There is some terminology confusion here... interpolation as implemented by
approx or spline usually means estimating values between known points. You
seem to have approximate (not known) points, and are looking to apply a linear
regression model to estimate missing data. Beware that mixing est
Dear all,
I am a newbie in interpolation using R and I would like to learn
better the procedure.
I am applying interpolation to quantify nucleic acid targets using an
assay known as PCR. To do this, I have two sets of variables: standard
of known concentrations and query for which I need to identif
Dear All,
An interpolation grid of spacing 100m by 100m was created for irregular
space spatial data. The plot of the grid and sample data shows some grid
points outside the data area. I guess the predicted values of those points
outside the data area will not be reliable. How do I modify my R scri
Hello R-helpers,
I have a data set of points in three dimensions (x, y, z), each with an
associated amplitude. These are data collected using a radar, and of tree
roots.
There are holes in the data, and I'm interested in interpolating the
missing values to create a root map. This could simply be
Jan,
There are a lot of packages that can help you, the best one depends on
your needs (with or without prediction uncertainty, format of results,
different options) and the size of your problem.
CRAN has a spatial Task View
http://cran.r-project.org/web/views/Spatial.html
with a short descrip
Hi
I have a data set with geo coordinates and values for each coordinate.
I want to interpolate the values to new positions on a finer grid,
also geo coordinates.
I have looked at the fields package (interp.surface) and the akima
package (interp) but cant quite figure what I am doing wrong, or if
On Sat, Jun 16, 2012 at 8:19 AM, stef salvez wrote:
> I have a panel data set (in MS excel) like the one below
>
>
> 1 "23/11/08" 2
> 1 "28/12/08" 3
> 1 "25/01/09" 4
> 1 "22/02/09" 5
> 1 "29/03/09"
Hi Ken, Stef,
We can make your script more elegant like below:
On Sun, Jun 17, 2012 at 12:52 AM, Ken wrote:
>
> stef salvez googlemail.com> writes:
[snip]
> #load library
> library(plyr)
>
> # utility function
> mean.var = function(df, var){ mean(df[[var]], na.rm = T)};
>
> # create example
stef salvez googlemail.com> writes:
>
> I would like to clarify that since each observation is obtained every
> 28 days, each such observation is a 4-week average
>
> thanks
>
> On 6/16/12, stef salvez googlemail.com> wrote:
> > I have a panel data set (in MS excel) like the one below
> >
>
I would like to clarify that since each observation is obtained every
28 days, each such observation is a 4-week average
thanks
On 6/16/12, stef salvez wrote:
> I have a panel data set (in MS excel) like the one below
>
>
> 1 "23/11/08"2
> 1 "28/12/08" 3
I have a panel data set (in MS excel) like the one below
1 "23/11/08"2
1 "28/12/08" 3
1"25/01/09" 4
1 "22/02/09" 5
1"29/03/09" 6
1 "26/04/09" 32
1 "24/05/09"
Dear R-users,
I am working on interpolating the station level temperature data to farm level
data. I have z vector consisting of station level temperature observations and
my x and y are latitude and longitude corresponding to a farm. My understanding
is I can use raster combined with tps.
On Wed, Apr 18, 2012 at 07:15:45AM -0700, uday wrote:
> hi Petr ,
> Thanks for replay and sorry for typo mistake
> approx(pres, sci.pre) its nothing but approx(pre2, pre1).
>
> so for more simplicity
> x <- c(10.34615 , 52.02116, 146.17357, 243.28644, 347.41504, 431.67105,
> 521.4
Your problem is that length(x) != length(y)
approx uses linear interpolation but there's no way to make sense of
that if you can't match up the x and y coordinates -- and you can't
match up the x and y coordinates if there aren't the same number of
them.
Michael
On Wed, Apr 18, 2012 at 10:15 AM,
hi Petr ,
Thanks for replay and sorry for typo mistake
approx(pres, sci.pre) its nothing but approx(pre2, pre1).
so for more simplicity
x <- c(10.34615 , 52.02116, 146.17357, 243.28644, 347.41504, 431.67105,
521.42712, 629.00446 ,729.95941, 827.86279, 921.55078,
956.6)
y <- c(
On Wed, Apr 18, 2012 at 03:55:07AM -0700, uday wrote:
> This moment I got stuck with one interpolation issue
> the sample data which I have is as follows
>
> pre1 <- c(10.34615 , 52.02116, 146.17357, 243.28644, 347.41504, 431.67105,
> 521.42712, 629.00446 ,729.95941,82
This moment I got stuck with one interpolation issue
the sample data which I have is as follows
pre1 <- c(10.34615 , 52.02116, 146.17357, 243.28644, 347.41504, 431.67105,
521.42712, 629.00446 ,729.95941,827.86279,
921.55078, 956.6)
pre2 <- c( 983.4477692, 973.619
On Jun 1, 2011, at 9:24 AM, Clement LAUZIN wrote:
Hello,
I have a x,y,z file.Z is not corresponding to a simple analytical
function of x and y (see below).
I am trying to find the minimum location of this surface and the z
value corresponding to this location by a spline interpolation or
Clement LAUZIN hotmail.com> writes:
> Hello,
>
> Hello,
>
> I have a x,y,z file.Z is not corresponding to a simple analytical function
> of x and y (see below). I am trying to find the minimum location of this
> surface and the z value corresponding to this location by a spline
> interpolation
Hello,
I have a x,y,z file.Z is not corresponding to a simple analytical function of x
and y (see below).
I am trying to find the minimum location of this surface and the z value
corresponding to this location by a spline interpolation or from a polynomial
fit.
I tried with the akima package
Hello,
I have a x,y,z file.
Z is not corresponding to a simple analytical function of x and y.
I am trying to find the minimum value of z by a spline interpolation or from a
polynomial fit.
I tried the akima package but as the location of the point I am looking for
(the minimum) is outside of t
Hi:
Look into the na.approx() function in package zoo. The discussion below may
be of help:
http://r.789695.n4.nabble.com/Filling-in-missing-time-samples-with-na-approx-td3063682.html
HTH,
Dennis
On Wed, Jan 5, 2011 at 10:17 PM, Rustamali Manesiya wrote:
> Hello,
>
>
> I am new to R and n
Hello,
I am new to R and need some help.
I have data in following format
DATA matrix
DateTime oh l c
2009-01-01 07:30:00 2 3 4 5
2009-01-01 07:33:00 4 2 5 7
I am able to fill the gap using combination of seq and chron
2009-0
try packages:
{yaImpute}, {impute}, etc.
--
View this message in context:
http://r.789695.n4.nabble.com/Interpolation-missing-data-tp2530871p2531288.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
ht
Hi R experts,
I have set of data consists of 50 data. some of them are missing. I would
need a function in R that can estimate missing data using interpolation
methods.
If you know this kind of function, write me the name of the function and its
library.
Thanks very much in advance!
abotaha
_personal_pages/Varadhan.h
tml
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of ogbos okike
Sent: Thursday, January 28, 2010 12:30 PM
To: r-help@r-project.org
Subject: [R] Interpolation
Why not look into the zoo package na.approx? And related functions.
On Thu, Jan 28, 2010 at 11:29 AM, ogbos okike wrote:
> Happy New Year.
> I have a data of four columns - year, month, day and count. The last column,
> count, contains some missing data which I have to replace with NA. I tried
>
On Jan 11, 2010, at 11:49 AM, René Mayer wrote:
My problem is that x values increas with y
x is mostly decreasing in the order you presented:
plot(x, type="l")
until some point then the pattern
reverses. The whole line
which line?
is a kind of U-shape with a right-buttom to middel-top d
My problem is that x values increas with y until some point then the pattern
reverses. The whole line is a kind of U-shape with a right-buttom to
middel-top diagonal at the end of it (a look at the plot makes it
clearer). The interpolation (approx, spline) makes a zick-zack aut of
it. What I
On Jan 11, 2010, at 7:44 AM, René Mayer wrote:
Dear R-users,
I have a complex line by xy-values (ordered by z).
And I would like to get interpolated y-values on the positions of x
= 0:600.
How do I get the correct points?
x
=
c
(790,790,790,790,790,786,783,778,778,766,763,761,761,761,715
Dear R-users,
I have a complex line by xy-values (ordered by z).
And I would like to get interpolated y-values on the positions of x = 0:600.
How do I get the correct points?
x=c(790,790,790,790,790,786,783,778,778,766,763,761,761,761,715,628,521,350,160,134,134,129,108,101,93,111,161,249,288,243
Hi R community
I need to interpolate precipitation data for a natural park. I have
precipitation data from some climate stationts. (I know the table is not
complete but I only need to show you X,Y, Altitude and PrepJul)
X
Y
Altitude
PrepJan
PrepFeb
PrepMar
PrepAp
PrepMay
It appears the answer to your goal after a discursive exploration of
"interpolation", which was really extrapolation, is that you need to
look at the predict methods for linear (and other sorts as well) models.
?predict
?predict.lm
> y <- c(16,45,77,101,125)
> x <- c(0,5,10,15,20)
>
> lmmo
On Jan 15, 2009, at 11:31 AM, e-letter wrote:
Perhaps a coding error on my part (or on your part). Perhaps
different
methods (none of which you describe)?
I suspect that my method only used the first two points (I just
checked by plotting and -2.7 is closer to the paper and pen result I
g
> Perhaps a coding error on my part (or on your part). Perhaps different
> methods (none of which you describe)?
>
> I suspect that my method only used the first two points (I just
> checked by plotting and -2.7 is closer to the paper and pen result I
> get than is -3.28. Perhaps you made an extra
On Jan 15, 2009, at 10:04 AM, e-letter wrote:
On 13/01/2009, David Winsemius wrote:
It's fairly clear from the documentation that approxfun() will not
extrapolate.
help.search("extrapolate")
library(Hmisc)
?approxExtrap
Some sort of minimization approach:
approxExtrap(x=c(0,5,10,15,20),
On 13/01/2009, David Winsemius wrote:
> It's fairly clear from the documentation that approxfun() will not
> extrapolate.
>
> help.search("extrapolate")
> library(Hmisc)
> ?approxExtrap
>
> Some sort of minimization approach:
>
> > approxExtrap(x=c(0,5,10,15,20), y=c(16,45,77,101,125),xout=c(-4,0
It's fairly clear from the documentation that approxfun() will not
extrapolate.
help.search("extrapolate")
library(Hmisc)
?approxExtrap
Some sort of minimization approach:
> approxExtrap(x=c(0,5,10,15,20), y=c(16,45,77,101,125),xout=c(-4,0,4))
$x
[1] -4 0 4
$y
[1] -7.2 16.0 39.2
> approxE
>
> What is the problem that you are trying to solve?
>
>From the data I provided: x=c(0,5,10,15,20) y=c(16,45,77,101,125); I
want to obtain the value of x when y=0.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
approxfun returns a function; that is not an error message:
> x=c(0,5,10,15,20)
> y=c(16,45,77,101,125)
>
> approx(x,y,method="linear")
$x
[1] 0.000 0.4081633 0.8163265 1.2244898 1.6326531 2.0408163
2.4489796 2.8571429 3.2653061
[10] 3.6734694 4.0816327 4.4897959 4.8979592 5.306
On 08/01/2009, Greg Snow wrote:
> If you want to just linearly interpolate, then use the functions approx or
> approxfun from the stats package (one of those that is loaded by default).
I have read the guide for approx and approxfun functions. Below is my data.
x=c(0,5,10,15,20)
y=c(16,45,77,101,
8, 2009 9:22 AM
> To: r-help@r-project.org
> Subject: [R] interpolation to abscissa
>
> Readers,
>
> I have looked at various documents hosted on the web site; I couldn't
> find anything on interpolation. So I started r and accessed the help
> (help.start()). (by the
Readers,
I have looked at various documents hosted on the web site; I couldn't
find anything on interpolation. So I started r and accessed the help
(help.start()). (by the way is it possible to configure r to open help
in opera instead of firefox?) Initially I read the help for the akima
package b
Thank you very much, You have helped me to resolve the problem.
Thank you!!
A greetings, Luismi
Henrique Dallazuanna wrote:
>
> I think that you can use the splinefun function:
>
> f <- splinefun(x, y)
>
> f(15)
>
> On Thu, Sep 4, 2008 at 1:52 PM, ermimi <[EMAIL PROTECTED]> wrote:
>
>>
I think that you can use the splinefun function:
f <- splinefun(x, y)
f(15)
On Thu, Sep 4, 2008 at 1:52 PM, ermimi <[EMAIL PROTECTED]> wrote:
>
> Hello friends!!!
>
> I have a list of values called y.
> The list is y=c(221.0, 212.0, 206.0, 202.7, 198.4, 195.1, 192.2, 189.7,
> 187.6, 185.8);
> y
Hello friends!!!
I have a list of values called y.
The list is y=c(221.0, 212.0, 206.0, 202.7, 198.4, 195.1, 192.2, 189.7,
187.6, 185.8);
y is f(x) and x=c(10,20,30,40,50,60,70,80,90,100).
I only have x and y. I don´t know f(x). I would like interpolate f(x) to
obtain other values as f(15), f(2
ueToCheck-x0)*(y1-y0)/(x1-x0)
}
}
return(y)
}
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Steve Murray
Sent: Monday, September 01, 2008 6:17 PM
To: r-help@r-project.org
Subject: [R] Interpolation Problems
Dear all,
I'm trying to interpolate a datase
Steve Murray hotmail.com> writes:
>
>
> Thanks Duncan - a couple of extra points... I should have perhaps pointed
> out that the data are on a *regular*
> 'box' grid (with each value currently spaced at 1 degree intervals). Also,
> I'm looking for something
> fairly simple, like a bilinear i
again,
Steve
> Date: Mon, 1 Sep 2008 18:45:35 -0400
> From: [EMAIL PROTECTED]
> To: [EMAIL PROTECTED]
> CC: r-help@r-project.org
> Subject: Re: [R] Interpolation Problems
>
> On 01/09/2008 6:17 PM, Steve Murray wrote:
>> Dear all,
>>
>> I'm trying to inte
On 01/09/2008 6:17 PM, Steve Murray wrote:
Dear all,
I'm trying to interpolate a dataset to give it twice as many values (I'm giving
the dataset a finer resolution by interpolating from 1 degree to 0.5 degrees)
to match that of a corresponding dataset.
I have the data in both a data frame for
Dear all,
I'm trying to interpolate a dataset to give it twice as many values (I'm giving
the dataset a finer resolution by interpolating from 1 degree to 0.5 degrees)
to match that of a corresponding dataset.
I have the data in both a data frame format (longitude column header values
along t
Does this thread solve your problem? ->
https://stat.ethz.ch/pipermail/r-help/2007-July/136814.html
On 10-Jul-08, at 3:15 AM, [EMAIL PROTECTED] wrote:
Hello,
I have the data whcih are not balanced (several missing observations),
and one possibility is t use interpolation method
to get the inf
Try
help.search('interpolate')
and
help.search('impute')
(most of the responses to the latter come from packages that you may
not have installed, such as Hmisc)
-Don
At 8:15 AM +0200 7/10/08, [EMAIL PROTECTED] wrote:
Hello,
I have the data whcih are not balanced (several missing observ
if it is a time series the interpolation methods in zoo are an option.
On Thu, Jul 10, 2008 at 6:41 AM, Daniel Malter <[EMAIL PROTECTED]> wrote:
>
> Please do read the posting guide. Please provide self-contained code (e.g.
> to
> randomly generate data) and illustrate (e.g. in a small table) wha
Please do read the posting guide. Please provide self-contained code (calls
to randomly generated data) and illustrate (e.g. in a small table) what you
want to do and also illustrate (with the self-contained code) where your
current approach (if any) fails. After reading your message, I have only
Hello,
I have the data whcih are not balanced (several missing observations),
and one possibility is t use interpolation method
to get the information missing in this series from other series.
Does anybody know how I can program interpolation of
series1 (which ahs missing observations) and series
Hello,
I tried the approx() and it worked. Now, I have a list named "interpol"
resulting from the interpolation. I would like to append the values in
interpol$y in the position specified by interpol$x in a existing vector
"spect1". I tried with append() and the following code:
spect1 <- c(1:109
check out the 'approx' function.
On Feb 19, 2008 12:44 PM, Dani Valverde <[EMAIL PROTECTED]> wrote:
> Hello,
> I have two vectors, one with 13112 points and the other one with 10909.
> I wonder if there is a way to interpolate the data so the shorter
> vectors has the same number of points as the
Hello,
I have two vectors, one with 13112 points and the other one with 10909.
I wonder if there is a way to interpolate the data so the shorter
vectors has the same number of points as the longer one.
Best,
Dani
--
Daniel Valverde Saubí
Grup de Biologia Molecular de Llevats
Facultat de Veteri
Maybe arima with the xreg= argument.
On 10/11/07, Creighton, Sean <[EMAIL PROTECTED]> wrote:
> Hi
>
> I have a collection of about 16 time series with occasional missing
> data. A few of these time-series start later than the rest. There is a
> relatively high correlation between them (they are ho
Hi
I have a collection of about 16 time series with occasional missing
data. A few of these time-series start later than the rest. There is a
relatively high correlation between them (they are hourly temps at
various locations around the UK). The longest series contains about
4 points
I have
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