the command 'spectrum' I read:
>
> The spectrum here is defined with scaling 1/frequency(x), following
S-PLUS.
> This makes the spectral density a density over the range
(-frequency(x)/2,
> +frequency(x)/2], whereas a more common sca
Hi
see from plot.boot help page
Side Effects
All screens are closed and cleared and a number of plots are produced on
the current graphics device. Screens are closed but not cleared at
termination of this function.
You need to change a behaviour of this function what shall not be simple.
Or m
Hi
I am not sure what you want to do as you speak about using different
function for different part of your data.
> head(data)
city year sex V
11 1975 F 25.3044
21 1975 M 16.5711
31 1976 M 16.6072
41 1976 F 24.2841
51 1977 M 14.8838
61 1977 F 24.8124
Hi
If you bother to look into posting guide it will tell you that providing a
reproducible example is a good way to obtain relevant answers to your
question.
And maybe while constructing reproducible example you will find yourself
what is wrong, learn some more from it and need not wait for a
Hi
AFAIK sometimes this can happen if your .Rdata was saved when some
packages had been in action. But in that case you should have your message
extended with someting like:
"cannot load package " or similar.
It helped me when I started plain R session invoked necessary packages and
open
Hi
[EMAIL PROTECTED] napsal dne 02.12.2007 23:33:47:
>
> I am having trouble getting a loop to work for the following problem.
Any
> help would be much appreciated. Thanks.
Lets try it without loop.
Does this
lapply(split(data[,4:3], data$Participant), function(x) coef(lm(x)))
give you resu
Hi
[EMAIL PROTECTED] napsal dne 24.11.2007 08:55:54:
> Dear list,
>
> I want to combine several plots in one graph.
> I did this: plot(a1); plot(a2, add=TRUE); ...plot(a5, add=TRUE)
> The problem is the more plot we put, the more complex the graph.
> Is there any way to label each line; or other
onas Ă–ster
Cc: r-help@r-project.org
Subject: [R] Odp: help with image function
Hi
[EMAIL PROTECTED] napsal dne 24.10.2007 12:01:09:
> Hallo!
> Have succeded in creating an image plot using image()
> Have failed to provide explanation to what the colors means in terms of
> values/nu
Hi
[EMAIL PROTECTED] napsal dne 24.10.2007 12:01:09:
> Hallo!
> Have succeded in creating an image plot using image()
> Have failed to provide explanation to what the colors means in terms of
> values/numbers (i.e. the height of the staples that the colors
represent).
> Wonder if anyone have any
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