date:20131210

Hi,

It's unlikely that anyone will be able to see your problem without a
reproducible example.

http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

Are you using the mclust package? That would be the first thing we need to know.

Then, have you worked through the examples in ?predict.Mclust, and do
you understand them?

Then, what do your data look like? str() and such are useful, but the
best option is to use dput() to create a reproducible subset, or to
reproduce your problem using one of the built-in datasets in R, as is
done in the above documentation examples.

Sarah

On Tue, Dec 10, 2013 at 6:41 AM, Welma Pereira welma.pere...@gmail.com wrote:
 Hi,

 I am trying to use mclust to cluster some data (train_pca_10), I get the
 clusters, but when I try to use the model to predict new data (test1) I get
 this error



 mClust2 - Mclust(train_pca_10,G=2)
 pred-predict.Mclust(mClust2,test1)

 Error in if (warn) warning(WARNING) : argument is of length zero

 Can anyone see the problem here?

 Thanks,
 Pereira.

-- 
Sarah Goslee
http://www.functionaldiversity.org

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Using assign with mapply

2013-12-10 Thread Julio Sergio Santana

David Winsemius dwinsemius at comcast.net writes:

 So what happens if you try this:
 
 mapply(assign,  kkk$vars, kkk$vals, MoreArgs = list(envir = .GlobalEnv)
 

Yes, it works in certain situations, as well as the equivalent code:

 kkk - data.frame(vars=c(var1, var2, var3), 
  vals=c(10, 20, 30), stringsAsFactors=F)

 mapply(assign,  kkk$vars, kkk$vals, MoreArgs = list(pos = 1))
 var1
[1] 10

See, however, the following example

: example - function () {
var1 - 250
kkk - data.frame(vars=c(var1, var2, var3), 
  vals=c(10, 20, 30), stringsAsFactors=F)
mapply(assign,  kkk$vars, kkk$vals, MoreArgs = list(pos = 1))
print (var2)
print (var1)
  }
 
  example()
 [1] 20
 [1] 250

My question is: how to get the combination of mapply and assign to affect 
the variables defined in the block where the construction is executed?

Maybe there is still something I don't understand.


Thanks,

  -Sergio.

__
R-help@r-project.org mailing list
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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.

Re: [R] 3-D interpretation

Hi Shane,

On Tue, Dec 10, 2013 at 10:35 AM, Shane Carey careys...@gmail.com wrote:
 Hi,

 im trying to create a 3-D interpretation of a geological fault using
 the akima package. But my code fails below highlighted in red:
 Does anyone have any ideas why this is.

This is a plain-text email list, so no highlighting in red is available.

More importantly, you should take a look at:

http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

and provide a simple example with data that demonstrates your problem.

Sarah


 Thanks

 dat-read.delim(D:\\fault.txt,header=T,sep=,)
 library(rgl)
 # data
 rgl.spheres(dat$X,dat$Z , dat$Y,1,color=red)
 rgl.bbox()
 # bivariate linear interpolation
 # interp:
 akima.li - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100))

 This is my error:
 Error in interp.old(x, y, z, xo = xo, yo = yo, ncp = 0, extrap = extrap,  :
 duplicate data points: need to set 'duplicate = ..'

 # interp surface:
 rgl.surface(akima.li$X,akima.li$Y,akima.li$Z,color=green,alpha=c(0.5))
 # interpp:
 akima.p - interpp(dat$X, dat$Y, dat$Z,
runif(200,min(dat$X),max(dat$X)),
runif(200,min(dat$Y),max(dat$Y)))
 # interpp points:
 rgl.points(akima.p$X,akima.p$Z , akima.p$Y,size=4,color=yellow)
 # bivariate spline interpolation
 # data
 rgl.spheres(dat$X,dat$Z ,dat$Y,0.5,color=red)
 rgl.bbox()
 # bivariate cubic spline interpolation
 # interp:
 akima.si - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100),
linear = FALSE, extrap = TRUE)




-- 
Sarah Goslee
http://www.functionaldiversity.org

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] 3-D interpretation

Hi, I have since solved that problem,

but now R keeps crashing. I have over 20,000 points. Can R handle that
amount?

Cheers


On Tue, Dec 10, 2013 at 3:35 PM, Shane Carey careys...@gmail.com wrote:

 Hi,

 im trying to create a 3-D interpretation of a geological fault using
 the akima package. But my code fails below highlighted in red:
 Does anyone have any ideas why this is.

 Thanks

 dat-read.delim(D:\\fault.txt,header=T,sep=,)
 library(rgl)
 # data
 rgl.spheres(dat$X,dat$Z , dat$Y,1,color=red)
 rgl.bbox()
 # bivariate linear interpolation
 # interp:
 akima.li - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100))

 This is my error:
 Error in interp.old(x, y, z, xo = xo, yo = yo, ncp = 0, extrap = extrap,
  :
 duplicate data points: need to set 'duplicate = ..'

 # interp surface:
 rgl.surface(akima.li$X,akima.li$Y,akima.li$Z,color=green,alpha=c(0.5))
 # interpp:
 akima.p - interpp(dat$X, dat$Y, dat$Z,
runif(200,min(dat$X),max(dat$X)),
runif(200,min(dat$Y),max(dat$Y)))
 # interpp points:
 rgl.points(akima.p$X,akima.p$Z , akima.p$Y,size=4,color=yellow)
 # bivariate spline interpolation
 # data
 rgl.spheres(dat$X,dat$Z ,dat$Y,0.5,color=red)
 rgl.bbox()
 # bivariate cubic spline interpolation
 # interp:
 akima.si - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100),
linear = FALSE, extrap = TRUE)






 --
 Shane




-- 
Shane

[[alternative HTML version deleted]]

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] 3-D interpretation

Hi,

im trying to create a 3-D interpretation of a geological fault using
the akima package. But my code fails below highlighted in red:
Does anyone have any ideas why this is.

Thanks

dat-read.delim(D:\\fault.txt,header=T,sep=,)
library(rgl)
# data
rgl.spheres(dat$X,dat$Z , dat$Y,1,color=red)
rgl.bbox()
# bivariate linear interpolation
# interp:
akima.li - interp(dat$X, dat$Y, dat$Z,
   xo=seq(min(dat$X), max(dat$X), length = 100),
   yo=seq(min(dat$Y), max(dat$Y), length = 100))

This is my error:
Error in interp.old(x, y, z, xo = xo, yo = yo, ncp = 0, extrap = extrap,  :
duplicate data points: need to set 'duplicate = ..'

# interp surface:
rgl.surface(akima.li$X,akima.li$Y,akima.li$Z,color=green,alpha=c(0.5))
# interpp:
akima.p - interpp(dat$X, dat$Y, dat$Z,
   runif(200,min(dat$X),max(dat$X)),
   runif(200,min(dat$Y),max(dat$Y)))
# interpp points:
rgl.points(akima.p$X,akima.p$Z , akima.p$Y,size=4,color=yellow)
# bivariate spline interpolation
# data
rgl.spheres(dat$X,dat$Z ,dat$Y,0.5,color=red)
rgl.bbox()
# bivariate cubic spline interpolation
# interp:
akima.si - interp(dat$X, dat$Y, dat$Z,
   xo=seq(min(dat$X), max(dat$X), length = 100),
   yo=seq(min(dat$Y), max(dat$Y), length = 100),
   linear = FALSE, extrap = TRUE)






-- 
Shane

[[alternative HTML version deleted]]

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] 3-D interpretation

Hi,

On Tue, Dec 10, 2013 at 10:42 AM, Shane Carey careys...@gmail.com wrote:
 Hi, I have since solved that problem,

It would be useful to post the solution, for the archives, so that
others with the same problem can benefit from your insight.


 but now R keeps crashing. I have over 20,000 points. Can R handle that
 amount?

R can. Your computer may not be able to.

But again, we don't have enough information to answer you, since that
depends on OS, available RAM, etc.

Sarah


 Cheers


 On Tue, Dec 10, 2013 at 3:35 PM, Shane Carey careys...@gmail.com wrote:

 Hi,

 im trying to create a 3-D interpretation of a geological fault using
 the akima package. But my code fails below highlighted in red:
 Does anyone have any ideas why this is.

 Thanks

 dat-read.delim(D:\\fault.txt,header=T,sep=,)
 library(rgl)
 # data
 rgl.spheres(dat$X,dat$Z , dat$Y,1,color=red)
 rgl.bbox()
 # bivariate linear interpolation
 # interp:
 akima.li - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100))

 This is my error:
 Error in interp.old(x, y, z, xo = xo, yo = yo, ncp = 0, extrap = extrap,
  :
 duplicate data points: need to set 'duplicate = ..'

 # interp surface:
 rgl.surface(akima.li$X,akima.li$Y,akima.li$Z,color=green,alpha=c(0.5))
 # interpp:
 akima.p - interpp(dat$X, dat$Y, dat$Z,
runif(200,min(dat$X),max(dat$X)),
runif(200,min(dat$Y),max(dat$Y)))
 # interpp points:
 rgl.points(akima.p$X,akima.p$Z , akima.p$Y,size=4,color=yellow)
 # bivariate spline interpolation
 # data
 rgl.spheres(dat$X,dat$Z ,dat$Y,0.5,color=red)
 rgl.bbox()
 # bivariate cubic spline interpolation
 # interp:
 akima.si - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100),
linear = FALSE, extrap = TRUE)



-- 
Sarah Goslee
http://www.functionaldiversity.org

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] 3-D interpretation

Hi Sarah,

here is some of the data:

structure(list(X = c(9816966.951, 9816963.08, 9816947.516, 9816939.51,
9816924.005, 9816916.096, 9816901.984, 9816896.967, 9816892.928,
9816890.743, 9816886.427, 9816884.006, 9816879.185, 9816876.468,
9816871.027, 9816868.276, 9816863.404, 9816860.712, 9816855.409,
9816852.487, 9816846.77, 9816843.635, 9816837.53, 9816834.196,
9816827.733, 9816824.214, 9816817.419, 9816811.926, 9816797.531,
9816789.588, 9816774.134, 9816765.716, 9816749.492, 9816740.717,
9816723.946, 9816714.931, 9816697.836, 9816675.781, 9816606.59,
9816578.983, 9816544.487, 9816531.179, 9816514.376, 9816505.891,
9816489.52, 9816481.288, 9816465.377, 9816457.374, 9816441.836,
9816434.026, 9816418.807, 9816411.161, 9816396.205, 9816385.052,
9816355.719, 9816341.037, 9816312.075, 9816304.796, 9816304.784,
9816916.962, 9816917.365, 9816918.982, 9816919.414, 9816919.519,
9816919.545, 9816599.694, 9816615.198, 9816677.27, 9816697.31,
9816714.671, 9816723.431, 9816740.464, 9816748.995, 9816765.474,
9816773.661, 9816789.36, 9816797.089, 9816811.764, 9816817.212,
9816824.113, 9816827.536, 9816834.1, 9816837.345, 9816843.545,
9816846.596, 9816852.404, 9816855.248, 9816860.635, 9816863.257,
9816868.198, 9816870.861, 9816876.39, 9816879.039, 9816883.937,
9816886.296, 9816890.68, 9816892.809, 9816896.829, 9816901.555,
9816915.878, 9816921.977, 9816931.382, 9816933.721, 9816619.581,
9816631.77, 9816680.572, 9816697.262, 9816714.648, 9816723.385,
9816740.441, 9816748.95, 9816765.452, 9816773.618, 9816789.339,
9816797.049, 9816811.749, 9816817.193, 9816824.103, 9816827.518,
9816834.092, 9816837.328, 9816843.537, 9816846.58, 9816852.396,
9816855.233, 9816860.628, 9816863.244, 9816868.191, 9816870.846,
9816876.383, 9816879.025, 9816883.93, 9816886.284, 9816890.674,
9816892.798, 9816896.817, 9816901.516, 9816915.859, 9816921.464,
9816929.009, 9816930.886, 9816639.468, 9816648.343, 9816683.875,
9816697.215, 9816714.624, 9816723.338, 9816740.418, 9816748.905,
9816765.43, 9816773.575, 9816789.318, 9816797.009, 9816811.734,
9816817.175, 9816824.094, 9816827.5, 9816834.083, 9816837.311,
9816843.528, 9816846.565, 9816852.389, 9816855.218, 9816860.621,
9816863.23, 9816868.184, 9816870.831, 9816876.376, 9816879.012,
9816883.924, 9816886.272, 9816890.669, 9816892.787, 9816896.805,
9816901.477, 9816915.839, 9816920.952, 9816926.637, 9816928.05,
9816659.354, 9816664.915, 9816687.177, 9816697.167, 9816714.6,
9816723.291, 9816740.395, 9816748.859, 9816765.407, 9816773.532,
9816789.298, 9816796.969, 9816811.719, 9816817.156, 9816824.085,
9816827.482, 9816834.074, 9816837.294, 9816843.52, 9816846.549,
9816852.381, 9816855.204, 9816860.614, 9816863.217, 9816868.176,
9816870.816, 9816876.369, 9816878.999, 9816883.918, 9816886.26,
9816890.663, 9816892.777, 9816896.792, 9816901.438, 9816915.819,
9816920.439, 9816924.264, 9816925.215, 9816679.241, 9816681.487,
9816690.48, 9816697.119, 9816714.577, 9816723.244, 9816740.372,
9816748.814, 9816765.385, 9816773.489, 9816789.277, 9816796.929,
9816811.705, 9816817.137, 9816824.076, 9816827.464, 9816834.065,
9816837.277, 9816843.512, 9816846.533, 9816852.373, 9816855.189,
9816860.607, 9816863.203, 9816868.169, 9816870.801, 9816876.362,
9816878.985, 9816883.911, 9816886.248, 9816890.657, 9816892.766,
9816896.78, 9816901.399, 9816915.799, 9816919.926, 9816921.892,
9816922.38, 9816541.016, 9816548.667, 9816579.391, 9816605.021,
9816675.293, 9816697.454, 9816714.742, 9816723.572, 9816740.533,
9816749.131, 9816765.54, 9816773.79, 9816789.422, 9816797.21,
9816811.808, 9816817.269, 9816824.14, 9816827.589, 9816834.126,
9816837.395, 9816843.569, 9816846.643, 9816852.427, 9816855.292,
9816860.656, 9816863.297, 9816868.219, 9816870.906, 9816876.411,
9816879.078, 9816883.956, 9816886.331, 9816890.697, 9816892.842,
9816896.867, 9816901.672, 9816915.938, 9816923.514, 9816938.5,
9816942.226, 9816560.486, 9816564.893, 9816582.587, 9816604.825,
9816675.232, 9816697.406, 9816714.719, 9816723.525, 9816740.51,
9816749.085, 9816765.518, 9816773.747, 9816789.401, 9816797.17,
9816811.793, 9816817.25, 9816824.131, 9816827.572, 9816834.118,
9816837.378, 9816843.561, 9816846.628, 9816852.419, 9816855.277,
9816860.649, 9816863.284, 9816868.212, 9816870.891, 9816876.404,
9816879.065, 9816883.949, 9816886.32, 9816890.692, 9816892.831,
9816896.854, 9816901.633, 9816915.918, 9816923.002, 9816936.127,
9816939.391, 9816579.957, 9816581.118, 9816585.783, 9816604.629,
9816675.171, 9816697.358, 9816714.695, 9816723.478, 9816740.487,
9816749.04, 9816765.496, 9816773.704, 9816789.381, 9816797.13,
9816811.778, 9816817.231, 9816824.122, 9816827.554, 9816834.109,
9816837.361, 9816843.553, 9816846.612, 9816852.411, 9816855.262,
9816860.642, 9816863.27, 9816868.205, 9816870.876, 9816876.397,
9816879.052, 9816883.943, 9816886.308, 9816890.686, 9816892.82,
9816896.842, 9816901.594, 9816915.898, 9816922.489, 9816933.755,
9816936.556, 9816521.879, 9816524.089, 9816532.945, 9816543.835,
9816578.476, 9816605.217, 9816675.354, 9816697.501,

Re: [R] 3-D interpretation

2013-12-10 Thread David Carlson

The error says you have duplicate points in dat so you need to
set duplicate= to tell interp() how to handle them.

?interp

-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352


-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Shane Carey
Sent: Tuesday, December 10, 2013 9:36 AM
To: r-help@r-project.org
Subject: [R] 3-D interpretation

Hi,

im trying to create a 3-D interpretation of a geological fault
using
the akima package. But my code fails below highlighted in red:
Does anyone have any ideas why this is.

Thanks

dat-read.delim(D:\\fault.txt,header=T,sep=,)
library(rgl)
# data
rgl.spheres(dat$X,dat$Z , dat$Y,1,color=red)
rgl.bbox()
# bivariate linear interpolation
# interp:
akima.li - interp(dat$X, dat$Y, dat$Z,
   xo=seq(min(dat$X), max(dat$X), length = 100),
   yo=seq(min(dat$Y), max(dat$Y), length = 100))

This is my error:
Error in interp.old(x, y, z, xo = xo, yo = yo, ncp = 0, extrap =
extrap,  :
duplicate data points: need to set 'duplicate = ..'

# interp surface:
rgl.surface(akima.li$X,akima.li$Y,akima.li$Z,color=green,alpha
=c(0.5))
# interpp:
akima.p - interpp(dat$X, dat$Y, dat$Z,
   runif(200,min(dat$X),max(dat$X)),
   runif(200,min(dat$Y),max(dat$Y)))
# interpp points:
rgl.points(akima.p$X,akima.p$Z ,
akima.p$Y,size=4,color=yellow)
# bivariate spline interpolation
# data
rgl.spheres(dat$X,dat$Z ,dat$Y,0.5,color=red)
rgl.bbox()
# bivariate cubic spline interpolation
# interp:
akima.si - interp(dat$X, dat$Y, dat$Z,
   xo=seq(min(dat$X), max(dat$X), length = 100),
   yo=seq(min(dat$Y), max(dat$Y), length = 100),
   linear = FALSE, extrap = TRUE)






-- 
Shane

[[alternative HTML version deleted]]

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible
code.

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] 3-D interpretation

Hey David,

I set it equal to the mean. duplicate=mean

but still no joy,
Thanks


On Tue, Dec 10, 2013 at 4:28 PM, David Carlson dcarl...@tamu.edu wrote:

 The error says you have duplicate points in dat so you need to
 set duplicate= to tell interp() how to handle them.

 ?interp

 -
 David L Carlson
 Department of Anthropology
 Texas AM University
 College Station, TX 77840-4352


 -Original Message-
 From: r-help-boun...@r-project.org
 [mailto:r-help-boun...@r-project.org] On Behalf Of Shane Carey
 Sent: Tuesday, December 10, 2013 9:36 AM
 To: r-help@r-project.org
 Subject: [R] 3-D interpretation

 Hi,

 im trying to create a 3-D interpretation of a geological fault
 using
 the akima package. But my code fails below highlighted in red:
 Does anyone have any ideas why this is.

 Thanks

 dat-read.delim(D:\\fault.txt,header=T,sep=,)
 library(rgl)
 # data
 rgl.spheres(dat$X,dat$Z , dat$Y,1,color=red)
 rgl.bbox()
 # bivariate linear interpolation
 # interp:
 akima.li - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100))

 This is my error:
 Error in interp.old(x, y, z, xo = xo, yo = yo, ncp = 0, extrap =
 extrap,  :
 duplicate data points: need to set 'duplicate = ..'

 # interp surface:
 rgl.surface(akima.li$X,akima.li$Y,akima.li$Z,color=green,alpha
 =c(0.5))
 # interpp:
 akima.p - interpp(dat$X, dat$Y, dat$Z,
runif(200,min(dat$X),max(dat$X)),
runif(200,min(dat$Y),max(dat$Y)))
 # interpp points:
 rgl.points(akima.p$X,akima.p$Z ,
 akima.p$Y,size=4,color=yellow)
 # bivariate spline interpolation
 # data
 rgl.spheres(dat$X,dat$Z ,dat$Y,0.5,color=red)
 rgl.bbox()
 # bivariate cubic spline interpolation
 # interp:
 akima.si - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100),
linear = FALSE, extrap = TRUE)






 --
 Shane

 [[alternative HTML version deleted]]

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible
 code.




-- 
Shane

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Re: [R] 3-D interpretation

it just keeps crashing on me,

It seems to crash on this line

rgl.surface(akima.li$x,akima.li$y,akima.li$z,color=green,alpha=c(0.5))


On Tue, Dec 10, 2013 at 4:32 PM, David Carlson dcarl...@tamu.edu wrote:

 I guess you will have to give us a reproducible example.
 Are you still getting the same error message?

 David

 From: Shane Carey [mailto:careys...@gmail.com]
 Sent: Tuesday, December 10, 2013 10:30 AM
 To: dcarl...@tamu.edu
 Cc: r-help@r-project.org
 Subject: Re: [R] 3-D interpretation

 Hey David,

 I set it equal to the mean. duplicate=mean

 but still no joy,
 Thanks

 On Tue, Dec 10, 2013 at 4:28 PM, David Carlson
 dcarl...@tamu.edu wrote:
 The error says you have duplicate points in dat so you need to
 set duplicate= to tell interp() how to handle them.

 ?interp

 -
 David L Carlson
 Department of Anthropology
 Texas AM University
 College Station, TX 77840-4352


 -Original Message-
 From: r-help-boun...@r-project.org
 [mailto:r-help-boun...@r-project.org] On Behalf Of Shane Carey
 Sent: Tuesday, December 10, 2013 9:36 AM
 To: r-help@r-project.org
 Subject: [R] 3-D interpretation

 Hi,

 im trying to create a 3-D interpretation of a geological fault
 using
 the akima package. But my code fails below highlighted in red:
 Does anyone have any ideas why this is.

 Thanks

 dat-read.delim(D:\\fault.txt,header=T,sep=,)
 library(rgl)
 # data
 rgl.spheres(dat$X,dat$Z , dat$Y,1,color=red)
 rgl.bbox()
 # bivariate linear interpolation
 # interp:
 akima.li - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100))

 This is my error:
 Error in interp.old(x, y, z, xo = xo, yo = yo, ncp = 0, extrap =
 extrap,  :
 duplicate data points: need to set 'duplicate = ..'

 # interp surface:
 rgl.surface(akima.li$X,akima.li$Y,akima.li$Z,color=green,alpha
 =c(0.5))
 # interpp:
 akima.p - interpp(dat$X, dat$Y, dat$Z,
runif(200,min(dat$X),max(dat$X)),
runif(200,min(dat$Y),max(dat$Y)))
 # interpp points:
 rgl.points(akima.p$X,akima.p$Z ,
 akima.p$Y,size=4,color=yellow)
 # bivariate spline interpolation
 # data
 rgl.spheres(dat$X,dat$Z ,dat$Y,0.5,color=red)
 rgl.bbox()
 # bivariate cubic spline interpolation
 # interp:
 akima.si - interp(dat$X, dat$Y, dat$Z,
xo=seq(min(dat$X), max(dat$X), length = 100),
yo=seq(min(dat$Y), max(dat$Y), length = 100),
linear = FALSE, extrap = TRUE)






 --
 Shane
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 and provide commented, minimal, self-contained, reproducible
 code.




 --
 Shane




-- 
Shane

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Re: [R] 3-D interpretation

2013-12-10 Thread David Carlson

I guess you will have to give us a reproducible example. 
Are you still getting the same error message?

David

From: Shane Carey [mailto:careys...@gmail.com] 
Sent: Tuesday, December 10, 2013 10:30 AM
To: dcarl...@tamu.edu
Cc: r-help@r-project.org
Subject: Re: [R] 3-D interpretation

Hey David,

I set it equal to the mean. duplicate=mean

but still no joy,
Thanks

On Tue, Dec 10, 2013 at 4:28 PM, David Carlson
dcarl...@tamu.edu wrote:
The error says you have duplicate points in dat so you need to
set duplicate= to tell interp() how to handle them.

?interp

-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352

-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Shane Carey
Sent: Tuesday, December 10, 2013 9:36 AM
To: r-help@r-project.org
Subject: [R] 3-D interpretation

Hi,

im trying to create a 3-D interpretation of a geological fault
using
the akima package. But my code fails below highlighted in red:
Does anyone have any ideas why this is.

Thanks

dat-read.delim(D:\\fault.txt,header=T,sep=,)
library(rgl)
# data
rgl.spheres(dat$X,dat$Z , dat$Y,1,color=red)
rgl.bbox()
# bivariate linear interpolation
# interp:
akima.li - interp(dat$X, dat$Y, dat$Z,
                   xo=seq(min(dat$X), max(dat$X), length = 100),
                   yo=seq(min(dat$Y), max(dat$Y), length = 100))

This is my error:
Error in interp.old(x, y, z, xo = xo, yo = yo, ncp = 0, extrap =
extrap,  :
duplicate data points: need to set 'duplicate = ..'

# interp surface:
rgl.surface(akima.li$X,akima.li$Y,akima.li$Z,color=green,alpha
=c(0.5))
# interpp:
akima.p - interpp(dat$X, dat$Y, dat$Z,
                   runif(200,min(dat$X),max(dat$X)),
                   runif(200,min(dat$Y),max(dat$Y)))
# interpp points:
rgl.points(akima.p$X,akima.p$Z ,
akima.p$Y,size=4,color=yellow)
# bivariate spline interpolation
# data
rgl.spheres(dat$X,dat$Z ,dat$Y,0.5,color=red)
rgl.bbox()
# bivariate cubic spline interpolation
# interp:
akima.si - interp(dat$X, dat$Y, dat$Z,
                   xo=seq(min(dat$X), max(dat$X), length = 100),
                   yo=seq(min(dat$Y), max(dat$Y), length = 100),
                   linear = FALSE, extrap = TRUE)

--
Shane
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__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible
code.

-- 
Shane 

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] rbind/timestamp problem

2013-12-10 Thread Georg Hörmann


Thank you,

now it works. I would never have never found the source of the problem. 
I am trying

to build a data frame with results from a sensitivity analysis
for a model. I use rbind because the number of rows is
unkown and i use the timestamp as a hint for the students
that they do not analyse data from last week :-).
Anyway, thank you very much, it helped a lot.

Greetings,
Georg


On 10.12.2013 15:40, jim holtman wrote:

try this:


sensitivity=rbind(sensitivity,data.frame(mtype=2,cdate=as.POSIXct(Sys.time(), 
origin=1970-01-01)))
sensitivity

   mtype   cdate
1 1 2013-12-10 09:35:53
2 2 2013-12-10 09:37:02

The 'c' function is coercing to numeric.  If you want to 'rbind' rows
to a dataframe, then make sure you use a dataframe.

Jim Holtman
Data Munger Guru



--
Georg Hoermann,
Department of Hydrology and Water Resources Management
Kiel University, Germany
+49/431/2190916, mo: +49/176/64335754, icq:348340729, skype: ghoermann

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Re: [R] growth curve estimation

2013-12-10 Thread Dániel Kehl

Dear Vito, Robert and David,

thank you for your replies, although I made some step forward on my own, your 
answers helped a lot and gave me more insight.
The only question I have left is why this code gives me an error (and that is 
what David asked for):

m1 - lms(BMI,age,data=adatok_fiu, cent=c(3,10,25,50,75,90,97), families=BCCG)
centiles.pred(m1, xname=age, xvalues=seq(10,20,.5), 
cent=c(3,10,25,50,75,90,97))
Error in X[onlydata, , drop = FALSE] : 
  (subscript) logical subscript too long

but

m3 - gamlss(BMI~pb(age), sigma.formula=~pb(age), nu.formula = ~pb(age), 
tau.formula = ~pb(age), family=BCT, data=adatok_fiu)
centiles.pred(m3, xname=age, xvalues=seq(10,20,.5), 
cent=c(3,10,25,50,75,90,97))

works as expected (although I get a warning message
Warning message:
In predict.gamlss(obj, what = tau, newdata = newx, type = response,  :
  There is a discrepancy  between the original and the re-fit 
 used to achieve 'safe' predictions 


I had the feeling centiles.pred should work for the result of an lms() function 
just like in case of gamlss().

Thank you all again!

daniel


Feladó: David Winsemius [dwinsem...@comcast.net]
Küldve: 2013. december 9. 21:40
To: Dániel Kehl
Cc: r-help@r-project.org
Tárgy: Re: [R] growth curve estimation

On Dec 8, 2013, at 7:45 AM, Dániel Kehl wrote:

 Dear Community,

 I am struggling with a growth curve estimation problem. It is a classic BMI 
 change with age calculation. I am not an expert of this field but have some 
 statistical experience in other fields.
 Of course I started reading classical papers related to the topic and 
 understood the concept of the LMS method. After that I thought it will be a 
 piece of cake, R must have a package related to the topic, so I just need 
 the data and I am just a few lines of code from the results.


You might want to look at a more recent discussion:

http://www.who.int/entity/childgrowth/standards/velocity/tr3chap_2.pdf

(The WHO centre has published their R code.)


 I encountered some problems:
 - found at least three packages related to LMS: gamlss, VGAM and an old one 
 lmsqreg (I did not try this because it does not support my R version (3.0.1.))
 - it was relatively easy to get plots of percentiles in both packages, 
 although they were far not the same (I also tried an other software, 
 LMSchartmaker it gave different results from the previous ones)
 - I tried to get tables of predicted values (with the predict function in 
 VGAM and with centiles.pre in gamlss) but without any success.

Don't see any code or error messages.

 - I tried to use the function gamlss() instead of lms() in gamlss but I could 
 not force them to give the same (or very similar results), but the 
 centiles.pred() function did work as expected for the model resulted from 
 galmss()
 - lms gives really different results if k is specified different ways, which 
 is best?

Won't that depend on the amount and distribution of the data?


 Also I have a general question: some publications state they estimated the 
 centiles so that aroun 18 years of age the curves pass through certain 
 points. How is that possible?

 Thank you for any suggestions or insights about the methods or preferred 
 package!

 Here is my code (without data):

 #gamlss
 library(gamlss)
 library(VGAM)
 library(foreign)
 adatok - read.spss(MDSZ adatok.sav, to.data.frame=TRUE)

 adatok_fiu - subset(adatok, adatok$gender == Fiúk)[,2:3]
 row.names(adatok_fiu) - NULL
 adatok_lany - subset(adatok, adatok$gender == L÷nyok)[,2:3]
 row.names(adatok_lany) - NULL

 m1 - lms(BMI,age,data=adatok_fiu, cent=c(3,10,25,50,75,90,97), 
 families=BCCG)
 fittedPlot(m1, x=adatok_fiu$age)
 m1 - lms(BMI,age,data=adatok_fiu, cent=c(3,10,25,50,75,90,97), 
 families=BCCG, method.pb=GAIC, k=log(1455))
 fittedPlot(m1, x=adatok_fiu$age)
 m1 - lms(BMI,age,data=adatok_fiu, cent=c(3,10,25,50,75,90,97), 
 families=BCCG, method.pb=GAIC)
 fittedPlot(m1, x=adatok_fiu$age)

 m2 - lms(BMI,age,data=adatok_lany, cent=c(3,10,25,50,75,90,97), 
 families=BCCG)
 m2 - lms(BMI,age,data=adatok_lany, cent=c(3,10,25,50,75,90,97), 
 families=BCCG, method.pb=GAIC, k=log(1144))
 m2 - lms(BMI,age,data=adatok_lany, cent=c(3,10,25,50,75,90,97), 
 families=BCCG, method.pb=GAIC)

 m3 - gamlss(BMI~age, family=BCT, data=adatok_fiu)
 centiles(m3,xvar=adatok_fiu$age, cent=c(3,10,25,50,75,90,97))

 newx - seq(12,20,.5)

 centiles.pred(m1, xname=age, xvalues=newx)
 centiles.pred(m3, xname=age, xvalues=newx)
 centiles(m1,adatok_fiu$age)

 #VGAM
 library(foreign)
 library(VGAM)
 library(VGAMdata)

 adatok - read.spss(MDSZ adatok.sav, to.data.frame=TRUE)

 adatok_fiu - subset(adatok, adatok$gender == Fiúk)
 adatok_lany - subset(adatok, adatok$gender == L÷nyok)

 fit1 - vgam(BMI ~ s(age), lms.bcn(percentiles = c(3, 10, 25, 50, 75, 90, 
 97)), adatok_fiu)
 fit2 - vgam(BMI ~ s(age), lms.bcn(percentiles = c(3, 10, 25, 50, 75, 90, 
 97)), adatok_lany)

 qtplot(fit1, percentiles = c(3, 10, 25,

[R] If-statement in for-loop

2013-12-10 Thread gncl dzgn

Hi everyone,

you might find my question elementary but I am a beginner and unfortunately I 
can't fix the problem. 

So, I simulate this following algorithm and some values of c are NA. Therefore, 
I should add these following two if-statements but I don't know how I should do 
it in a for-loop. 

Thank in advance if someone helps me!

The conditions: 

If there is no input greater or equal to d, then ALG = b*T
If  max(s +  b*(0:T))  b*T , then OPT = b*T


The algorithm:

a - 0.0008
b - 0.0001 
T - 100 
t - 0:T 
alpha - 1 

d- sqrt(a * b) * T - b * t

N - 100
c - rep(0, N)
for (j in 1:N) {

e - rnorm(T, mean = 0, sd = 0.001) 
s- c( runif(1,0, a*T), rep(0, T-1) ) 
minx - 0 
for(i in 2:T) { 
x - alpha * s[i-1] + e[i] 
maxx - a*(T-i) 
if(x  minx) { 
 s[i] - minx 
} else { 
if(x  maxx) { 
 s[i] - maxx 
} else s[i] - x 
} 
} 

p- which(s = d)[1]
ALG- s[p] + b*(p-1)
OPT - max(s +  b*(0:T))

c[j] -  OPT / ALG

}

head(c)
mean(c)
plot(c)



  
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Re: [R] adding tables

2013-12-10 Thread Eik Vettorazzi

How about this:
t2 - table(c(10,11,12,13))
t1 -table(c(1,1,2,4,5))
t12-(rbind(as.data.frame(t1),as.data.frame(t2)))
xtabs(Freq~.,t12)

it works for overlapping tables

t2 - table(c(10,11,12,13,1,2))
t1 -table(c(1,1,2,4,5,11,12))
t12-(rbind(as.data.frame(t1),as.data.frame(t2)))
xtabs(Freq~.,t12)

and it will work for two-dimensional tables as well:

t2 - table(c(10,11,12,13),letters[rep(1:2,each=2)])
t1 -table(c(1,1,2,4,5),letters[rep(c(1,3),length.out=5)])
t12a-(rbind(as.data.frame(t1),as.data.frame(t2)))
xtabs(Freq~.,t12a)

cheers.

Am 10.12.2013 00:59, schrieb Ross Boylan:
 Can anyone recommend a good way to add tables?
 Ideally I would like
 t1 - table(x1)
 t2 - table(x2)
 t1+t2
 
 It t1 and t2 have the same levels this works fine, but I need something
 that will work even if they differ, e.g.,
   t1
 
  1 2 4 5
  2 1 1 1
   t2 - table(c(10, 11, 12, 13))
   t1+t2  # apparently does simple vector addition
 
  1 2 4 5
  3 2 2 2
 whereas I want
 1 2 4 5 10 11 12 13
 2 1 1 1  1  1  1  1
 
 __
 R-help@r-project.org mailing list
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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.
 

-- 
Eik Vettorazzi

Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790
--

Besuchen Sie uns auf: www.uke.de
_

Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; 
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden), 
Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik
_

SAVE PAPER - THINK BEFORE PRINTING

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Re: [R] Problem to predict new data with mclust

2013-12-10 Thread Welma Pereira

Btw I am using mclust, and basically wanted to reproduce the mstep and
estep of the Mclust algorithm for new data.

Any help is appreaciated!


On 10 December 2013 21:23, Welma Pereira welma.pere...@gmail.com wrote:

 Hi,

 Thanks for the hint Sarah. So here is my problem again: (files train_pca
 is in train_pca and test is in test_pca attached)

 mclust2_pca - Mclust(train_pca,G=2, modelNames= c(EII, VII, EEI,
 EVI, VEI, VVI))
 pred-predict.Mclust(mclust2_pca,test)
 Warning message:
 In cdensVVI(data = c(2.80217508052409, 2.75071740560707, 2.6175058179515,
 :
   cannot compute E-step

 Thanks!


 On 10 December 2013 16:30, Sarah Goslee sarah.gos...@gmail.com wrote:

 Hi,

 It's unlikely that anyone will be able to see your problem without a
 reproducible example.


 http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

 Are you using the mclust package? That would be the first thing we need
 to know.

 Then, have you worked through the examples in ?predict.Mclust, and do
 you understand them?

 Then, what do your data look like? str() and such are useful, but the
 best option is to use dput() to create a reproducible subset, or to
 reproduce your problem using one of the built-in datasets in R, as is
 done in the above documentation examples.

 Sarah

 On Tue, Dec 10, 2013 at 6:41 AM, Welma Pereira welma.pere...@gmail.com
 wrote:
  Hi,
 
  I am trying to use mclust to cluster some data (train_pca_10), I get the
  clusters, but when I try to use the model to predict new data (test1) I
 get
  this error
 
 
 
  mClust2 - Mclust(train_pca_10,G=2)
  pred-predict.Mclust(mClust2,test1)
 
  Error in if (warn) warning(WARNING) : argument is of length zero
 
  Can anyone see the problem here?
 
  Thanks,
  Pereira.
 
 --
 Sarah Goslee
 http://www.functionaldiversity.org




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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Problem to predict new data with mclust

Many people do not want to download and open random attachments.
That's why I suggested dput() with part of your data.

dput(head(yourdata, 20))

is enough for many problems, though it's good to try it out.

Or use one of the datasets built into R.

Sarah


On Tue, Dec 10, 2013 at 3:23 PM, Welma Pereira welma.pere...@gmail.com wrote:
 Hi,

 Thanks for the hint Sarah. So here is my problem again: (files train_pca is
 in train_pca and test is in test_pca attached)

 mclust2_pca - Mclust(train_pca,G=2, modelNames= c(EII, VII, EEI,
 EVI, VEI, VVI))
 pred-predict.Mclust(mclust2_pca,test)
 Warning message:
 In cdensVVI(data = c(2.80217508052409, 2.75071740560707, 2.6175058179515,  :
   cannot compute E-step

 Thanks!


 On 10 December 2013 16:30, Sarah Goslee sarah.gos...@gmail.com wrote:

 Hi,

 It's unlikely that anyone will be able to see your problem without a
 reproducible example.


 http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

 Are you using the mclust package? That would be the first thing we need to
 know.

 Then, have you worked through the examples in ?predict.Mclust, and do
 you understand them?

 Then, what do your data look like? str() and such are useful, but the
 best option is to use dput() to create a reproducible subset, or to
 reproduce your problem using one of the built-in datasets in R, as is
 done in the above documentation examples.

 Sarah

 On Tue, Dec 10, 2013 at 6:41 AM, Welma Pereira welma.pere...@gmail.com
 wrote:
  Hi,
 
  I am trying to use mclust to cluster some data (train_pca_10), I get the
  clusters, but when I try to use the model to predict new data (test1) I
  get
  this error
 
 
 
  mClust2 - Mclust(train_pca_10,G=2)
  pred-predict.Mclust(mClust2,test1)
 
  Error in if (warn) warning(WARNING) : argument is of length zero
 
  Can anyone see the problem here?
 
  Thanks,
  Pereira.
 

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[R] Get average model after dredge function ran in a loop

2013-12-10 Thread Catarina Ferreira

Dear all

I'm a beginner in R and I'm trying to get the final model after I run the
dredge function for 10 times (with a loop):

###Building the Model
Coy.glm0-glm(pa ~  shrub + snowdep + tree + bio5 + bio6 + bio12 +
log(human+1), data=Coy.pa, family=binomial)
summary(Coy.glm0)

install.packages('MuMIn')
library(MuMIn)
Coy.dredge-dredge(Coy.glm0)
head(Coy.dredge) ##Look in which colum is AIC
###Building a simulation
Coy.models-Coy.dredge[1,c(1:13)]
Coy.models


###Turn a loop who will create 10 models
run=1
while(run11) #11 means 10 models.
{
  Coy.abs$Long-runif(300,498,2579440)
  Coy.abs$Lat-runif(300,-51483,1377669)
  Coy.pa-rbind(Coy.train, Coy.abs) train ou prSS
  Coy.ppp-ppp(Coy.pa$Long,Coy.pa$Lat, window=win, unitname=meters)
  Coy.pa$snowdep-snowdepz.im[Coy.ppp, drop=F]
  Coy.pa$tree-treez.im[Coy.ppp, drop=F]
  Coy.pa$bio5-bio5z.im[Coy.ppp, drop=F]
  Coy.pa$bio6-bio6z.im[Coy.ppp, drop=F]
  Coy.pa$bio12-bio12z.im[Coy.ppp, drop=F]
  Coy.pa$human-humanz.im[Coy.ppp, drop=F]
  Coy.pa$shrub-shrub.im[Coy.ppp, drop=F]

  Coy.glm0-glm(pa ~ shrub + snowdep + tree + bio5 +  bio6 + bio12+
log(human+1), data=Coy.pa, family=binomial)
  Coy.dredge-dredge(Coy.glm0)
  Coy.models-rbind(Coy.models, Coy.dredge[1,c(1:13)])
  run=run+1
}

I do get a best model for each run which I then hand pick and add to a
table. The problem is that I have 11 models now in this table and I want
their average to get the final model. I don't know how to do it from the
table (as the model.avg() will tell me I only have one model in the table,
because it's not recognizing the different rows as different models), but
on the other hand there must be a way to do it directly in the loop, only
I'm not sure at what point of the script I should be asking for it and how
the code should be written.

I would very much appreciate any help you can give me.

Thank you.

Cat

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Re: [R] Need help figuring out sapply (and similar functions) with multiple parameter user defined function

2013-12-10 Thread Walter Anderson

Petr,

Thank you for you assistance.  Particularly this last bit.  It really 
helped me understand exactly how your solution worked and why I was 
confused by rapply processing rows.  It was apply that could process 
rows (and/or columns) of a matrix,


Anyway for anyone who wants the version of my script that incorporates 
all of the suggested changes, here it is.

survey.results - read.csv(SamplePairedComparisonData.csv)

# The following function evaluate which, if any, preference the
respondent has shown
get.pref - function(x)
{
   tb - table(x)
   if ((length(tb) == 3)  max((tb) == min(tb)))
 retVal - None  # This opton checks for a couple of special
cases where the simple which.max function fails to
# detect that each of the three items has a single response.
   else
 # For all other conditions the following detects which response
received the most responses (2+) and records a NONE
 # in any case that doesn't achieve that level of response.
 retVal - names(which.max(table(x)))
   return(retVal)
}

# The data is expected to be organized as so:
#  ID, Q1, Q2, Q3
# and that the questions are in the following format
#  Q.1) Which do you prefer?
#  1) Option 1
#  2) Option 2
#  Q.2) Which do you prefer?
#  1) Option 1
#  2) Option 3
#  Q.3) Which do you prefer?
#  1) Option 2
#  2) Option 3
# And the next three lines reprocess the data table to organize the
questions so that the values for Q.2) are 1 or 3 and the values for
Q.3) are 2 or 3
survey.results[survey.results[,3]==2,3]-3 # Convert column 3 (Q2)
to a value of 3 if the existing value is 2
# The order of the following two commands is important, don't change
survey.results[survey.results[,4]==2,4]-3 # Convert column 4 (Q3)
to a value of 3 if the existing value is 1
survey.results[survey.results[,4]==1,4]-2 # Convert column 4 (Q4)
to a value of 2 if the existing value is 1

# The following lines convert the numerical values in the fields to
text (factors)
# If the questions don't have empty (zero) values than the None
should be removed from those questions where zero isn't possible
# The labels 'Option 1' and 'Option 2' can be replaced with more
appropriate text as needed.
survey.results$Q1 - factor(survey.results$Q1,
labels=c(None,Option 1,Option 2))
survey.results$Q2 - factor(survey.results$Q2,
labels=c(None,Option 1,Option 3))
survey.results$Q3 - factor(survey.results$Q3,
labels=c(None,Option 2,Option 3))

# Take the survey data table and get rid of the first column (ID)
and convert the remaining three columns
# to a matrix. Then apply the get.pref function to each of the rows
in the matrix
survey.results$Preference -
apply(as.matrix(survey.results[,-1]),1,get.pref)

On 12/10/2013 05:02 AM, PIKAL Petr wrote:

 In that case you need correct those cases inside the function. 
 Everything else stays same.

 Something like

 fff - function(x) {

 tb-table(x)

 if(max(tb)==min(tb)) res - None else res - names(which.max(table(x)))

 }

 Regards

 Petr




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[R] Add column to DF based on 2 columns in another DF

2013-12-10 Thread bcrombie

LookupTable - read.table(header = TRUE, 
   stringsAsFactors = FALSE, 
   text=Str IND13 IND12 IND11 IND07 IND06
 1   517   529   562   562   567
 2   517   529   562   562   568
 3   517   529   562   562   567
 4   517   529   562   562   569
 5   517   529   562   562   567
 6   517   529   562   562   567
 7   517   529   562   562   560
 8   517   529   562   562   567
 9   517   529   562   562   567
 10   517   529   562   562   567)


MainDataFrame - read.table(header = TRUE, 
  stringsAsFactors = FALSE, 
  text=Pid YEAR MONTH  Fips Str index
600250 2006 7  6037  1 IND06
600250 2006 7  6037  2 IND06
600250 2006 7  6037  3 IND06
600250 2006 7  6037  4 IND06
600250 2006 7  6037  5 IND06
600353 2007 9 48097  6 IND07
600772 2006 2  6039  7 IND06
600947 2007 1 13207  7 IND07
601055 2007 9 13315  8 IND07
601103 2006 5 21093  10 IND06)

MainDataFrame_New - ??

#What is the best way to add a new column index_num to MainDataFrame that
is populated with the
#number corresponding to 'Str' and 'index' in LookupTable:

# Pid YEAR MONTH  Fips Str index index_num
# 600250 2006 7  6037  1 IND06 567
# 600250 2006 7  6037  2 IND06 568
# 600250 2006 7  6037  3 IND06 567
# 600250 2006 7  6037  4 IND06 569
# 600250 2006 7  6037  5 IND06 567
# 600353 2007 9 48097  6 IND07 562
# 600772 2006 2  6039  7 IND06 560
# 600947 2007 1 13207  7 IND07 562
# 601055 2007 9 13315  8 IND07 562
# 601103 2006 5 21093  10 IND06 567



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Re: [R] Fwd: Datatable manipulation

2013-12-10 Thread Nitisha jha

Hi,
I have a doubt after a long time :) . I have a function which has sqldf
statements and it works fine when I call it from console but when I am
calling that function within another function, it gives me the error that
*Error in sqliteExecStatement(con, statement, bind.data) : *
*  RS-DBI driver: (error in statement: no such table: table_name)*

What is wrong here?


On Fri, Nov 22, 2013 at 7:27 PM, arun smartpink...@yahoo.com wrote:



 Hi,
 Assuming that this is the case:
 dat1 - read.table(text=a b c d   e
 1 2 3 4 5
 10 9 8 7 6,sep=,header=TRUE)

 Names1- read.table(text=Original  New
 e ee
 ggg
 a aa
 c cc
 f ff,sep=,header=TRUE,stringsAsFactors=FALSE)

  indx - match(names(dat1),Names1[,1])
  names(dat1)[names(dat1) %in% Names1[,1]] - Names1[,2][indx[!is.na
 (indx)]]
  dat1
 #  aa b cc d ee
 #1  1 2  3 4  5
 #2 10 9  8 7  6


 A.K.

 On Friday, November 22, 2013 4:46 AM, Nitisha jha nitisha...@gmail.com
 wrote:

 Hey! I got this one. :)
 For the match function, actually I just want the ones that are matching to
 be replaced. Rest should stay the same. How do I do that? When I tried your
 command, if there is no match, it writes var2 or something.



 On Fri, Nov 22, 2013 at 12:38 AM, arun smartpink...@yahoo.com wrote:
 
 
 
 Hi,
 Try:
 
 dat1 - read.table(text=a b c d   e
 
 1 2 3 4 5
 10 9 8 7 6,sep=,header=TRUE)
 
 Names1- read.table(text=Original  New
 
 e ee
 b bb
 a aa
 c cc
 d dd,sep=,header=TRUE,stringsAsFactors=FALSE)
 
 It is better to dput() your dataset.  For example:
  dput(Names1)
 structure(list(Original = c(e, b, a, c, d), New = c(ee,
 bb, aa, cc, dd)), .Names = c(Original, New), class =
 data.frame, row.names = c(NA,
 -5L))
 
 
  names(dat1) - Names1[,2][match(names(dat1), Names1[,1])] ##
  dat1
 #  aa bb cc dd ee
 #1  1  2  3  4  5
 #2 10  9  8  7  6
 A.K.
 
 
 
 
 
 
 On Thursday, November 21, 2013 1:45 PM, Nitisha jha 
 nitisha...@gmail.com wrote:
 
 Hi,
 
 Thanks. I used as.character() and got the right strings.
 
 Btw, I have lots of handicaps regarding R.
 
 I have to rename the columns(I have 22 columns here). I have the new
 names along with the original names in another dataset. Right now, I am
 going hardcoding all the  19 name changes(tedious and not optimum). 1st 3
 names remain the same. I will give u a sample dataset. Let me know if there
 is any easy way of doing this.  Pardon the displaced column labels.
 
 
 
 Original dataset.
 
 
 
a b c d
 e
 1 2 3 4 5
 10 9 8 7 6
 
 
 
 
 
 Dataset for name change
 
 
 Original  New
 
 
 
 e ee
 
 
 
 b bb
 
 
 
 a aa
 
 
 
 c cc
 
 
 
 d dd
 
 
 
 
 
 
 
 
 I want my final dataset to be like this:
 
 aa bb cc dd ee
 1 2 3 4 5
 10 9 8 7 6
 
 
 
  Could u tell me an optimal way to do it. My method is tedious and not
 good.
 
 Also, is there a way to import .xls without perl (windows)?
 
 
 Thanks for being patient. :)
 
 
 
 


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[R] data distribution for lme

2013-12-10 Thread peyman

Hi folks,

I am using the lme package of R, and am wondering if it is assumed that
the dependent factor (what we fit for; y in many relevant texts) has to
have a normal Gaussian distribution? Is there any margins where some
skewness in the data is accepted and how within R itself one could check
distribution of the data?

Thanks,
Peyman

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Re: [R] Fwd: Datatable manipulation

2013-12-10 Thread arun

Hi,

Check these links:
http://stackoverflow.com/questions/19626534/r-issue-error-in-sqliteexecstatementcon-statement-bind-data-no-such-tabl


http://r.789695.n4.nabble.com/Problem-with-SQLDF-Error-in-sqliteExecStatement-con-statement-bind-data-RS-DBI-driver-error-in-state-td4621931.html


A.K.






On Tuesday, December 10, 2013 6:54 AM, Nitisha jha nitisha...@gmail.com wrote:

Hi,
I have a doubt after a long time :) . I have a function which has sqldf 
statements and it works fine when I call it from console but when I am calling 
that function within another function, it gives me the error that 
Error in sqliteExecStatement(con, statement, bind.data) : 
  RS-DBI driver: (error in statement: no such table: table_name)

What is wrong here?



On Fri, Nov 22, 2013 at 7:27 PM, arun smartpink...@yahoo.com wrote:



Hi,
Assuming that this is the case:

dat1 - read.table(text=a b c d   e
1 2 3 4 5
10 9 8 7 6,sep=,header=TRUE)

Names1- read.table(text=Original  New  
e ee
g    gg
a aa
c cc
f ff,sep=,header=TRUE,stringsAsFactors=FALSE)
 
 indx - match(names(dat1),Names1[,1])
 names(dat1)[names(dat1) %in% Names1[,1]] - Names1[,2][indx[!is.na(indx)]]
 dat1
#  aa b cc d ee

#1  1 2  3 4  5
#2 10 9  8 7  6


A.K.


On Friday, November 22, 2013 4:46 AM, Nitisha jha nitisha...@gmail.com wrote:

Hey! I got this one. :)
For the match function, actually I just want the ones that are matching to be 
replaced. Rest should stay the same. How do I do that? When I tried your 
command, if there is no match, it writes var2 or something. 




On Fri, Nov 22, 2013 at 12:38 AM, arun smartpink...@yahoo.com wrote:



Hi,
Try:

dat1 - read.table(text=a     b     c     d   e

1     2     3     4     5
10     9     8     7     6,sep=,header=TRUE)

Names1- read.table(text=Original      New   

e     ee
b     bb   
a     aa
c     cc
d     dd,sep=,header=TRUE,stringsAsFactors=FALSE)

It is better to dput() your dataset.  For example:
 dput(Names1)
structure(list(Original = c(e, b, a, c, d), New = c(ee,
bb, aa, cc, dd)), .Names = c(Original, New), class = 
data.frame, row.names = c(NA,
-5L))


 names(dat1) - Names1[,2][match(names(dat1), Names1[,1])] ##
 dat1
#  aa bb cc dd ee
#1  1  2  3  4  5
#2 10  9  8  7  6
A.K.






On Thursday, November 21, 2013 1:45 PM, Nitisha jha nitisha...@gmail.com 
wrote:

Hi,

Thanks. I used as.character() and got the right strings.

Btw, I have lots of handicaps regarding R.

I have to rename the columns(I have 22 columns here). I have the new names 
along with the original names in another dataset. Right now, I am going 
hardcoding all the  19 name changes(tedious and not optimum). 1st 3 names 
remain the same. I will give u a sample dataset. Let me know if there is 
any easy way of doing this.  Pardon the displaced column labels.



Original dataset.



   a b c d     
e
1 2 3 4 5
10 9 8 7 6





Dataset for name change


Original  New



e ee



b bb



a aa



c cc



d dd








I want my final dataset to be like this:

aa bb cc dd ee
1 2 3 4 5
10 9 8 7 6



 Could u tell me an optimal way to do it. My method is tedious and not 
good.

Also, is there a way to import .xls without perl (windows)?


Thanks for being patient. :)






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Re: [R] Add column to DF based on 2 columns in another DF

2013-12-10 Thread bcrombie

Thanks, AK. That gets the job done and I learned 2 ways to do it. Much
appreciated as always.



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Re: [R] data distribution for lme

2013-12-10 Thread Bert Gunter

This is not really an R question -- it is statistics.
In any case, you should do better posting this on the
R-Sig-Mixed-Models list, which concerns itself with matters like this.

However, I'll hazard a guess at an answer: maybe.  (Vague questions
elicit vague answers).

Cheers,
Bert

On Tue, Dec 10, 2013 at 6:55 AM, peyman zira...@gmail.com wrote:
 Hi folks,

 I am using the lme package of R, and am wondering if it is assumed that
 the dependent factor (what we fit for; y in many relevant texts) has to
 have a normal Gaussian distribution? Is there any margins where some
 skewness in the data is accepted and how within R itself one could check
 distribution of the data?

 Thanks,
 Peyman

 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

(650) 467-7374

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[R] radial.plot shaded region

2013-12-10 Thread Hodge, Steven

I'm working with radial.plot in the plotrix package.

Is there a way to show a shaded region for a specific range of values in 
radial.lim?

For example:
# test data from radial.plot{plotrix}
testlen-c(sin(seq(0,1.98*pi,length=100))+2+rnorm(100)/10)
testpos-seq(0,1.98*pi,length=100)

# the original plot
radial.plot(testlen,testpos,rp.type=p,main=Test Polygon,line.col=blue)

# my attempt to shade the region between 3 and 3.5:
radial.plot(
matrix(c(3.5, 3), byrow = TRUE),
matrix(c(testpos, testpos), byrow = TRUE, nrow = 2),
rp.type=p,
main=Test Polygon,
poly.col = c('light blue', 'white'),
radial.lim = c(0, 3.5)
)
# In effect, draw a polygon at 3.5 filled with light blue, then another at 3 
filled with white.
# Now overplot the values of interest:
radial.plot(testlen,testpos,rp.type=p,main=Test Polygon,line.col=blue, 
radial.lim = c(0, 3.5), add = TRUE)

Is there an easier way? How can I re-draw the grid lines that are in the 
original plot?

Thanks,

Steve

. . . . . . . . . . . . . . .
Steven M. Hodge
Child and Adolescent Neurodevelopment Initiative S3-301
University of Massachusetts Medical School
55 Lake Avenue North
Worcester, MA 01655
steven.ho...@umassmed.edu

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Re: [R] radial.plot shaded region

2013-12-10 Thread Jim Lemon


On 12/11/2013 10:33 AM, Hodge, Steven wrote:

I'm working with radial.plot in the plotrix package.

Is there a way to show a shaded region for a specific range of values in 
radial.lim?

For example:
# test data from radial.plot{plotrix}
testlen-c(sin(seq(0,1.98*pi,length=100))+2+rnorm(100)/10)
testpos-seq(0,1.98*pi,length=100)

# the original plot
radial.plot(testlen,testpos,rp.type=p,main=Test Polygon,line.col=blue)

# my attempt to shade the region between 3 and 3.5:
radial.plot(
matrix(c(3.5, 3), byrow = TRUE),
matrix(c(testpos, testpos), byrow = TRUE, nrow = 2),
rp.type=p,
main=Test Polygon,
poly.col = c('light blue', 'white'),
radial.lim = c(0, 3.5)
)
# In effect, draw a polygon at 3.5 filled with light blue, then another at 3 
filled with white.
# Now overplot the values of interest:
radial.plot(testlen,testpos,rp.type=p,main=Test Polygon,line.col=blue, 
radial.lim = c(0, 3.5), add = TRUE)

Is there an easier way? How can I re-draw the grid lines that are in the 
original plot?


Hi Steve,
That is a tricky one. The best I can do at the moment is to suggest the 
following be added:


radial.grid(radial.lim=c(0,3.5),
 grid.pos=seq(0,3.5,length.out=8))
radial.plot(testlen,testpos,rp.type=p,
 main=,line.col=blue,add=TRUE)

There may be a solution using the radial.pie function, and if I find it, 
I'll post it.


Jim

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Re: [R] data distribution for lme

2013-12-10 Thread Andrew Robinson

No - it is assumed to be conditionally normal, that is, normal conditional
on the model.  So you should be looking at the distributions of the
residuals rather than of the response variable, as an indicator for whether
or not the model assumptions are satisfied.   Skewness in the residuals may
or may not affect the outcome; it depends on what the purpose of the model
is.  Skewness in residuals may arise from a number of different problems
with the model.

I hope that this helps,

Andrew




On Wed, Dec 11, 2013 at 1:55 AM, peyman zira...@gmail.com wrote:

 Hi folks,

 I am using the lme package of R, and am wondering if it is assumed that
 the dependent factor (what we fit for; y in many relevant texts) has to
 have a normal Gaussian distribution? Is there any margins where some
 skewness in the data is accepted and how within R itself one could check
 distribution of the data?

 Thanks,
 Peyman

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




-- 
Andrew Robinson
Deputy Director, CEBRA
Senior Lecturer in Applied Statistics  Tel:
+61-3-8344-6410
Department of Mathematics and StatisticsFax: +61-3-8344 4599
University of Melbourne, VIC 3010 Australia
Email: a.robin...@ms.unimelb.edu.auWebsite: http://www.ms.unimelb.edu.au

FAwR: http://www.ms.unimelb.edu.au/~andrewpr/FAwR/
SPuR: http://www.ms.unimelb.edu.au/spuRs/

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Re: [R] fisher.test - can I use non-integer expected values?

On Dec 10, 2013, at 2:04 PM, bakerwl wrote:

I seem to be able to use expected values that are decimal (e.g., 1.33) when
using chisq.test but not when using fisher.test.

There are no expected values in the input to fisher.test.

This happens when using an
array/matrix as input. Fisher.test returns: Error in sprintf(gettext(fmt,
domain = domain), ...) : invalid format '%d'; use format %s for character
objects.

Thus, it appears fisher.test is looking for integers only.

That would seem to be a very reasonable assumption.

I tried putting the data in x and y factor objects, but that does not work
either.

Is there another way to use non-integer expected values with fisher.test or
is that a limitation of fisher.test?

If I must use integer expected values, I suppose one option would be round
the expected value down or up to an integer. But, which? I tried that, but
they produce different p values.

Well, of course. First, you tell us why you need `fisher.test` at all. It says
very clearly it is for count data and you clearly want to do something with
input that is not counts. `prop.test` will test a distribution of counts
against expected proportions and `binom.test` will do an exact test of a
Bernoulli experiment against (one) proportion.

Thanks for any help!

View this message in context:
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--
David Winsemius
Alameda, CA, USA

Re: [R] data distribution for lme

2013-12-10 Thread Rolf Turner



See inline below.

On 12/11/13 11:28, Bert Gunter wrote:

This is not really an R question -- it is statistics.
In any case, you should do better posting this on the
R-Sig-Mixed-Models list, which concerns itself with matters like this.

However, I'll hazard a guess at an answer: maybe.  (Vague questions
elicit vague answers).


No! Nay! Never!  Well, hardly ever.   The ***y*** values will rarely be 
Gaussian.

(Think about a simple one-way anova, with 3 levels, and N(0,sigma^2) errors.
The y values will have a distribution which is a mixture of 3 
independent Gaussian

distributions.)

You *may* wish to worry about whether the ***errors*** have a Gaussian
distribution.  Some inferential results depend on this, but in many cases
these results are quite robust to non-Gaussianity.

There.  I have exhausted my knowledge of the subject.

cheers,

Rolf


Cheers,
Bert

On Tue, Dec 10, 2013 at 6:55 AM, peyman zira...@gmail.com wrote:

Hi folks,

I am using the lme package of R, and am wondering if it is assumed that
the dependent factor (what we fit for; y in many relevant texts) has to
have a normal Gaussian distribution? Is there any margins where some
skewness in the data is accepted and how within R itself one could check
distribution of the data?


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Re: [R] radial.plot shaded region


On Dec 10, 2013, at 4:24 PM, Jim Lemon wrote:

 On 12/11/2013 10:33 AM, Hodge, Steven wrote:
 I'm working with radial.plot in the plotrix package.
 
 Is there a way to show a shaded region for a specific range of values in 
 radial.lim?
 
 For example:
 # test data from radial.plot{plotrix}
 testlen-c(sin(seq(0,1.98*pi,length=100))+2+rnorm(100)/10)
 testpos-seq(0,1.98*pi,length=100)
 
 # the original plot
 radial.plot(testlen,testpos,rp.type=p,main=Test Polygon,line.col=blue)
 
 # my attempt to shade the region between 3 and 3.5:
 radial.plot(
  matrix(c(3.5, 3), byrow = TRUE),
  matrix(c(testpos, testpos), byrow = TRUE, nrow = 2),
  rp.type=p,
  main=Test Polygon,
  poly.col = c('light blue', 'white'),
  radial.lim = c(0, 3.5)
  )
 # In effect, draw a polygon at 3.5 filled with light blue, then another at 3 
 filled with white.
 # Now overplot the values of interest:
 radial.plot(testlen,testpos,rp.type=p,main=Test Polygon,line.col=blue, 
 radial.lim = c(0, 3.5), add = TRUE)
 
 Is there an easier way? How can I re-draw the grid lines that are in the 
 original plot?
 
 Hi Steve,
 That is a tricky one. The best I can do at the moment is to suggest the 
 following be added:
 
 radial.grid(radial.lim=c(0,3.5),
 grid.pos=seq(0,3.5,length.out=8))
 radial.plot(testlen,testpos,rp.type=p,
 main=,line.col=blue,add=TRUE)
 
 There may be a solution using the radial.pie function, and if I find it, I'll 
 post it.

Looking at the code I would have thought it would not be too difficult to 
de-couple the plot setup and the grid plotting in this section:

if (!add) {
par(mar = mar, pty = s)
plot(c(-maxlength, maxlength), c(-maxlength, maxlength), 
type = n, axes = FALSE, main = main, xlab = xlab, 
ylab = ylab)
if (show.grid) {
for (i in seq(length(grid.pos), 1, by = -1)) {
xpos - cos(angles) * (grid.pos[i] - radial.lim[1])
ypos - sin(angles) * (grid.pos[i] - radial.lim[1])
polygon(xpos, ypos, border = grid.col, col = grid.bg)
}
}

But you are the Plotmeister and I am a humble Student.

-- 

David Winsemius
Alameda, CA, USA

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Re: [R] fisher.test - can I use non-integer expected values?

2013-12-10 Thread bakerwl

David,

Thanks for your reply--I appreciate your thoughts. I will look at prop.test.

The reason I chose fisher.test over chisq.test is that fisher.test is more
appropriate when observed counts are not numerous--empty cells and cells
with counts  5 are less a problem. 

Expected values are needed to test a null hypothesis against observed
counts, but if total observed counts are 20 for 3 categories, then a null
hypothesis of a random effect would use expected values = 6.67 in each of
the 3 categories (20/3). 

Yes, fisher.test is for count data and so is chisq.test, but chisq.test
allows 6.67 to be input as expected values in each of 3 categories, while
fisher.test does not seem to allow this? 

I don't think it is inherent in Fisher's exact test itself that expected
values must be integers, but not sure.





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Re: [R] radial.plot shaded region

2013-12-10 Thread Jim Lemon


On 12/11/2013 11:24 AM, Jim Lemon wrote:

...
There may be a solution using the radial.pie function, and if I find it,
I'll post it.


Hi Steve,
Here it is. Just call radial.pie twice to get the annulus, then call 
radial.grid with the appropriate arguments, and then you can add 
whatever radial.plot you want, using the add argument.


radial.pie(3.5,sector.colors=lightblue)
radial.pie(3,sector.colors=white,add=TRUE)
radial.grid(radial.lim=c(0,3.5),grid.pos=seq(0,3.5,length.out=8))

Jim

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Re: [R] data distribution for lme

2013-12-10 Thread Bert Gunter

Thanks Rolf and Andrew. I was entirely too careless and should take a
trip to the woodshed (google David Stockman woodshed  for the
reference).

The correct answer therefore is: maybe for the residuals, for the
right model, of course.

But I still think the crowd on r-sig-mixed-models is the right place
to hash it out, if anything meaningful can indeed be made of it.

Cheers,
Bert

On Tue, Dec 10, 2013 at 5:33 PM, Rolf Turner r.tur...@auckland.ac.nz wrote:

 See inline below.

 On 12/11/13 11:28, Bert Gunter wrote:

 This is not really an R question -- it is statistics.
 In any case, you should do better posting this on the
 R-Sig-Mixed-Models list, which concerns itself with matters like this.

 However, I'll hazard a guess at an answer: maybe.  (Vague questions
 elicit vague answers).


 No! Nay! Never!  Well, hardly ever.   The ***y*** values will rarely be
 Gaussian.
 (Think about a simple one-way anova, with 3 levels, and N(0,sigma^2) errors.
 The y values will have a distribution which is a mixture of 3 independent
 Gaussian
 distributions.)

 You *may* wish to worry about whether the ***errors*** have a Gaussian
 distribution.  Some inferential results depend on this, but in many cases
 these results are quite robust to non-Gaussianity.

 There.  I have exhausted my knowledge of the subject.

 cheers,

 Rolf


 Cheers,
 Bert

 On Tue, Dec 10, 2013 at 6:55 AM, peyman zira...@gmail.com wrote:

 Hi folks,

 I am using the lme package of R, and am wondering if it is assumed that
 the dependent factor (what we fit for; y in many relevant texts) has to
 have a normal Gaussian distribution? Is there any margins where some
 skewness in the data is accepted and how within R itself one could check
 distribution of the data?



-- 

Bert Gunter
Genentech Nonclinical Biostatistics

(650) 467-7374

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] fisher.test - can I use non-integer expected values?

On Dec 10, 2013, at 6:55 PM, bakerwl wrote:

David,

Thanks for your reply--I appreciate your thoughts. I will look at prop.test.

The reason I chose fisher.test over chisq.test is that fisher.test is more
appropriate when observed counts are not numerous--empty cells and cells
with counts 5 are less a problem.

Expected values are needed to test a null hypothesis against observed
counts, but if total observed counts are 20 for 3 categories, then a null
hypothesis of a random effect would use expected values = 6.67 in each of
the 3 categories (20/3).

Yes, fisher.test is for count data and so is chisq.test, but chisq.test
allows 6.67 to be input as expected values in each of 3 categories, while
fisher.test does not seem to allow this?

I don't think it is inherent in Fisher's exact test itself that expected
values must be integers, but not sure.

I see it differently, although I could be further educated on the subject and
I've been wrong on Rhelp before. I think it _is_ inherent in Fisher's Exact
Test. FET is essentially a permutation test built on the hypergeometric
distribution (a discrete distribution) and it is unclear what to do with 1.33
of an entity under conditions of permutation.

The chi-square test (one of many so-called chi-square tests) is a pretty good
approximation to the discrete counterparts despite the fact that the chi-square
distribution takes continuous arguments and generally holds well down to
expected counts of 5. The link between the chi-square and binomial
distributions is through there variances: npq vs sum(o-e)^2/n. You can develop
arguments in the limit that converge fairly quickly.

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View this message in context:
http://r.789695.n4.nabble.com/fisher-test-can-I-use-non-integer-expected-values-tp4681976p4681989.html
Sent from the R help mailing list archive at Nabble.com.

David Winsemius
Alameda, CA, USA

Re: [R] fisher.test - can I use non-integer expected values?