[R] [R-pkgs] New package - LGDtoolkit

[Andrija Djurovic]

Dear all,

my new R package LGDtoolkit is now on CRAN.

The goal of this package is to cover the most common steps in Loss Given
Default (LGD) rating model development. The main procedures available are
those that refer to bivariate and multivariate analysis.
In particular two statistical methods for multivariate analysis are
currently implemented – OLS regression and fractional logistic regression.
Both methods are also available within different blockwise model designs,
and both have customized stepwise algorithms. To cover other important
steps for LGD model development, it is recommended to use 'LGDtoolkit'
package along with 'PDtoolkit', and 'monobin' (or 'monobinShiny') packages.

With the next releases package will be extended with a new features. In
particular, validation procedures will be added, and significant part of
the package will be dedicated to the usage of survival analysis for LGD
modeling (extrapolation of LGD for incomplete recoveries, development of
risk differentiation function and collateral haircut modeling).

To check out the current package functionalities run
install.packages("LGDtoolkit") and help(package = "LGDtoolkit").
To stay up to date with new development features follow the github page:
https://github.com/andrija-djurovic/LGDtoolkit.

I hope some of you will find the package useful.

Kind regards,
Andrija

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[R] [R-pkgs] PDtoolkit: Collection of Tools for PD Rating Model Development

[Andrija Djurovic]

Dear R users,

I am happy to announce that my new package PDtoolkit is now available on
CRAN.

The goal of this package is to cover the most common steps in probability
of default (PD) rating model development and validation. The main
procedures available are those that refer to univariate, bivariate,
multivariate analysis, calibration and validation. Along with accompanied
monobin and monobinShiny packages, PDtoolkit provides functions which are
suitable for different data transformation and modeling tasks such as:
imputations, monotonic binning of numeric risk factors, binning of
categorical risk factors, weights of evidence (WoE) and information value
(IV) calculations, WoE coding (replacement of risk factors modalities with
WoE values), risk factor clustering, area under curve (AUC) calculation and
others. Additionally, package provides set of validation functions for
testing homogeneity, heterogeneity, discriminatory and predictive power of
the model.

For details and examples please check out the github page
https://github.com/andrija-djurovic/PDtoolkit  or README file on CRAN page
https://CRAN.R-project.org/package=PDtoolkit

Best regards,
Andrija Djurovic

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[R] [R-pkgs] monobin: new version 0.2.0

[Andrija Djurovic]

Dear R users,

the new version of monobin package is now on CRAN.
Additional binning algorithm is implemented - monotonic binning driven by
decision tree (mdt.bin).

For details and examples, install the new version of the monobin
(install.packages("monobin")) and check the help page of the function
(?mdt.bin) or visit the github page:
https://github.com/andrija-djurovic/monobin

Soon I will upload to the CRAN updated version of monobinShiny package in
order to reflect changes from the monobin. Until then users can install
github version (https://github.com/andrija-djurovic/monobinShiny) where new
binning algorithm is already implemented.

BR,
Andrija Djurovic

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[R] [R-pkgs] monobinShiny: Shiny User Interface for 'monobin' Package

[Andrija Djurovic]

Dear R users,

I am happy to announce that my new package monobinShiny is now available on
CRAN.
This is an add-on package to the 'monobin' package that simplifies its use.
As reminder, the goal of 'monobin' is to perform monotonic binning of
numeric risk factor in credit rating models (PD, LGD, EAD) development. All
functions handle both binary and continuous target variable. Missing values
and other possible special values are treated separately from so-called
complete cases.

'monobinShiny' provides shiny-based user interface to 'monobin' package and
it can be especially handy for less experienced R users as well as for
those who intend to perform quick scanning of numeric risk factors when
building credit rating models. The additional functions implemented in
'monobinShiny' that do no exist in 'monobin' package are: descriptive
statistics, special case and outliers imputation. The function descriptive
statistics is exported and can be used in R sessions independently from the
user interface, while the special case and the outlier imputation functions
are written to be used with shiny UI.

Here is, also, the link to github page:
https://github.com/andrija-djurovic/monobinShiny

Hope that some of you will find the package useful.

Best regards,
Andrija Djurovic

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[R] [R-pkgs] monobin: Monotonic Binning for Credit Rating Models

[Andrija Djurovic]

Dear R users,

I am happy to announce that my package monobin is now available on CRAN.

The package contains functions that perform monotonic binning of numeric
risk factor in credit rating models (PD, LGD, EAD) development. All
functions handle both binary and continuous target variable. Functions that
use isotonic regression in the first stage of binning process have an
additional feature for correction of minimum percentage of observations and
minimum target rate per bin. Additionally, monotonic trend can be
identified based on raw data or, if known in advance, forced by functions'
argument. Missing values and other possible special values are treated
separately from so-called complete cases.
Close attention should be paid when modeling continuous target due to
possible negative values.

Hope that some of you will find the package useful.

Best regards,
Andrija Djurovic

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Re: [R] Zoo rolling window with increasing window size

[Andrija Djurovic]

Something like this?

set.seed(123)
y <- rnorm(20)
sapply(1:length(y),  function(x) sum(y[1:x]))

or this, depending what is the output of your custom function

lapply(1:length(y),  function(x) sum(y[1:x]))

On Thu, Aug 10, 2017 at 8:39 PM, Christofer Bogaso <
bogaso.christo...@gmail.com> wrote:

> Hi Joshua, thanks for your prompt reply. However as I said, sum()
> function I used here just for demonstrating the problem, I have other
> custom function to implement, not necessarily sum()
>
> I am looking for a generic solution for above problem.
>
> Any better idea? Thanks,
>
> On Fri, Aug 11, 2017 at 12:04 AM, Joshua Ulrich 
> wrote:
> > Use a `width` of integer index locations.  And you likely want =
> > "right" (or rollapplyr(), as I used).
> >
> > R> set.seed(21)
> > R> x <- rnorm(10)
> > R> rs <- rollapplyr(x, seq_along(x), sum)
> > R> cs <- cumsum(x)
> > R> identical(rs, cs)
> > [1] TRUE
> >
> >
> > On Thu, Aug 10, 2017 at 1:28 PM, Christofer Bogaso
> >  wrote:
> >> Hi again,
> >>
> >> I am wondering there is any function for 'zoo' time series, where I
> >> can apply a user defined function rolling window basis, wherein window
> >> size is ever increasing i.e. not fixed. For example, let say I have
> >> below user defined function and a zoo time series :
> >>
> >>> library(zoo)
> >>
> >>> UDF = function(x) sum(x)
> >>
> >>> TS = zoo(rnorm(10), seq(as.Date('2017-01-01'), as.Date('2017-01-10'),
> by = '1 day'))
> >>
> >>>
> >>
> >> Now I want to apply UDF (this can be any custom function, however here
> >> I put it just quick example) rolling window basis like :
> >>
> >> 1st data point = 1st data point of TS
> >> 2nd data point = sum of 1st and 2nd data points of TS
> >> 3rd data point = sum of 1st 2nd and 3rd data points of TS
> >>
> >> so on
> >>
> >> I am aware of the rollapply() function from zoo, however, appears like
> >> it is only for fixed window size.
> >>
> >> Appreciate any pointer how to achieve above strategy of implementing
> >> rolling calculation based on increased window size.
> >>
> >> Thanks for your time.
> >>
> >> __
> >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> 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.
> >
> >
> >
> > --
> > Joshua Ulrich  |  about.me/joshuaulrich
> > FOSS Trading  |  www.fosstrading.com
> > R/Finance 2017 | www.rinfinance.com
>
> __
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> posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] Creating a new column with the number of the observation

[Andrija Djurovic]

Hi. Here is one approach:

x - rep(c(A, B, C), c(3,1,2))
DF - data.frame(x, stringsAsFactors=FALSE)
cbind(DF, new_c=c(lapply(rle(DF$x)[[1]], function(x) 1:x), recursive=T))

Andrija


On Wed, May 28, 2014 at 10:46 PM, Marcos Santos mmsantos...@gmail.comwrote:

 I have a data frame like this:

   Category Observed_value  A 100  A 130  A 140  B 90  C 80  D 120  D 130

 I need to create an index column that show the number of the observation
 for each category. I have three observations in the category A, one in
 category B, one in category C and two in category D. The new matrix should
 be like this:


   Category Observed_value Index  A 100 1  A 130 2  A 140 3  B 90 1  C 80 1
 D 120 1  D 130 2

 Please, could anyone help me with this problem?

 Thank you.
 --
 Marcos Martins Santos
 mmsantos...@gmail.com
 --
 Skype: marcosmartinssantos

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Re: [R] Creating a new column with the number of the observation

[Andrija Djurovic]

Here is another approach:

x - rep(c(A, B, C), c(3,1,2))
DF1 - data.frame(x)
cbind(DF1, new_c=ave(as.numeric(DF1$x), DF1$x, FUN=function(x) 1:length(x)))

Note the difference between DF (in previous solution) and DF1
str(DF)
str(DF1)


Andrija


On Thu, May 29, 2014 at 12:03 AM, Andrija Djurovic djandr...@gmail.comwrote:

 Hi. Here is one approach:

 x - rep(c(A, B, C), c(3,1,2))
 DF - data.frame(x, stringsAsFactors=FALSE)
 cbind(DF, new_c=c(lapply(rle(DF$x)[[1]], function(x) 1:x), recursive=T))

 Andrija


 On Wed, May 28, 2014 at 10:46 PM, Marcos Santos mmsantos...@gmail.comwrote:

 I have a data frame like this:

   Category Observed_value  A 100  A 130  A 140  B 90  C 80  D 120  D 130

 I need to create an index column that show the number of the observation
 for each category. I have three observations in the category A, one in
 category B, one in category C and two in category D. The new matrix should
 be like this:


   Category Observed_value Index  A 100 1  A 130 2  A 140 3  B 90 1  C 80 1
 D 120 1  D 130 2

 Please, could anyone help me with this problem?

 Thank you.
 --
 Marcos Martins Santos
 mmsantos...@gmail.com
 --
 Skype: marcosmartinssantos

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Re: [R] adding a frequency variable to a data frame

[Andrija Djurovic]

Hi. Here is one solution:

table(Donner$family)[Donner$family]

Andrija


On Fri, Apr 11, 2014 at 10:40 PM, Michael Friendly frien...@yorku.cawrote:

 I'm sure this is pretty simple, but it's Friday afternoon, and I just
 don't see it...

 In a data frame with a categorical/character factor, I want to add another
 column giving, for each observation, the frequency of that factor level.

 An example, where the variable of interest is family:

  data(Donner, package=vcdExtra)
  str(Donner)
 'data.frame':   90 obs. of  5 variables:
  $ family  : Factor w/ 10 levels Breen,Donner,..: 9 1 1 1 1 1 1 1 1 1
 ...
  $ age : int  23 13 1 5 14 40 51 9 3 8 ...
  $ sex : Factor w/ 2 levels Female,Male: 2 2 1 2 2 1 2 2 2 2 ...
  $ survived: int  0 1 1 1 1 1 1 1 1 1 ...
  $ death   : POSIXct, format: 1846-12-29 NA ...
  table(Donner$family)

 BreenDonner  Eddy  FosdWolfGraves  Keseberg McCutchen
 MurFosPik
 914 4 410 4 312
 Other  Reed
23 7
 

 Here, I want to create a new variable, family.size,   where all the Breens
 have 9,
 the Donners, 14,  and so on...

 --
 Michael Friendly Email: friendly AT yorku DOT ca
 Professor, Psychology Dept.  Chair, Quantitative Methods
 York University  Voice: 416 736-2100 x66249 Fax: 416 736-5814
 4700 Keele StreetWeb:   http://www.datavis.ca
 Toronto, ONT  M3J 1P3 CANADA

 __
 R-help@r-project.org mailing list
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 posting-guide.html
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Re: [R] adding a frequency variable to a data frame

[Andrija Djurovic]

another:

ave(as.numeric(Donner$family), Donner$family, FUN=length)


On Fri, Apr 11, 2014 at 10:51 PM, Andrija Djurovic djandr...@gmail.comwrote:

 Hi. Here is one solution:

 table(Donner$family)[Donner$family]

 Andrija


 On Fri, Apr 11, 2014 at 10:40 PM, Michael Friendly frien...@yorku.cawrote:

 I'm sure this is pretty simple, but it's Friday afternoon, and I just
 don't see it...

 In a data frame with a categorical/character factor, I want to add another
 column giving, for each observation, the frequency of that factor level.

 An example, where the variable of interest is family:

  data(Donner, package=vcdExtra)
  str(Donner)
 'data.frame':   90 obs. of  5 variables:
  $ family  : Factor w/ 10 levels Breen,Donner,..: 9 1 1 1 1 1 1 1 1 1
 ...
  $ age : int  23 13 1 5 14 40 51 9 3 8 ...
  $ sex : Factor w/ 2 levels Female,Male: 2 2 1 2 2 1 2 2 2 2 ...
  $ survived: int  0 1 1 1 1 1 1 1 1 1 ...
  $ death   : POSIXct, format: 1846-12-29 NA ...
  table(Donner$family)

 BreenDonner  Eddy  FosdWolfGraves  Keseberg McCutchen
 MurFosPik
 914 4 410 4 312
 Other  Reed
23 7
 

 Here, I want to create a new variable, family.size,   where all the
 Breens have 9,
 the Donners, 14,  and so on...

 --
 Michael Friendly Email: friendly AT yorku DOT ca
 Professor, Psychology Dept.  Chair, Quantitative Methods
 York University  Voice: 416 736-2100 x66249 Fax: 416 736-5814
 4700 Keele StreetWeb:   http://www.datavis.ca
 Toronto, ONT  M3J 1P3 CANADA

 __
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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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Re: [R] From list to dataframe

[andrija djurovic]

Hi. Here is one way:
l - list(structure(list(BP_A = c(27001689L, 27001689L, 27001689L,
27001689L, 27001689L, 27001689L), SNP_A = c(rs4822747, rs4822747,
rs4822747, rs4822747, rs4822747, rs4822747), BP_B = c(27002392L,
27004298L, 27004902L, 27004964L, 27005122L, 27005158L), SNP_B =
c(rs4820690,
rs5761627, rs5752355, rs4822748, rs4820691, rs4822749
), R2 = c(0.695642, 0.695642, 0.687861, 0.682715, 0.695642, 0.695642
)), .Names = c(BP_A, SNP_A, BP_B, SNP_B, R2), row.names = 2:7,
class = data.frame),
structure(list(BP_A = c(27002392L, 27002392L, 27002392L,
27002392L, 27002392L, 27002392L, 27002392L, 27002392L), SNP_A =
c(rs4820690,
rs4820690, rs4820690, rs4820690, rs4820690, rs4820690,
rs4820690, rs4820690), BP_B = c(27001689L, 27004298L,
27004902L, 27004964L, 27005122L, 27005158L, 27005194L, 27005470L
), SNP_B = c(rs4822747, rs5761627, rs5752355, rs4822748,
rs4820691, rs4822749, rs4822750, rs5761629), R2 = c(0.695642,
1, 0.988681, 0.988655, 1, 1, 0.988655, 0.695642)), .Names = c(BP_A,
SNP_A, BP_B, SNP_B, R2), row.names = c(9L, 11L, 12L,
13L, 14L, 15L, 16L, 17L), class = data.frame), structure(list(
BP_A = c(27004298L, 27004298L, 27004298L, 27004298L,
27004298L, 27004298L), SNP_A = c(rs5761627, rs5761627,
rs5761627, rs5761627, rs5761627, rs5761627),
BP_B = c(27001689L, 27002392L, 27004902L, 27004964L,
27005122L, 27005158L), SNP_B = c(rs4822747, rs4820690,
rs5752355, rs4822748, rs4820691, rs4822749),
R2 = c(0.695642, 1, 0.988681, 0.988655, 1, 1)), .Names = c(BP_A,
SNP_A, BP_B, SNP_B, R2), row.names = c(19L, 20L,
22L, 23L, 24L, 25L), class = data.frame))

do.call(rbind, l)


Andrija


On Thu, Nov 14, 2013 at 9:41 AM, Hermann Norpois hnorp...@gmail.com wrote:

 Hello,

 having a list like testlist I would like to transform it in dataframe. How
 does it work?

 Thanks
 Hermann

  testlist
 [[1]]
   BP_A SNP_A BP_B SNP_B   R2
 2 27001689 rs4822747 27002392 rs4820690 0.695642
 3 27001689 rs4822747 27004298 rs5761627 0.695642
 4 27001689 rs4822747 27004902 rs5752355 0.687861
 5 27001689 rs4822747 27004964 rs4822748 0.682715
 6 27001689 rs4822747 27005122 rs4820691 0.695642
 7 27001689 rs4822747 27005158 rs4822749 0.695642

 [[2]]
BP_A SNP_A BP_B SNP_B   R2
 9  27002392 rs4820690 27001689 rs4822747 0.695642
 11 27002392 rs4820690 27004298 rs5761627 1.00
 12 27002392 rs4820690 27004902 rs5752355 0.988681
 13 27002392 rs4820690 27004964 rs4822748 0.988655
 14 27002392 rs4820690 27005122 rs4820691 1.00
 15 27002392 rs4820690 27005158 rs4822749 1.00
 16 27002392 rs4820690 27005194 rs4822750 0.988655
 17 27002392 rs4820690 27005470 rs5761629 0.695642

 [[3]]
BP_A SNP_A BP_B SNP_B   R2
 19 27004298 rs5761627 27001689 rs4822747 0.695642
 20 27004298 rs5761627 27002392 rs4820690 1.00
 22 27004298 rs5761627 27004902 rs5752355 0.988681
 23 27004298 rs5761627 27004964 rs4822748 0.988655
 24 27004298 rs5761627 27005122 rs4820691 1.00
 25 27004298 rs5761627 27005158 rs4822749 1.00

 #The aim is the dataframe
BP_A SNP_A BP_B SNP_B   R2
 2  27001689 rs4822747 27002392 rs4820690 0.695642
 3  27001689 rs4822747 27004298 rs5761627 0.695642
 4  27001689 rs4822747 27004902 rs5752355 0.687861
 5  27001689 rs4822747 27004964 rs4822748 0.682715
 6  27001689 rs4822747 27005122 rs4820691 0.695642
 7  27001689 rs4822747 27005158 rs4822749 0.695642
 9  27002392 rs4820690 27001689 rs4822747 0.695642
 11 27002392 rs4820690 27004298 rs5761627 1.00
 12 27002392 rs4820690 27004902 rs5752355 0.988681
 13 27002392 rs4820690 27004964 rs4822748 0.988655
 14 27002392 rs4820690 27005122 rs4820691 1.00
 15 27002392 rs4820690 27005158 rs4822749 1.00
 16 27002392 rs4820690 27005194 rs4822750 0.988655
 17 27002392 rs4820690 27005470 rs5761629 0.695642
 19 27004298 rs5761627 27001689 rs4822747 0.695642
 20 27004298 rs5761627 27002392 rs4820690 1.00
 22 27004298 rs5761627 27004902 rs5752355 0.988681
 23 27004298 rs5761627 27004964 rs4822748 0.988655
 24 27004298 rs5761627 27005122 rs4820691 1.00
 25 27004298 rs5761627 27005158 rs4822749 1.00
 
  dput (testlist)
 list(structure(list(BP_A = c(27001689L, 27001689L, 27001689L,
 27001689L, 27001689L, 27001689L), SNP_A = c(rs4822747, rs4822747,
 rs4822747, rs4822747, rs4822747, rs4822747), BP_B = c(27002392L,
 27004298L, 27004902L, 27004964L, 27005122L, 27005158L), SNP_B =
 c(rs4820690,
 rs5761627, rs5752355, rs4822748, rs4820691, rs4822749
 ), R2 = c(0.695642, 0.695642, 0.687861, 0.682715, 0.695642, 0.695642
 )), .Names = c(BP_A, SNP_A, BP_B, SNP_B, R2), row.names = 2:7,
 class = data.frame),
 structure(list(BP_A = c(27002392L, 27002392L, 27002392L,
 27002392L, 27002392L, 27002392L, 27002392L, 27002392L), SNP_A =
 c(rs4820690,
 rs4820690, rs4820690, rs4820690, rs4820690, rs4820690,
 rs4820690, rs4820690), BP_B = c(27001689L, 27004298L,
 27004902L, 

Re: [R] repeating values in an index two by two

[andrija djurovic]

Hi. Here are two approaches:

c(mapply(function(x,y) rep(c(x,y), 2), (1:10)[c(T,F)], (1:10)[c(F,T)]))

c(tapply(1:10, rep(1:(10/2), each=2), rep, 2), recursive=T)

Andrija





On Mon, Nov 11, 2013 at 1:11 PM, Federico Calboli
f.calb...@imperial.ac.ukwrote:

 Hi All,

 I am trying to create an index that returns something like

 1,2,1,2,3,4,3,4,5,6,5,6,7,8,7,8

 and so on and so forth until a predetermined value (which is obviously
 even).  I am trying very hard to avoid for loops or for loops front ends.

 I'd be obliged if anybody could offer a suggestion.

 BW

 F

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Re: [R] R function to locate Excel sheet?

[andrija djurovic]

Hi here is the solutions using XLConnect package:

library(XLConnect)
wb - loadWorkbook(path to your Excel file)
 c(particular sheet name)%in%getSheets(wb)

Andrija






On Tue, Oct 29, 2013 at 9:26 PM, Ron Michael ron_michae...@yahoo.comwrote:

 Hi,

 I am looking for some R function which will tell me, whether a particular
 sheet in an Excel file (.xlsx/.xls) exists or not. I just need to get some
 TRUE/FALSE type of answer.

 Can somebody give me any pointer if such function exists or not?

 Thanks and regards,

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Re: [R] resampling

[andrija djurovic]

Hi.
See ?sample, ?replicate,?colMeans, ?plot..

Here is the simple example:

sample(1:1000,20)
replicate(5, sample(1:1000,20))
colMeans(replicate(5, sample(1:1000,20)))

Andrija


On Wed, Jul 31, 2013 at 1:23 PM, Rita Gamito rslo...@fc.ul.pt wrote:

 Could anyone tell me how,from a pool of 1002 observations (one variable),
  can I resample 1000 samples of 20 observations?
 And then calculate the mean and standard deviation between 2, 3, 4, ...,
 1000 samples and plot them?
 Thank you!

 _

 Rita Gamito
 Centro de Oceanografia
 Faculdade de Ciências, Universidade de Lisboa
 Campo Grande, 1749-016 Lisboa, Portugal
 e-mail: rgam...@fc.ul.pt
 Tel: + 351 21 750 00 00 - ext. 22575
 Fax: + 351 21 750 02 07
 www.co.fc.ul.pt
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Re: [R] calculate time from dates

[andrija djurovic]

Hi.

See
http://stackoverflow.com/questions/1995933/number-of-months-between-two-dates

Andrija


On Thu, Jul 11, 2013 at 11:56 AM, Gallon Li gallon...@gmail.com wrote:

 My data are from 2008 to 2010, with repeated measures for same subjects. I
 wish to compute number of months since january 2008.

 The data are like the following:

 ID date
 1 4/12/2008
 1 4/13/2008
 1 5/11/2008
 2 3/21/2009
 2 4/22/2009
 2 8/05/2009
 ...

 the date column are in the format %m/%d/%y. i wish to obtain

 ID month
 1 4
 1 4
 1 5
 2 15
 2 16
 2 20
 ...

 also, for the same ID with two identical month, I only want to keep the
 last one. can any expert help with this question?

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Re: [R] Write an Excel workbook?

[andrija djurovic]

XLConnect package
Andrija
On Jun 27, 2013 9:15 PM, Spencer Graves 
spencer.gra...@structuremonitoring.com wrote:

 Hello:


 Is there a fully transportable way to write an Excel workbook?


 The writeFindFn2xls function in the sos package attempts to write
 search results to an Excel workbook consisting of (1) a summary of the
 packages identified in the search, (2) the individual help pages found, and
 (3) the search that produced 1 and 2. Unfortunately, for some time now, it
 only works with 32-bit R; the 64-bit R writes this information to three
 separate csv files.


 The current code first tries WriteXLS{WriteXLS}. Unfortunately, this fails
 on my computer, because it can't find the Perl module Text::CSV_XS. I
 don't see an easy fix to this.


 If WriteXLS fails, the code then tries odbcConnectExcel{ODBC}. From this,
 I get, odbcConnectExcel is only usable with 32-bit Windows. That's OK,
 but it means that I need to run all my searches on 32-bit R, and I usually
 use 64-bit R.


 I tried write.xls{dataframes2xls}. This put quotes around everything,
 which I found unacceptable.


 findFn('write.xls') identified a function by that name in the marray
 package. Unfortunately, install.packages(marray) produced, Warning
 message: package ‘marray’ is not available (for R version 3.0.1) ???xls
 failed to produce anything else that looked to me like it would write a
 complete workbook.


 Suggestions?
 Thanks,
 Spencer Graves


  sessionInfo()
 R version 3.0.1 (2013-05-16)
 Platform: x86_64-w64-mingw32/x64 (64-bit)

 locale:
 [1] LC_COLLATE=English_United States.1252
 [2] LC_CTYPE=English_United States.1252
 [3] LC_MONETARY=English_United States.1252
 [4] LC_NUMERIC=C
 [5] LC_TIME=English_United States.1252

 attached base packages:
 [1] stats graphics grDevices utils datasets methods base

 other attached packages:
 [1] RODBC_1.3-6 WriteXLS_2.3.1 sos_1.3-7 brew_1.0-6

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Re: [R] Choosing subset of data.frame

[andrija djurovic]

Hi.
Try this:

beta_results[beta_results$instrument%in%instru, ]

and  see help page ?%in%

Hope this helps

Andrija


On Thu, Jun 20, 2013 at 12:45 PM, Katherine Gobin katherine_go...@yahoo.com
 wrote:

 Dear R Forum

 I have a data frame as

 beta_results = data.frame(instrument = c(ABC, DEF, JKL,  LMN,
 PQR, STU, UVW, XYZ),

 beta_values = c(1.27, -0.22, 0.529, 0.011, 2.31, -1.08, -2.7, 0.42))

  beta_results
   instrument beta_values
 1ABC   1.270
 2DEF  -0.220
 3JKL   0.529
 4LMN   0.011
 5PQR   2.310
 6STU  -1.080
 7UVW  -2.700
 8XYZ   0.420


 Through some other process, I am getting instrument names as say (which
 may change each time I run this process
 and hence I can't hard code it).


 instru = c(JKL, STU, XYZ)

 Now I want the subset of beta_results, (say beta_results_A)  pertaining to
 only instru i.e

 beta_results_A =


   instrument beta_values
 3JKL   0.529
 6STU  -1.080
 8XYZ   0.420


 I did try

 beta_results_A = beta_results[instru]
 or
 beta_results_A = subset(beta_results, beta_results$instrument = instru]

 but I guess it's failing.

 Kindly guide

 Regards

 Katherine
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Re: [R] Trouble sorting the matrix

[andrija djurovic]

Hi. Here is an example of sorting matrix columns:

 mat - matrix(10:1, ncol=2, byrow=TRUE)
 mat
 [,1] [,2]
[1,]   109
[2,]87
[3,]65
[4,]43
[5,]21
 apply(mat, 2, function(x) x[order(x)])
 [,1] [,2]
[1,]21
[2,]43
[3,]65
[4,]87
[5,]   109

You should read help page  [, apply and order (?[, ?apply, ?order)

Hope this helps

Andrija


On Sat, Jun 15, 2013 at 12:36 PM, Ola Cabaj olaca...@gmail.com wrote:

 I would like to sort matrix3, so that every column in output matrix is
 sorted. Also I have to unsort it later on.

 matrix1-matrix(rnorm(100,354,78),ncol=10)
 matrix2-matrix(rnorm(100,225,102),ncol=10)
 matrix3-cbind(matrix1,matrix2)
 nrCol-length(matrix3[1,])
 class1-1:10
   for(i in 1:nrCol)
   {
 sorted.matrix-matrix3[order(matrix3[class1,i]),]
   }
  for(i in 1:liczbaKolumn)
   {
output-sorted.matrix[order(matrix3[class1,i]),]
   }

 --
 Aleksandra Cabaj

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Re: [R] SQL queries in R

[andrija djurovic]

?sqlQuery
as.is - argument
Andrija
On Jun 7, 2013 9:10 PM, Sneha Bishnoi sneha.bish...@gmail.com wrote:

 Hey all!

 I am trying to select a bunch of id's  (data type -character) from a table
 and store them in a variable in R
 But when i do this, it automatically truncates the leading zero's in id's
 even though they are of character type.

 code is :-

 myconn-odbcConnect(testdata)
 sql.select-paste(select UNIT_ID from UNITS where (UNIT_TYPE=',unit,'
 and COMMUNITY=',property,'),sep=)
  unit_ids-sqlQuery(myconn,sql.select)
 print(unit_ids)
 is there anyway i can retain the UNIT_ID's as they are.

 Thanks!
 Sneha

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Re: [R] SQL queries in R

[andrija djurovic]

myconn-odbcConnect(testdata)
sql.select-paste(select UNIT_ID from UNITS where (UNIT_TYPE=',unit,'
and COMMUNITY=',property,'),sep=)

unit_ids-sqlQuery(myconn,sql.select,as.is=TRUE)

This should works if myconn and sql.select are defined properly

Andrija
On Jun 7, 2013 9:58 PM, Sneha Bishnoi sneha.bish...@gmail.com wrote:

 tried as.is ,gives an error,

 [1] 01000 10054 [Microsoft][ODBC SQL Server
 Driver][DBNETLIB]ConnectionWrite (send()).
 [2] [RODBC] ERROR: Could not SQLExecDirect 'select UNIT_ID from UNITS
 where (UNIT_TYPE='1X1' and COMMUNITY='SAN1193')'


 On Fri, Jun 7, 2013 at 3:21 PM, andrija djurovic djandr...@gmail.comwrote:

 ?sqlQuery
 as.is - argument
 Andrija
 On Jun 7, 2013 9:10 PM, Sneha Bishnoi sneha.bish...@gmail.com wrote:

 Hey all!

 I am trying to select a bunch of id's  (data type -character) from a
 table
 and store them in a variable in R
 But when i do this, it automatically truncates the leading zero's in id's
 even though they are of character type.

 code is :-

 myconn-odbcConnect(testdata)
 sql.select-paste(select UNIT_ID from UNITS where (UNIT_TYPE=',unit,'
 and COMMUNITY=',property,'),sep=)
  unit_ids-sqlQuery(myconn,sql.select)
 print(unit_ids)
 is there anyway i can retain the UNIT_ID's as they are.

 Thanks!
 Sneha

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 --
 Sneha Bishnoi
 +14047235469
 H. Milton Stewart School of Industrial   Systems Engineering
 Georgia Tech


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Re: [R] numeric not equal its value

[andrija djurovic]

Hi. Check R FAQ, section 7.31 Why doesn't R think these numbers are equal?

http://cran.r-project.org/doc/FAQ/R-FAQ.html

all.equal(a,b)

Andrija


On Mon, May 20, 2013 at 2:36 PM, Jannis bt_jan...@yahoo.de wrote:

 Dear R users,



 I ran into the strange situation where a number does not seem to equal its
 value. Try this:



 a - seq(0.05,0.7,0.05)[3]

 b - 0.15

 a == b


 Should this not be TRUE? a-b yields a very small number (and not 0) so
 this most probably is a numerical error, but why does seq create such
 numbers?


 Thanks a lot
 Jannis

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Re: [R] Questions on function readNamedRegionFromFile in XLConnect pacakge

[andrija djurovic]

Hi Miao.

1.  Calendar, Iris, IQ are the named region in excel file
multiregion.xlsx
(demoFiles/multiregion.xlsx). Open this file (demoFiles/multiregion.xlsx),
go to Formulas tab and click Name manager to see reference regions.

2. Check following functions:
?readWorksheet   #arguments startCol, startRow, endCol, endRow
?idx2col
?col2idx

Example:
 col2idx(AK)
[1] 37
 idx2col(37)
[1] AK

Hope this helps.

Andrija


On Tue, Apr 23, 2013 at 8:49 AM, jpm miao miao...@gmail.com wrote:

 Hi,

I have two questions on the function readNamedRegionFromFile in
 XLConnect pacakge.

 1.

In the documentation,

 # multiregion xlsx file from demoFiles subfolder of package XLConnect
 demoExcelFile - system.file(demoFiles/multiregion.xlsx,
  package = XLConnect)

 # Load a single named region into a single data.frame.
 df - readNamedRegionFromFile(demoExcelFile, name=Iris)

 # Load multiple regions at once - returns a (named) list
 # of data.frames.
 df - readNamedRegionFromFile(demoExcelFile,
   name=c(Calendar, Iris, IQ))

What are the names Calendar, Iris, IQ? I just couldn't find them
 from the file.

 2. Since my xlsx file is big, I might want to read data like AK9:AK18, for
 example. If my computation is right, AK is the 37th column. Then I need to
 tell the computer to read the 37th column, row 9 to row 18. Can I ask the
 function (any function in the package) to read the column by the header
 name (say, the name saved in AK1) instead of 37th?

 Thanks,

 Miao

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Re: [R] Questions on function readNamedRegionFromFile in XLConnect pacakge

[andrija djurovic]

Here is the concrete example:

 # multiregion xlsx file from demoFiles subfolder of package XLConnect
 demoExcelFile - system.file(demoFiles/multiregion.xlsx,
+  package = XLConnect)

 # Load a single named region into a single data.frame.
 DF1 - readNamedRegionFromFile(demoExcelFile, name=Iris)


 DF2 - readWorksheetFromFile(demoExcelFile, sheet=SecondSheet,
+
 startCol=col2idx(B),
+
 endCol=col2idx(F),
+
 startRow=3,
+header=TRUE)
 head(DF1)
  Sepal.Length Sepal.Width Petal.Length Petal.Width   Species
1  5.1 3.5  1.4 0.2setosa
2  4.9 3.0  1.4 0.2setosa
3  4.7 3.2  1.3 0.2 barbarica
4  4.6 3.1  1.5 0.2setosa
5  5.0 3.6  1.4 0.2setosa
6  5.4 3.9  1.7 0.4setosa
 head(DF2)
  Sepal.Length Sepal.Width Petal.Length Petal.Width   Species
1  5.1 3.5  1.4 0.2setosa
2  4.9 3.0  1.4 0.2setosa
3  4.7 3.2  1.3 0.2 barbarica
4  4.6 3.1  1.5 0.2setosa
5  5.0 3.6  1.4 0.2setosa
6  5.4 3.9  1.7 0.4setosa
 identical(DF1, DF2)
[1] TRUE


Andrija


On Tue, Apr 23, 2013 at 9:04 AM, andrija djurovic djandr...@gmail.comwrote:

 Hi Miao.

 1.  Calendar, Iris, IQ are the named region in excel file  
 multiregion.xlsx
 (demoFiles/multiregion.xlsx). Open this file (demoFiles/multiregion.xlsx),
 go to Formulas tab and click Name manager to see reference regions.

 2. Check following functions:
 ?readWorksheet   #arguments startCol, startRow, endCol, endRow
 ?idx2col
 ?col2idx

 Example:
  col2idx(AK)
 [1] 37
  idx2col(37)
 [1] AK

 Hope this helps.

 Andrija


 On Tue, Apr 23, 2013 at 8:49 AM, jpm miao miao...@gmail.com wrote:

 Hi,

I have two questions on the function readNamedRegionFromFile in
 XLConnect pacakge.

 1.

In the documentation,

 # multiregion xlsx file from demoFiles subfolder of package XLConnect
 demoExcelFile - system.file(demoFiles/multiregion.xlsx,
  package = XLConnect)

 # Load a single named region into a single data.frame.
 df - readNamedRegionFromFile(demoExcelFile, name=Iris)

 # Load multiple regions at once - returns a (named) list
 # of data.frames.
 df - readNamedRegionFromFile(demoExcelFile,
   name=c(Calendar, Iris, IQ))

What are the names Calendar, Iris, IQ? I just couldn't find them
 from the file.

 2. Since my xlsx file is big, I might want to read data like AK9:AK18, for
 example. If my computation is right, AK is the 37th column. Then I need to
 tell the computer to read the 37th column, row 9 to row 18. Can I ask the
 function (any function in the package) to read the column by the header
 name (say, the name saved in AK1) instead of 37th?

 Thanks,

 Miao

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Re: [R] Automation of R input

[andrija djurovic]

Hi.
If you want to do it completely in R you can create R gui using gWidgets
package.

Here is an example of creating gui for function arguments (thanks to John
Verzani):

library(gWidgets)
options(guiToolkit=tcltk)

##' An editor for an argument
##' depends on calss of object
arg_editor - function(x, cont, ...) UseMethod(arg_editor)
arg_editor.call - function(x, cont, ...) arg_editor(eval(x), cont, ...)
arg_editor.name - function(x, cont, ...) arg_editor(eval(x), cont, ...)
arg_editor.numeric - function(x, cont, ...) gcombobox(sort(unique(x)),
editable=TRUE, coerce.with=as.numeric, cont=cont)
arg_editor.character - function(x, cont, ...) gcombobox(sort(unique(x)),
editable=TRUE, coerce.with=as.character, cont=cont)
arg_editor.factor -  function(x, cont, ...) gcombobox(sort(unique(x)),
cont=cont)
arg_editor.logical - function(x, cont, ...) gcombobox(c(TRUE, FALSE),
selected=2 - as.numeric(x[1]), cont=cont)

## This replaces ggenericwidget
make_gui - function(FUN, parent) {
  args - formals(FUN)
  if(missing(parent)) {
parent - gwindow(Some GUI, visible=FALSE)
  }

  g - ggroup(cont=parent, horizontal=FALSE)

  lyt - glayout(cont=g)
  n - length(args)

  f - function(i, nm, x) {
lyt[i,1] - nm
lyt[i,2] - arg_editor(args[[i]], cont=lyt)
  }
  mapply(f, i=seq_along(args), nm=names(args), x=args)

  bg - ggroup(cont=g)
  gbutton(ok, cont=bg, handler=function(h, ...) {
values - lapply(lyt[,2], svalue)
nms - sapply(lyt[,1], svalue)
names(values) - nms
do.call(FUN, values)
dispose(parent)
  })

  visible(parent) - TRUE
  invisible(lyt)
}

## Try it out
s - function(x=1:4, y = letters, z=TRUE) {
  print(list(x, y, z))
  assign(x,x, envir=globalenv())
  assign(y,y, envir=globalenv())
  assign(z,z, envir=globalenv())
}

make_gui(s)

You can easily adjust this example to your needs.

If you want you can use Excel, Access in order to create front end
application and using commandArgs to supply  arguments to an R script when
is invoked.
Examples of calling an R scripts from Excel and Access you can find here:
http://www.slideshare.net/djandrija/excelr
http://www.slideshare.net/djandrija/r-and-access-2007

Also, shiny package can be useful here (http://www.rstudio.com/shiny/),
integration of html, php and R...

Hope this helps.

Andrija







On Tue, Apr 23, 2013 at 9:41 AM, Vahe nr vne...@gmail.com wrote:

 Hi all,

 I have R script which during its run require an input like this:
 choose between one of the grouping factor available :
 c(Village, Country)

 can I automate this part, in other word to pass for example Village when I
 am running the script.

 One more thing the script is the TimeSeriesAnalysis {ndvits}.

 Thanks in advance for any help or suggestion.

 Regards,
  Vahe

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Re: [R] Using different function (parameters) with apply

[andrija djurovic]

Hi.

It would be easier to help you if you show us the result you are trying to
obtain.

Here are my attempts:

 sweep(a,2,div, FUN=/)
 [,1] [,2] [,3]
[1,]1  2.0 2.33
[2,]2  2.5 2.67
[3,]3  3.0 3.00

or

 a/matrix(rep(div,3),ncol=3, byrow=TRUE)
 [,1] [,2] [,3]
[1,]1  2.0 2.33
[2,]2  2.5 2.67
[3,]3  3.0 3.00

Hope this helps.

Andrija




On Thu, Apr 18, 2013 at 7:20 AM, Sachinthaka Abeywardana 
sachin.abeyward...@gmail.com wrote:

 Hi All,

 I have the following problem (read the commented bit below):

 a-matrix(1:9,nrow=3)


 a

  [,1] [,2] [,3]
 [1,]147
 [2,]258
 [3,]369


 div-1:3

 apply(a,2,function(x)x/div) ##want to divide each column by div-
 instead each row is divided##


  [,1] [,2] [,3]
 [1,]1  4.07
 [2,]1  2.54
 [3,]1  2.03


 apply(a,1,function(x)x/div) ##Changing Margin from 2 to 1 does
 something completele weird## [,1] [,2] [,3]
 [1,] 1.00 2.003
 [2,] 2.00 2.503
 [3,] 2.33 2.673


 Any thoughts?


 Thanks,

 Sachin

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Re: [R] How can I ask R to skip the title when reading the data?

[andrija djurovic]

Hi.

You can try with argument skip in read.csv function(check ?read.csv).

Also, if you want directly to import an Excel file you can use
readWorksheet function from XLConnect package and use argument startRow to
set up the first row to read from.

Hope this helps.

Andrija


On Thu, Apr 18, 2013 at 9:46 AM, jpm miao miao...@gmail.com wrote:

 I have many xls grade report sheets with the same format



 XXX High School Grade Report

 Confidential Yes



Math English Science

 John 90 85 90

 Mary 75 88 93

 ……



 Since the reports are prepared on a regular basis, I have many reports with
 identical format. Without the title “XXX High School Grade Report

 Confidential Yes”, I can just change the xls to csv and read them via
 read.csv. How can I read the data in the presence of the title? How can I
 ask R to skip the title and just read the data?



 Thanks,



 Miao

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

Re: [R] Subsetting a large number into smaller numbers and find the largest product

[andrija djurovic]

Hi.

Here is one approach:

options(scipen=300)
numb -
73167176531330624919225119674426574742355349194934969835203127745063262395783180169848018694788518438586156078911294949545950173795833195285320880551112540698747158523863050715693290963295227443043557
strsplit(as.character(numb), )
blocks - rep(1:(nchar(numb)/5), each=5)
tapply(as.numeric(unlist(strsplit(as.character(numb), ))),
   blocks,
   prod)

Andrija


On Thu, Apr 18, 2013 at 10:47 AM, Janesh Devkota
janesh.devk...@gmail.comwrote:

 Hello,

 I have a big number lets say of around hundred digits. I want to subset
 that big number into consecutive number of 5 digits and find the product of
 those 5 digits. For example my first 5 digit number would be 73167. I need
 to check the product of the individual numbers in 73167 and so on.

 The sample number is as follows:



 73167176531330624919225119674426574742355349194934969835203127745063262395783180169848018694788518438586156078911294949545950173795833195285320880551112540698747158523863050715693290963295227443043557

 I have a problem subsetting the small numbers out of the big number.

 Any help is highly appreciated.

 Best Regards,
 Janesh Devkota

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

Re: [R] dividing a long column to many short ones by a condition

[andrija djurovic]

Hi Igor.

Here is one way:

DF - read.table(textConnection(90.1194354
87.94788274
80.34744843
64.06080347
30.40173724
0
0
0
0
0
16.28664495
23.88707926
29.31596091
48.85993485
13.02931596
0
0
0
7.600434311
20.62975027
29.31596091
32.5732899), header=FALSE)

a - DF$V1[which(DF$V1!=0)]
indx - which(DF$V1!=0)

blocks - cut(1:length(a), breaks=c(0,which(diff(indx)!=1), length(a)))
n_levels - length(levels(blocks))

l - vector(list, n_levels)
for(i in 1:n_levels)
{
l[[i]] - a[blocks==levels(blocks)[i]]
}
l

Andrija


On Thu, Apr 18, 2013 at 11:33 AM, Igor Mintz igormi...@gmail.com wrote:

 hello
 i have a very long column of numbers. i want R to make a new column every
 time the value changes from zero.
 example for the column:
 90.1194354
 87.94788274
 80.34744843
 64.06080347
 30.40173724
 0
 0
 0
 0
 0
 16.28664495
 23.88707926
 29.31596091
 48.85993485
 13.02931596
 0
 0
 0
 7.600434311
 20.62975027
 29.31596091
 32.5732899

 for this example i want to get 3 columns.

 thanks!

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


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

Re: [R] dividing a long column to many short ones by a condition

[andrija djurovic]

Hi.
I completely forgot split function so loop:

l - vector(list, n_levels)
for(i in 1:n_levels)
{
l[[i]] - a[blocks==levels(blocks)[i]]
}
l

could be substitute with:

split(a, blocks),

but anyway Rui's solution is better.

Andrija


On Thu, Apr 18, 2013 at 1:18 PM, andrija djurovic djandr...@gmail.comwrote:

 Hi Igor.

 Here is one way:

 DF - read.table(textConnection(90.1194354
 87.94788274
 80.34744843
 64.06080347
 30.40173724
 0
 0
 0
 0
 0
 16.28664495
 23.88707926
 29.31596091
 48.85993485
 13.02931596
 0
 0
 0
 7.600434311
 20.62975027
 29.31596091
 32.5732899), header=FALSE)

 a - DF$V1[which(DF$V1!=0)]
 indx - which(DF$V1!=0)

 blocks - cut(1:length(a), breaks=c(0,which(diff(indx)!=1), length(a)))
 n_levels - length(levels(blocks))

 l - vector(list, n_levels)
 for(i in 1:n_levels)
 {
 l[[i]] - a[blocks==levels(blocks)[i]]
 }
 l

 Andrija


 On Thu, Apr 18, 2013 at 11:33 AM, Igor Mintz igormi...@gmail.com wrote:

 hello
 i have a very long column of numbers. i want R to make a new column every
 time the value changes from zero.
 example for the column:
 90.1194354
 87.94788274
 80.34744843
 64.06080347
 30.40173724
 0
 0
 0
 0
 0
 16.28664495
 23.88707926
 29.31596091
 48.85993485
 13.02931596
 0
 0
 0
 7.600434311
 20.62975027
 29.31596091
 32.5732899

 for this example i want to get 3 columns.

 thanks!

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




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

Re: [R] Creating a vector with repeating dates

[andrija djurovic]

?rep


On Wed, Apr 17, 2013 at 11:11 AM, Katherine Gobin katherine_go...@yahoo.com
 wrote:

 Dear R forum

 I have a data.frame

 df = data.frame(dates = c(4/15/2013, 4/14/2013, 4/13/2013,
 4/12/2013), values = c(47, 38, 56, 92))

 I need to to create a vector by repeating the dates as

 Current_date, 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013,
 Current_date, 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, Current_date,
 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013

 i.e. I need to create a new vector as given below which I need to use for
 some other purpose.

 Current_date
 4/15/2013
 4/14/2013
 4/13/2013
 4/12/2013
 Current_date
 4/15/2013
 4/14/2013
 4/13/2013
 4/12/2013
 Current_date
 4/15/2013
 4/14/2013
 4/13/2013
 4/12/2013

 Is it possible to construct such a
  column?

 Regards

 Katherine



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

Re: [R] Creating a vector with repeating dates

[andrija djurovic]

Hi.

Here are some examples that can maybe help you:

a - Current date
b - Sys.Date()-1:5
a
b

class(a)
class(b)

c(a,b)
mode(b)
as.numeric(b)
class(c(a,b))

c(a, as.character(b))
class(c(a,b))
class(c(a,as.character(b)))

Hope this helps.


On Wed, Apr 17, 2013 at 11:21 AM, Katherine Gobin katherine_go...@yahoo.com
 wrote:

 Dear Andrija Djurovic,

 Thanks for the suggestion. Ia m aware of rep. However, here I need to
 repeat not only dates, but a string Current_date. Thus, I need to create
 a vector ( to be included in some other data.frame) with the name say dt
 which will contain

 dt
 Current_date
 4/15/2013
 4/14/2013
 4/13/2013
 4/12/2013
 Current_date
 4/15/2013
 4/14/2013
 4/13/2013
 4/12/2013
 Current_date
 4/15/2013
 4/14/2013
 4/13/2013
 4/12/2013

 So this is combination of dates and a string. Hence, I am just wondering
 if it is possible to create such a vector or not?

 Regards

 Katherine


 --- On *Wed, 17/4/13, andrija djurovic djandr...@gmail.com* wrote:


 From: andrija djurovic djandr...@gmail.com
 Subject: Re: [R] Creating a vector with repeating dates
 To: Katherine Gobin katherine_go...@yahoo.com
 Cc: r-help@r-project.org r-help@r-project.org
 Date: Wednesday, 17 April, 2013, 10:14 AM

 ?rep


 On Wed, Apr 17, 2013 at 11:11 AM, Katherine Gobin 
 katherine_go...@yahoo.com http://mc/compose?to=katherine_go...@yahoo.com
  wrote:

 Dear R forum

 I have a data.frame

 df = data.frame(dates = c(4/15/2013, 4/14/2013, 4/13/2013,
 4/12/2013), values = c(47, 38, 56, 92))

 I need to to create a vector by repeating the dates as

 Current_date, 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013,
 Current_date, 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, Current_date,
 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013

 i.e. I need to create a new vector as given below which I need to use for
 some other purpose.

 Current_date
 4/15/2013
 4/14/2013
 4/13/2013
 4/12/2013
 Current_date
 4/15/2013
 4/14/2013
 4/13/2013
 4/12/2013
 Current_date
 4/15/2013
 4/14/2013
 4/13/2013
 4/12/2013

 Is it possible to construct such a
  column?

 Regards

 Katherine



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Re: [R] speedometer charts in R

[andrija djurovic]

Hi.
Thanks once more for helping me out.
Beside that you gave me a nice idea and here is the sample code of dashboard
for an R-powered race car that you can't wait to see  :)

library(animation)
library(jpeg)

# Original code by Gaston Sanchez
# http://www.r-bloggers.com/gauge-chart-in-r/
# Modified by Jeff Hemsley

dial.plot - function(label = UseR!, value = 78, dial.radius = 1
  , value.cex = 3, value.color = black
  , label.cex = 3, label.color = black
  , gage.bg.color = white
  , yellowFrom = 75, yellowTo = 90, yellow.slice.color = #FF9900
  , redFrom = 90, redTo = 100, red.slice.color = #DC3912
  , needle.color = red, needle.center.color = black,
needle.center.cex = 1
  , dial.digets.color = grey50
  , heavy.border.color = gray85, thin.border.color = gray20,
minor.ticks.color = gray55, major.ticks.color = gray45) {

  whiteFrom = min(yellowFrom, redFrom) - 2
  whiteTo = max(yellowTo, redTo) + 2

  # function to create a circle
  circle - function(center=c(0,0), radius=1, npoints=100)
  {
r = radius
tt = seq(0, 2*pi, length=npoints)
xx = center[1] + r * cos(tt)
yy = center[1] + r * sin(tt)
return(data.frame(x = xx, y = yy))
  }

  # function to get slices
  slice2xy - function(t, rad)
  {
t2p = -1 * t * pi + 10*pi/8
list(x = rad * cos(t2p), y = rad * sin(t2p))
  }

  # function to get major and minor tick marks
  ticks - function(center=c(0,0), from=0, to=2*pi, radius=0.9, npoints=5)
  {
r = radius
tt = seq(from, to, length=npoints)
xx = center[1] + r * cos(tt)
yy = center[1] + r * sin(tt)
return(data.frame(x = xx, y = yy))
  }

  # external circle (this will be used for the black border)
  border_cir = circle(c(0,0), radius=dial.radius, npoints = 100)

  # open plot
  plot(border_cir$x, border_cir$y, type=n, asp=1, axes=FALSE,
   xlim=c(-1.05,1.05), ylim=c(-1.05,1.05),
   xlab=, ylab=)
  # gray border circle
  external_cir = circle(c(0,0), radius=( dial.radius * 0.97 ), npoints =
100)
# initial gage background
  polygon(external_cir$x, external_cir$y,
  border = gage.bg.color, col = gage.bg.color, lty = NULL)

  # add gray border
  lines(external_cir$x, external_cir$y, col=heavy.border.color, lwd=18)
  # add external border
  lines(border_cir$x, border_cir$y, col=thin.border.color, lwd=2)

  # yellow slice (this will be used for the yellow band)
  yel_ini = (yellowFrom/100) * (12/8)
  yel_fin = (yellowTo/100) * (12/8)
  Syel = slice2xy(seq.int(yel_ini, yel_fin, length.out = 30), rad=
(dial.radius * 0.9) )
  polygon(c(Syel$x, 0), c(Syel$y, 0),
  border = yellow.slice.color, col = yellow.slice.color, lty = NULL)

  # red slice (this will be used for the red band)
  red_ini = (redFrom/100) * (12/8)
  red_fin = (redTo/100) * (12/8)
  Sred = slice2xy(seq.int(red_ini, red_fin, length.out = 30), rad=
(dial.radius * 0.9) )
  polygon(c(Sred$x, 0), c(Sred$y, 0),
  border = red.slice.color, col = red.slice.color, lty = NULL)

  # white slice (this will be used to get the yellow and red bands)
  white_ini = (whiteFrom/100) * (12/8)
  white_fin = (whiteTo/100) * (12/8)
  Swhi = slice2xy(seq.int(white_ini, white_fin, length.out = 30), rad=
(dial.radius * 0.8) )
  polygon(c(Swhi$x, 0), c(Swhi$y, 0),
  border = gage.bg.color, col = gage.bg.color, lty = NULL)



  # calc and plot minor ticks
  minor.tix.out - ticks(c(0,0), from=5*pi/4, to=-pi/4, radius=(
dial.radius * 0.89 ), 21)
  minor.tix.in - ticks(c(0,0), from=5*pi/4, to=-pi/4, radius=( dial.radius
* 0.85 ), 21)
  arrows(x0=minor.tix.out$x, y0=minor.tix.out$y, x1=minor.tix.in$x, y1=
minor.tix.in$y,
 length=0, lwd=2.5, col=minor.ticks.color)

  # coordinates of major ticks (will be plotted as arrows)
  major_ticks_out = ticks(c(0,0), from=5*pi/4, to=-pi/4, radius=(
dial.radius * 0.9 ), 5)
  major_ticks_in = ticks(c(0,0), from=5*pi/4, to=-pi/4, radius=(
dial.radius * 0.77 ), 5)
  arrows(x0=major_ticks_out$x, y0=major_ticks_out$y, col=major.ticks.color,
 x1=major_ticks_in$x, y1=major_ticks_in$y, length=0, lwd=3)

  # calc and plot numbers at major ticks
  dial.numbers - ticks(c(0,0), from=5*pi/4, to=-pi/4, radius=( dial.radius
* 0.70 ), 5)
  dial.lables - c(0, 25, 50, 75, 100)
  text(dial.numbers$x, dial.numbers$y, labels=dial.lables,
col=dial.digets.color, cex=.8)


  # Add dial lables
  text(0, (dial.radius * -0.65), value, cex=value.cex, col=value.color)
  # add label of variable
  text(0, (dial.radius * 0.43), label, cex=label.cex, col=label.color)

  # add needle
  # angle of needle pointing to the specified value
  val = (value/100) * (12/8)
  v = -1 * val * pi + 10*pi/8 # 10/8 becuase we are drawing on only %80 of
the cir
  # x-y coordinates of needle
  needle.length - dial.radius * .67
  needle.end.x = needle.length * cos(v)
  needle.end.y = needle.length * sin(v)

  needle.short.length - dial.radius * .1
  needle.short.end.x = needle.short.length * -cos(v)
  needle.short.end.y = 

Re: [R] speedometer charts in R

[andrija djurovic]

Hi. Thanks for help.
In meanwhile some of contributors already send me the same link with
examples.
I agree with you and I am not trying to drive a race car just to do what
was asked from me :)

Andrija



On Wed, Apr 3, 2013 at 11:24 AM, Barry Rowlingson 
b.rowling...@lancaster.ac.uk wrote:

 On Tue, Apr 2, 2013 at 4:00 PM, R. Michael Weylandt
 michael.weyla...@gmail.com wrote:
  Look at the R GoogleVis package.

  Or read what Hadley W had to say on a similar question first:

 The question would why would you want to?  You are trying to
 understand your data, not driving a race car or aeroplane.  

  -
 http://r.789695.n4.nabble.com/Graphical-output-dials-and-meters-for-a-dashboard-td845090.html

 But maybe you *are* creating a dashboard for an R-powered race car, in
 which case here's an R-native version of the google vis speedometers:

 http://gastonsanchez.wordpress.com/2013/01/10/gauge-chart-in-r/

 Can't wait to see the full source code for your race car:

 require(engine)
 block = engine(cc=2000,cylinders=6)
 require(downforce)
 ...

 Barry


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[R] speedometer charts in R

[andrija djurovic]

Hi useRs.

Does anybody know if there is some function that creates speedometer chart
in R? Or if  anybody has proposals where to start looking and which
functions I can modify in order to create this kind of chart?


Thanks for any help

Andrija

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[R] highlight overlapping region of two densities

[andrija djurovic]

Hi all.

I would like to highlight overlapping regions of two densities and I could
not find a way to do it.

Here is the sample code:

myd - c(2,4,5, 4,3,2,2,3,3,3,2,3,3,4,2,4,3,3,3,2,2.5,
 2, 3,3, 2.3, 3, 3, 2, 3)
myd1 - myd-2
plot(range(density(myd)$x, density(myd1)$x), range(density(myd)$y,
density(myd1)$y), type = n)
lines(density(myd), col=1, lwd=4)
lines(density(myd1), col=2, lwd=4, lty=2)

So, I am trying to highlight the region from 1 to 4, on x axis, taking the
minimum value of corresponding y values.

I am aware of polygon function but don't know how to define correctly x and
y coordinate vectors for this function.

Could someone help me to solve this?

Thank in advance.

Andrija

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Re: [R] highlight overlapping region of two densities

[andrija djurovic]

Hi Pierrick,

thanks for reply. This is the closest solution that I got.

I also found something like this:

dens1 - density(myd)
dens2 - density(myd1)
plot(range(dens1$x, dens2$x), range(dens1$y, dens2$y), type = n)
polygon(dens1$x, dens1$y, col = rgb(1,0,0, .5),lwd=4, lty=2)
polygon(dens2$x, dens2$y, col = rgb(0,0,1, .5),lwd=4)


Your solution is closer to what I want but also solution above can be,
maybe,  useful.

Thank you

Andrija

On Wed, Mar 20, 2013 at 2:20 PM, Pierrick Bruneau pbrun...@gmail.comwrote:

 Hi Andrija,

 As far as I understood, your problem comes from the density() function -
 the domains of myd and myd1 are different, which causes slightly shifted x
 values in the density() output.

 A simple workaround (which you maybe already came up with):

 dens1 - density(myd, from=-1, to=6) # force both densities on same
 domains,
 dens2 - density(myd1, from=-1, to=6) # which facilitates drawing
 plot(range(dens1$x, dens2$x), range(dens1$y, dens2$y), type = n)
 lines(dens1, col=1, lwd=4)
 lines(dens2, col=2, lwd=4, lty=2)
 polygon(dens1$x, pmin(dens1$y, dens2$y), col=grey)

 But doing so causes inelegant distribution tails... Maybe somebody will
 have a more elegant suggestion ?

 Pierrick Bruneau
 CRP Gabriel Lippmann


 On Wed, Mar 20, 2013 at 1:32 PM, andrija djurovic djandr...@gmail.comwrote:

 Hi all.

 I would like to highlight overlapping regions of two densities and I could
 not find a way to do it.

 Here is the sample code:

 myd - c(2,4,5, 4,3,2,2,3,3,3,2,3,3,4,2,4,3,3,3,2,2.5,
  2, 3,3, 2.3, 3, 3, 2, 3)
 myd1 - myd-2
 plot(range(density(myd)$x, density(myd1)$x), range(density(myd)$y,
 density(myd1)$y), type = n)
 lines(density(myd), col=1, lwd=4)
 lines(density(myd1), col=2, lwd=4, lty=2)

 So, I am trying to highlight the region from 1 to 4, on x axis, taking the
 minimum value of corresponding y values.

 I am aware of polygon function but don't know how to define correctly x
 and
 y coordinate vectors for this function.

 Could someone help me to solve this?

 Thank in advance.

 Andrija

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Re: [R] highlight overlapping region of two densities

[andrija djurovic]

Hi Rui.

Thank you very much. I had similar idea like yours but your is more elegant.

Andrija


On Wed, Mar 20, 2013 at 4:03 PM, Rui Barradas ruipbarra...@sapo.pt wrote:

 Hello,

 Try the following.



 d0 - density(myd)
 d1 - density(myd1)

 idx0 - d0$x = 1  d0$x = 4
 idx1 - d1$x = 1  d1$x = 4
 yy - apply(cbind(d0$y[idx0], d1$y[idx1]), 1, min)
 xx - d0$x[idx0]
 xx - c(xx[1], xx, xx[1])
 yy - c(0, yy, 0)

 polygon(xx, yy, col = blue)


 Hope this helps,

 Rui Barradas

 Em 20-03-2013 12:32, andrija djurovic escreveu:

 Hi all.

 I would like to highlight overlapping regions of two densities and I could
 not find a way to do it.

 Here is the sample code:

 myd - c(2,4,5, 4,3,2,2,3,3,3,2,3,3,4,2,4,3,3,**3,2,2.5,
   2, 3,3, 2.3, 3, 3, 2, 3)
 myd1 - myd-2
 plot(range(density(myd)$x, density(myd1)$x), range(density(myd)$y,
 density(myd1)$y), type = n)
 lines(density(myd), col=1, lwd=4)
 lines(density(myd1), col=2, lwd=4, lty=2)

 So, I am trying to highlight the region from 1 to 4, on x axis, taking the
 minimum value of corresponding y values.

 I am aware of polygon function but don't know how to define correctly x
 and
 y coordinate vectors for this function.

 Could someone help me to solve this?

 Thank in advance.

 Andrija

 [[alternative HTML version deleted]]

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

Re: [R] top 10 (n values) for each classes

[andrija djurovic]

Hi,

see function ?head or [.

If DF is you data frame, top n values of DF you can select with head(DF, n)
or DF[1:n, ]...

Andrija

On Sun, Feb 10, 2013 at 11:31 AM, catalin roibu catalinro...@gmail.comwrote:

 Dear R users,
 I have a problem. I don't know how to select the top 10 (n) values for each
 classes.
 Thank you!
 My data is like this:
  row.names proc cls  7271 568,03338 0,5  7270 554,68458 0,5  7269 510,20638
 0,5  7268 485,59969 0,5  7267 421,92852 0,5  7272 410,12101 0,5  3414
 409,71429 0,5  3452 402,78699 0,5  3451 401,28114 0,5  3450 361,80607 0,5
 3413 360,29883 0,5  3449 352,90043 0,5  3453 339,36457 0,5  7266 331,38278
 0,5  3415 327,89374 0,5  3454 324,4335 0,5  3412 291,95631 0,5  4770
 274,12649 0,5  3455 274,09326 0,5  6030 264,72652 0,5  4772 264,17632 0,5
 6031 261,19403 0,5  3411 255,50239 0,5  4773 241,50886 0,5  3447 237,78135
 0,5  3448 232,9554 0,5  4774 232,92985 0,5  7273 229,18868 0,5  6028
 225,24249 0,5  11839 220,61201 0,5  4771 218,94856 0,5  6026 218,7 0,5
 11838 213,44438 0,5  3410 209,35622 0,5  6027 206,68399 0,5  6029 206,10424
 0,5  11837 205,28621 0,5  6025 201,21227 0,5  1042 199,78883 0,5  4775
 197,29348 1  7265 195,07017 1  4776 175,90948 1  10725 175,60525 1  10724
 173,2342 1  3456 172,44641 1  11589 170,2784 1  7190 169,53567 1  11799
 169,19097 1  6024 169,09314 1  3416 168,39827 1  10726 165,25617 1  11840
 164,43032 1  11800 162,7957 1  11590 155,31572 1  1043 154,6438 1  11841
 153,80017 1  7189 152,26293 1  11588 151,07692 1  11798 151,0658 1  3446
 147,0405 1  11836 147,02381 1  7188 142,01954 1  11835 140,95302 1  5994
 138,06452 1  3523 137,98974 1  11797 137,57633 1  11728 137,49488 1  7192
 136,87805 1  11731 135,10162 1  5995 132,41401 1  3096 131,11658 1  7191
 131
 1  3525 128,74521 1  11801 128,57672 1  7264 128,24427 1  3099 128,20339
 1,5
 11377 127,53474 1,5  10948 126,81159 1,5  11834 126,67371 1,5  11378
 126,42474 1,5  11842 126,27561 1,5  4777 126,26175 1,5  10949 124,39545 1,5
 3524 123,64685 1,5  5993 123,39782 1,5  10943 123,27824 1,5  3409 122,84964
 1,5  10942 121,28378 1,5  11730 118,13305 1,5  10702 117,91925 1,5  11796
 117,5239 1,5  4769 117,38123 1,5  11587 116,49326 1,5  7223 114,19009 1,5
 3396 114,16026 1,5  3097 113,99856 1,5  3526 113,97446 1,5  11376 113,58453
 1,5  11591 113,46819 1,5  7163 110,99291 1,5  1044 110,76636 1,5  3457
 110,35221 1,5  10944 109,87002 1,5  11802 109,64554 1,5  3100 109,2207 1,5
 6023 109,01308 1,5  10947 108,45018 1,5  3098 108,20547 1,5  11593 108,0705
 1,5  3458 108,03632 1,5  11592 107,53228 1,5  11729 107,35615 1,5  3095
 107,24396 1,5  7222 106,02489 1,5  10945 105,24515 1,5  3101 104,75914 1,5
 10946 103,92964 1,5  5199 103,81953 1,5  11832 103,6405 1,5  11333
 103,53049
 1,5  7220 102,87019 1,5  11379 102,66542 1,5  3104 102,5464 1,5  10941
 101,4321 1,5  5198 100,62386 1,5  11375 99,731904 1,5  11803 99,08347 1,5
 7187 98,076923 1,5  7221 96,853367 1,5  11356 96,797409 1,5  7167 96,326062
 1,5  11715 95,902613 1,5  11727 95,79491 1,5  7166 95,747126 1,5  7253
 95,56314 2  10703 95,379713 2  3102 94,146825 2  11732 93,961076 2  5992
 93,93088 2  11833 93,902859 2  7224 93,114144 2  3103 93,022202 2  7164
 92,4972 2  12180 91,891892 2  11355 90,585242 2  10679 90,491487 2  7252
 89,664613 2  3105 89,112093 2  5200 88,798701 2  5203 86,954782 2  11354
 86,686174 2  11357 86,655518 2  10940 86,422977 2  11594 86,189863 2  12179
 86,122178 2  7263 84,991274 2  10520 84,532374 2  11733 83,963012 2  7225
 83,766234 2  11586 83,194864 2


 Catalin-Constantin ROIBU
 Forestry engineer, PhD
 Forestry Faculty of Suceava
 Str. Universitatii no. 13, Suceava, 720229, Romania
 office phone +4 0230 52 29 78, ext. 531
 mobile phone   +4 0745 53 18 01
+4 0766 71 76 58
 FAX:+4 0230 52 16 64
 silvic.usv.ro

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Re: [R] Script for conditional sums of vectors

[andrija djurovic]

Here are some examples of data aggregation functions in R:

http://www.slideshare.net/djandrija/data-aggregation-in-r

http://www.psychwire.co.uk/2011/04/data-aggregation-in-r-plyr-sqldf-and-data-table/

Andrija



On Mon, Feb 4, 2013 at 10:29 AM, Benjamin Gillespie gy...@leeds.ac.ukwrote:

 Hi guys,

 I hope you can help me with this (probably) simple query:

 I have a data frame:

 --

 a=c(1,1,1,1,1,1,2,2,2,2,2,2)
 b=c(1,1,1,2,3,4,1,1,2,2,3,4)
 c=c(400,200,300,100,500,300,200,100,500,400,200,100)


 data=data.frame(a=a,b=b,c=c)

 --

 And I would like to get the following output:

 --

 b
 a   1   2   3   4
 1   900 100 500 300
 2   300 900 200 100

 --

 The values in the output represent the sum of values c in data frame
 data, for each a and b combination.

 For example, where a = 1 and b = 1, the output is 400+200+300 = 900.

 Please would anyone be able to provide a script to create my desired
 output?

 Many thanks in advance,

 Ben Gillespie
 Research Postgraduate

 School of Geography
 University of Leeds
 Leeds
 LS2 9JT


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Re: [R] export figure by pdf command

[andrija djurovic]

set.seed(2211)
x - rnorm(100)

#get your working directory
getwd()

#save it as pdf
pdf(hist.pdf)
hist(x)
dev.off()

On Tue, Jan 29, 2013 at 10:22 PM, hp wan huaping@gmail.com wrote:

 Can you implement it using my provided example? I read the user guide
 about dev.copy2pdf
 but I still failed.

 Thanks

 2013/1/30 ilai ke...@math.montana.edu

  ?dev.copy2pdf
 
  On Tue, Jan 29, 2013 at 1:48 PM, hp wan huaping@gmail.com wrote:
 
  Dear R mailing listers,
 
 
  After plotting, I wanna save it as file in pdf format using
  pdf(name.pdf)
  command. It failed, but I can do it by GUI operation (file-save as-pdf).
 
  e.g.
 
  x11()
  hist(x, breaks = 50, probability = FALSE)
  pdf(hist.pdf)
 
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Re: [R] apply -- data.frame

[andrija djurovic]

hi
try with plyr library and function ddply

Andrija

On 30 Aug 2012 12:58, Sam Steingold s...@gnu.org wrote:

 Is there a way for an apply-type function to return a data frame?
 the closest thing I think of is

   foo - as.data.frame(sapply(...))
   names(foo) - c()

 is there a more elegant way?
 Thanks!
 --
 Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X
11.0.11103000
 http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com
 http://honestreporting.com http://ffii.org http://mideasttruth.com
 Lisp: it's here to save your butt.

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

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Re: [R] apply -- data.frame

[andrija djurovic]

I forgot to mention data.table package and also function aggregate as part
of base R functions could be useful here

Andrija

On 30 Aug 2012 13:09, andrija djurovic djandr...@gmail.com wrote:

 hi
 try with plyr library and function ddply

 Andrija

 On 30 Aug 2012 12:58, Sam Steingold s...@gnu.org wrote:
 
  Is there a way for an apply-type function to return a data frame?
  the closest thing I think of is
 
foo - as.data.frame(sapply(...))
names(foo) - c()
 
  is there a more elegant way?
  Thanks!
  --
  Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X
11.0.11103000
  http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com
  http://honestreporting.com http://ffii.org http://mideasttruth.com
  Lisp: it's here to save your butt.
 
  __
  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.

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Re: [R] Getting warning message

[andrija djurovic]

Hi.
I don't know how did you create data frame X but if you check str(X) you
will see that you have one or more factors inside.
Try using stringsAsFactors=FALSE options while creating data frame.

Hope this helps.

Andrija

On Thu, Jul 26, 2012 at 3:23 PM, namit saileshchowd...@gmail.com wrote:

 Hi Friends,


 I  have  a data frame X, and I want to add “%”  “$” in row  4 and 5
 respectively. when I’m trying using below logic, I’m getting warning
 message.
 Can anyone  help me out on this.


 X:
 Summary   G Y   R   T
 Accts582644 0   1226
 AcctCov  230165 0   395
 Cov%  4026  0   32
 UnCov%   60 74  0   68
 EqVol11$MM8.5   10.60   19.1

 Using this logic:

 Z16[5,2:5]-paste($,Z16[5,2:5],sep=)
 Z16[3,2:5]-paste(Z16[3,2:5],%,sep=)
 Z16[4,2:5]-paste(Z16[4,2:5],%,sep=)

 Getting this Warning: In `[-.factor`(`*tmp*`, iseq, value = c(3L, 1L, 2L,
 4L, NA))
 invalid factor level, NAs
 generate


 Final result:

 Summary   G Y   R   T
 Accts   582 644 0   1226
 AcctCov 230 165 0   395
 Cov%40% 26% 0%  32%
 UnCov%  60% 74% 0%  68%
 EqVol11$MM$8.50 $10.60  $0  $19.10


 Thanks in Advance.

 Thanks,
 Namit




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Re: [R] comparing three vectors

[andrija djurovic]

Hi. Try this:

a=c  cb

Andrija

On Tue, Jul 17, 2012 at 9:02 AM, arunkumar akpbond...@gmail.com wrote:

 Hi

 I 've a data

 a=c(10,20,30)
 b=c(100,200,300)
 c=c(50,60,70)

 I want to compare a[1]=c[1]b[1],..

 How to compare for all the records

 -
 Thanks in Advance
 Arun
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Re: [R] Table/Frame - output

[andrija djurovic]

Hi.

You should check function addtable2plot from plotrix package.

Also, here is an example of modified addtable2plot function that I
sometimes use.
This function is probably to far from perfect but, maybe, it will give you
some ideas how to solve your problem:

t2p - function(x=par(usr)[1],y= par(usr)[4],
table, cex=par(cex), bg=par(bg), text.col=par(fg))
{
plot(1,type=n, axes=F, xlab=, ylab=)
column.names-colnames(table)
mwidth-strwidth(M,cex=cex)
cellwidth - max(strwidth(c(column.names,
 as.vector(unlist(table))),cex=cex))+mwidth
cellheight - max(strheight(c(column.names,as.vector(unlist(table))),
  cex=cex))*1.5
tabdim-dim(table)

xleft - x
ytop - y

for(row in 1:tabdim[1])
 {
for(column in 1:tabdim[2])
   {
if(row==1) par(font=4) else
par(font=1)
   rect(xleft+(column-0.9)*cellwidth,
ytop-(row-0.9)*cellheight,
xleft+(column)*cellwidth,
ytop-row*cellheight,
col=bg[row,column],border=NA)
   text(xleft+(column-0.45)*cellwidth,
ytop-(row-0.45)*cellheight,
table[row,column],cex=cex,col=text.col)
}
 }
}
DF - data.frame(x1=1:5, x2=5:1, x3=rep(1,5), x4=rep(5,5))
DF1 - rbind(names(DF), DF)
mat -  matrix(NA,ncol=ncol(DF), nrow=nrow(DF1)+1)
mat[1,] - lightblue
mat[seq(2,nrow(mat),2), ] - gray
mat[seq(3,nrow(mat),2), ] - white

t2p(table=DF1, bg=mat)
t2p(table=DF1, bg=mat, x=0.7, y=1.2)

As you can see t2p function has some arguments that you can play with in
order to get better formatted table.
After plotting a table you can save it as bitmap.

I hope this helps.

Andrija



On Tue, Jul 17, 2012 at 9:11 PM, jcrosbie ja...@crosb.ie wrote:

 I would like to output a nicely formatted data frame to a bitmap.

 Is this possible in R?

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Re: [R] Subset based on multiple values

[andrija djurovic]

Hi. Maybe this:

ct - table(test)
as.numeric(names(ct[ct==max(ct)]))
test[test[,1]%in%as.numeric(names(ct[ct==max(ct)])),,drop=FALSE]

?

Andrija

On Wed, Jul 11, 2012 at 8:33 PM, Amanduh320 aadam...@uwo.ca wrote:

 I'm stuck on a seemingly simple problem. I'm trying to subset the data by
 several numbers and it cuts out half of the rows. Here is the sample code:

 test - as.matrix(c(1,1,1,1,3,3,7,7,7,7))
 Count - tapply(test[,1], test[,1], length)   # count for each value
 spp - unique(test[,1])
 Count1 - as.data.frame(cbind(Count,spp))
 Max - max(Count)
 Count1$sppMax - ifelse(Count1$Count = Max, Count1$spp, 0)  # only keep
 values that =Max
 Count2 - subset(Count1, sppMax  0)#get rid of values that are less
 than Max
 AllMax - unique(Count2$sppMax)
 test2 - subset(test, test[,1] == AllMax)


 this works where there is only one value for AllMax, how to make it work
 when there are multiple?

 Thank you!

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Re: [R] convert a table

[andrija djurovic]

Hi. Here are some examples that could be useful here:

set.seed(1)
x - sample(1:2, 100, rep=T)
y - sample(1:3, 100, rep=T)
TAB - table(x,y)

#check the object class

class(TAB)
apply(TAB, 2, max)
apply(TAB[,2:3], 2, max)
apply(TAB, 1, max)

DF - as.data.frame.matrix(TAB)
class(DF)
sapply(DF, max)

Look at ?class, ?table, ?[ , ?apply, ?as.data.frame.matrix and read
posting guide.

Andrija


On Fri, Jul 6, 2012 at 5:03 PM, Amanduh320 aadam...@uwo.ca wrote:

 I have my data in a table
 table - table(test2$Filename, test2$PREDICT)

 I need to convert this table so it keeps the same structure, but is a
 different format. The current output is count data by Filename and I want
 to
 get the max for each Filename.

 Columns are:
 Filename, 1, 2, 3, 4, 5, 6, 7

 When I try to convert it to a data.frame it reverts to Var1(Filename),
 Var2(1:7), Freq.

 My end goal is to find the max by row (Filename), then do ifelse(xmax, 0,
 max) for each value in columns 2:8

 My problem is that I don't understand what format the table is in.
 Thank you!
 Amanda

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Re: [R] Intersection

[andrija djurovic]

Hi. Try with following functions:

?intersection
?%in%
?[

Perhaps someone will provide you more help if you read and follow posting
guide  
http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html

Andrija

On Tue, Jun 26, 2012 at 5:03 PM, ÷ÁÓÉÌØÞÅÎËÏ áÌÅËÓÁÎÄÒ 
vasilchenko@gmail.com wrote:

 Hello.
 I have a problem with 2 dataframes. There are 2 columns - value and
 dates. These dataframes have different dimension. Some dates coincide.
 And I need to intersect them by dates and have on output two dataframes
 with identical columns dates and new dimension . value have to
 recieve in compliance with dates.
 Regards, Aleksander.

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Re: [R] flagging values without a loop

[andrija djurovic]

Hi. Yes it is possible.
Here is one approach:

DF - read.table(textConnection(
Unit  DayHour Price Flag
afd11/2/20031   1   N
afd11/2/20031   2   N
afd11/2/20031   3   N
afd11/2/20031   4   Y
dcf11/2/20032   2   N
dcf11/2/20032   3   Y
dcf11/2/20032   1   N
dcf11/2/20032   2   N
dcf11/2/20032   3   Y
ghg21/2/20033   1   N
afd11/2/20033   2   N
),header=TRUE)

cbind(DF, flag = ave(DF$Price, DF$Hour, FUN=function(x) ifelse(x==max(x),
1, 0)))

   Unit  Day Hour Price Flag flag
1  afd1 1/2/20031 1N0
2  afd1 1/2/20031 2N0
3  afd1 1/2/20031 3N0
4  afd1 1/2/20031 4Y1
5  dcf1 1/2/20032 2N0
6  dcf1 1/2/20032 3Y1
7  dcf1 1/2/20032 1N0
8  dcf1 1/2/20032 2N0
9  dcf1 1/2/20032 3Y1
10 ghg2 1/2/20033 1N0
11 afd1 1/2/20033 2N1


On Thu, Jun 7, 2012 at 4:17 PM, jcrosbie ja...@crosb.ie wrote:

 For a given hour I want to be able to  add a new column called flag.  The
 flag column will flag the highest price in a given hour.  Is there a way to
 do this without a loop?

 matrix:
 Unit,   Day,Hour,   Price,  Flag
 afd11/2/20031   1   N
 afd11/2/20031   2   N
 afd11/2/20031   3   N
 afd11/2/20031   4   Y
 dcf11/2/20032   2   N
 dcf11/2/20032   3   Y
 dcf11/2/20032   1   N
 dcf11/2/20032   2   N
 dcf11/2/20032   3   Y
 ghg21/2/20033   1   N
 afd11/2/20033   2   N
 .


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Re: [R] Removing Double Quotations After Using Cbind

[andrija djurovic]

Hi. Try:

as.data.frame(cbind(a[,1],a[,2]))

Andrija

On Wed, Jun 6, 2012 at 9:51 PM, Joshua Budman josh.bud...@gmail.com wrote:

 Hi,
 I am trying to process genomics data and the presence of both
 characters and integers in an array is giving issues. The following is
 an example:
   a-array(c(2,2,X,1:3,2:4),dim=c(3,3))
   b-cbind(a[,1],a[,2])
 With the output being:
  [,1] [,2]
 [1,] 2  1
 [2,] 2  2
 [3,] X  3

 Is there any way for me to remove the quotation marks from every
 integer/character in the new array? Or, is there a way to create the
 new array without getting the quotation marks?

 Regards,
 Josh
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Re: [R] how to remove part of the string

[andrija djurovic]

Hi. You can do something like this:

gsub(\\|\\|Leukotriene A4 hydrolase,,LTA4H||Leukotriene A4 hydrolase)

Andrija

On Wed, Jun 6, 2012 at 10:45 PM, Bill Hyman billhym...@yahoo.com wrote:

 Dear all,

 Does any one know how to remove part of the string?

 For example, LTA4H||Leukotriene A4 hydrolase is a gene name plus gene
 description. I hope to remove ||Leukotriene A4 hydrolase. What would be
 the R code to do that using gsub()? Many thanks!

 Bill

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Re: [R] how to remove part of the string

[andrija djurovic]

Hi. Rui already gave you a solution.
Beside that you can, also, use substr function in this concrete example:

substr(LTA4H||Leukotriene A4 hydrolase, 1, 5)

This can be adjusted to rest of your data also, but you haven't provided
enough information.

Andrija

On Wed, Jun 6, 2012 at 11:27 PM, Rui Barradas ruipbarra...@sapo.pt wrote:

 Hello,

 Try

 txt - LTA4H||Leukotriene A4 hydrolase
 pattern - \\|\\|.*$

 gsub(pattern, , txt)


 Hope this helps,

 Rui Barradas

 Em 06-06-2012 21:45, Bill Hyman escreveu:

 Dear all,

 Does any one know how to remove part of the string?

 For example, LTA4H||Leukotriene A4 hydrolase is a gene name plus gene
 description. I hope to remove ||Leukotriene A4 hydrolase. What would be
 the R code to do that using gsub()? Many thanks!

 Bill

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Re: [R] Rolling Sample VAR

[andrija djurovic]

Hi.

rollapply function for zoo package could be a useful here.

library(zoo)
?rollapply

Andrija

On Fri, May 25, 2012 at 5:22 PM, bantex bantexmutat...@hotmail.com wrote:

 hi guys,

 I am using trivariate VAR model to get 10 step ahead orthogonalized impulse
 response functions. I want to use rolling sample analysis on the
 coefficients of the irf  but I have no idea how to do that. I looked
 through
 the forums but I can't seem to find any solutions.

 Any suggestions would be helpful.

 B

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Re: [R] count number of groups

[andrija djurovic]

Hi.
try using table function:
 ID=c(1,1,1,1,2,2,2,2,3,3,3,3)
 table(IF)
ID
1 2 3
4 4 4

Also check ?tapply function

Andrija



On Fri, May 25, 2012 at 5:38 PM, Charles Determan Jr deter...@umn.eduwrote:

 Hello,

 Simple question that I am stuck on and can't seem to find an answer in the
 help files currently.  I have a list which contains repeated ID's.  I would
 like to have R count the number of ID's.  For example:

 ID=c(1,1,1,1,2,2,2,2,3,3,3,3)
 as.data.frame(ID)

 Clearly, there are 3 groups.  How would I have R give me the summary:

 ID
 3

 Many thanks,
 Charles

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Re: [R] odbcConnectExcel() fails to fetch all columns

[andrija djurovic]

Hi.
I use RODBC for importing Excel files quiet often and never got the
similar problem.

Have you tried with sqlQuery?

z - odbcConnectExcel(./BBaselinePtQaires_apr2011.xls)

BQ - sqlQuery(z, select * from [BBaselinePtQaires$])

Andrija

On Fri, Apr 20, 2012 at 11:57 PM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
 Excel is not a database, and the Excel ODBC driver is extremely limited. Put 
 your data in a CSV file or a SQL database (even a Jet database is a step up 
 from Excel).

  http://www.stata.com/support/faqs/data/odbc_excel.html
 ---
 Jeff Newmiller                        The     .       .  Go Live...
 DCN:jdnew...@dcn.davis.ca.us        Basics: ##.#.       ##.#.  Live Go...
                                      Live:   OO#.. Dead: OO#..  Playing
 Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
 /Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k
 ---
 Sent from my phone. Please excuse my brevity.

 Andrew Roberts and...@thinkingbone.org wrote:

Folks,

Is there a parameter somewhere in RODBC that enables more columns to be

retrieved from an Excel worksheet?

# This next bit uses an undocumented call in RODBC
z - odbcConnectExcel(./BBaselinePtQaires_apr2011.xls)
BQ - sqlFetch(z, BBaselinePtQaires)

Gives me:

z RODBC[1]

And

BQ 134 obs. of 59 variables

I have all the rows in the worksheet but only the first 59 out of a
total of 70 columns. I’m in RStudio 0.95.263 using RODBC 1.3-3 and R
version 2.12.2 (2011-02-25).

I'm puzzled - the worksheet seems ok. If the worst comes to the worst I

will have to split the worksheet and cbind to put it back together but
that seems inelegant. The worksheet contains 134 rows, 70 columns and
is
in a spreadsheet that weighs in at 154 KB in total.

Can you help unbaffle me?

Andrew

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Re: [R] Database connectivity

[andrija djurovic]

RODBC

On Mon, Apr 16, 2012 at 11:40 AM, Partha Sinha pnsinh...@gmail.com wrote:
 Dear All
 please let me know how to connect SQL server from R? which all
 packages I will require for this?
 Thanks
 Parth

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[R] Error: R for Windows GUI front-end has stopped working

[andrija djurovic]

Hi all.

I found one situation, on my OS - Windows 7,  where R stops working
with reported error R for Windows GUI front-end has stopped working.

Here is the example:

library(plyr)

DF - data.frame(x=c(1:3, NA, NA), y=factor(sample(1:3,5,rep=T),levels=1:5))

DF[DF$x3, ]

#this works properly
ddply(DF, .(y), nrow, .drop=FALSE)

#this causes the problem
ddply(DF[DF$x3, ], .(y), nrow, .drop=FALSE)


Sometimes R deals with this without closing a program and with reported error:

Error in split_indices(seq_along(splitv), as.integer(splitv), attr(splitv,  :
  INTEGER() can only be applied to a 'integer', not a 'char'

but in the most of cases just stops working.

Does anyone know if this is happening also with other OS or only with Windows 7.

Here is my session info:

 sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: i386-pc-mingw32/i386 (32-bit)

locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base

other attached packages:
[1] plyr_1.7.1

Andrija

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Re: [R] Error: R for Windows GUI front-end has stopped working

[andrija djurovic]

Peter,

thanks for clarification.

Andrija

On Sat, Apr 14, 2012 at 12:11 PM, peter dalgaard pda...@gmail.com wrote:
 Yes, that causes a segmentation fault on OSX too
 --
 ddply(DF[DF$x3, ], .(y), nrow, .drop=FALSE)

  *** caught segfault ***
 address 0x0, cause 'memory not mapped'

 Traceback:
  1: .Call(split_indices, index, group, as.integer(n))
  2: split_indices(seq_along(splitv), as.integer(splitv), attr(splitv,     
 n))
  3: splitter_d(.data, .variables, drop = .drop)
  4: ddply(DF[DF$x  3, ], .(y), nrow, .drop = FALSE)

 Possible actions:
 1: abort (with core dump, if enabled)
 2: normal R exit
 3: exit R without saving workspace
 4: exit R saving workspace
 
 The problem is in a C function from the plyr package, so you need to take it 
 up the maintainer.

 -pd


 On Apr 14, 2012, at 11:28 , andrija djurovic wrote:

 Hi all.

 I found one situation, on my OS - Windows 7,  where R stops working
 with reported error R for Windows GUI front-end has stopped working.

 Here is the example:

 library(plyr)

 DF - data.frame(x=c(1:3, NA, NA), y=factor(sample(1:3,5,rep=T),levels=1:5))

 DF[DF$x3, ]

 #this works properly
 ddply(DF, .(y), nrow, .drop=FALSE)

 #this causes the problem
 ddply(DF[DF$x3, ], .(y), nrow, .drop=FALSE)


 Sometimes R deals with this without closing a program and with reported 
 error:

 Error in split_indices(seq_along(splitv), as.integer(splitv), attr(splitv,  :
  INTEGER() can only be applied to a 'integer', not a 'char'

 but in the most of cases just stops working.

 Does anyone know if this is happening also with other OS or only with 
 Windows 7.

 Here is my session info:

 sessionInfo()
 R version 2.15.0 (2012-03-30)
 Platform: i386-pc-mingw32/i386 (32-bit)

 locale:
 [1] LC_COLLATE=English_United States.1252
 [2] LC_CTYPE=English_United States.1252
 [3] LC_MONETARY=English_United States.1252
 [4] LC_NUMERIC=C
 [5] LC_TIME=English_United States.1252

 attached base packages:
 [1] stats     graphics  grDevices utils     datasets  methods   base

 other attached packages:
 [1] plyr_1.7.1

 Andrija

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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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 --
 Peter Dalgaard, Professor,
 Center for Statistics, Copenhagen Business School
 Solbjerg Plads 3, 2000 Frederiksberg, Denmark
 Phone: (+45)38153501
 Email: pd@cbs.dk  Priv: pda...@gmail.com









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Re: [R] R equivalent for SQL query

[andrija djurovic]

Hi,

here are some solutions:

DF - read.table(textConnection(
A   B   C
1   1   3
1   1   4
1   1   5
1   2   6
1   2   7
1   3   8 ), header=TRUE)

#using sqldf package
library(sqldf)
sqldf(select A, B, count(*)
  from DF
  group by A, B
  order by count(*) desc)

#using function table
as.data.frame(table(DF$A, DF$B))

As you can see, you can use sqldf package for performing sql queries
on R data frames.

Andrija



On Tue, Apr 3, 2012 at 8:26 PM, Steven Raemaekers s.raemaek...@sig.eu wrote:
 Hi,

 I have a query which I would like to translate into R, but I do not know how 
 to do it in an easy way.
 Assume a data frame has columns A, B and C:

 A       B       C
 1       1       3
 1       1       4
 1       1       5
 1       2       6
 1       2       7
 1       3       8

 The query is as follows:

 select A, B, count(*)
 from data.frame
 group by A, B
 order by count(*) desc

 How do I translate this into R statements in such way that the result is a 
 data frame structured as follows:

 A       B       count(*)
 1       1       3
 1       2       2
 1       3       1

 Thanks,

 Steven
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Re: [R] subset problem

[andrija djurovic]

Hi. You can try this:

df[type%in%type_list, ]

You can  also use sqldf package and subset data frames usign sql statements:

library(sqldf)

df - data.frame(x1=1:10, type=10:1)
type_list - data.frame(index=seq(1,10,by=2))

sqldf(select df.*
  from df
  where df.type in (select * from type_list))


Andrija



On Thu, Mar 29, 2012 at 7:25 AM, reeyarn reey...@gmail.com wrote:
 Hi,

 If my data frame df has a index type, I want to get a subset such
 that the type belongs to a type_list;
 using sql, I want
  SELECT name, type FROM df
    WHERE type in type_list;

 Now in R I have to write a loop like
  mysubset- df [ df$type == type_list[1], ]
  for (type1 in type_list[ 2: length (type_list) ] ) {
    mysubset-cbind (mysubset, df [ df$type == type1, ])
  }

 What is the natural way of doing this in R? Is it possible to use
 subset() to attain this?
 Thanks!

 Best,
 Reeyarn

 On Fri, Dec 3, 2010 at 11:26 AM,  William Dunlap [hidden email] wrote:
 HI,
 I have a dataframe like this:
 name    type
 A          t1
 B           t2
 C          t1
 D           t4
 E           t3
 F            t2
 how can I have a sub dataframe based with the column type like this:
 (for type = t1)
 name    type
 A          t1
 C          t1
 D           t1

 Hi,

 Let's say your data.frame is stored in a variable named df:

 R subset(df, type == 't1')

 Read the help files:

 R ?subset

 Also take a look at ?split

 -steve

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Re: [R] function for filtering and deleting vector entries

[andrija djurovic]

Hi.

Maybe this will help you:

set.seed(1)
temp - 1:100
v - rnorm(100)

temp[temp16 | temp38]
v[temp16 | temp38]

Andrija

On Wed, Feb 29, 2012 at 7:09 PM, babyluck madr...@gmx.ch wrote:
 Dear helpers

 I have two data sets saved as vectors (temperature and velocity). Now I need
 to take out a span of temperature and its corresponding velocity in the
 other vector. How can I achieve that?

 I tried to write a function,which takes a vector entry and then decides
 wether to delete the temperature entry or not and  simultaneously doing so
 with same entry in the velocity vector..
 But somehow it's not working...could somebody please help me?
 Thanks a lot..


 norm = function(Temp,v){
        for (i in 1:length(Temp)){

                if (Temp[i]=16 || Temp[i] = 38.5)
                 {Temp[-i];v[-i]}

                return(Temp,v)
        }
 }


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Re: [R] How to replace the values in a column

[andrija djurovic]

Hi.

You can do something like this:

df - read.table(textConnection(
GenRep
A_1  1
A_1   2
A_2 1
A_2 2
B_1   1
B_1   2
B_3   1
B_3   2
OP1_1   1
OP1_1   2
OP1_5   1
OP1_5   2),header=TRUE,stringsAsFactors=FALSE)

str(df)

#adding a new column
cbind(df,ncol =
ifelse(df$Gen==A_1,Wynda,ifelse(df$Gen==B_3,Wynda,df$Gen)))

#replacing exisiting values
df$Gen[df$Gen==A_1 | df$Gen==B_3] - Wynda
df


Andrija




On Wed, Feb 29, 2012 at 7:06 AM, ilai ke...@math.montana.edu wrote:
 Hannah,
 If Gen is a factor you can simply build the new factor on top of it:

 dataframe$Gen- factor( c('Wynda' , 'A_2' , 'B_1' , 'Wynda' , 'Wynda'
 , 'OP1_5')[Gen] )

 Just make sure the replacement labels are in the same order as levels(Gen).

 Cheers

 On Tue, Feb 28, 2012 at 8:39 PM, hannahmaohuang
 hannahmaohu...@gmail.com wrote:
 Dear All,
 I've been searching relevant topics about replacing values, none seemed to
 be applicable to me...

 I have a file with many many varieties, and want to replace some of them
 into different names.
 I tried various of ways, still don't know how to do that most efficiently..
 Here is part of the example data:


 Gen    Rep
 A_1      1
 A_1       2
 A_2     1
 A_2     2
 B_1       1
 B_1       2
 B_3       1
 B_3       2
 OP1_1   1
 OP1_1   2
 OP1_5   1
 OP1_5   2

 For example, I want to replace  A_1,  B_3,  OP1_1 into different name
 Wynda

 So that the expected file should become:

 Gen          Rep
 Wynda      1
 Wynda        2
 A_2           1
 A_2           2
 B_1              1
 B_1             2
 Wynda        1
 Wynda        2
 Wynda        1
 Wynda        2
 OP1_5          1
 OP1_5         2


 I have created a link file, which contains two rows, translating which Gen
 correlating to which Name. Not sure if this file helps or not, example as
 below:

 Column1(Gen)        Column2(Name)
 A_1                               Wynda
 A_2                                 A_2
 B_1                                 B_1
 B_3                             Wynda
 OP1_1                       Wynda
 OP1_5                        OP1_5


 Though I can replace one by one in excel, since there are too many files and
 too many reps, it'll be very time-consuming also easy to make mistakes.

 Please give me any guidance or help in terms of finish this with R.

 Thanks so much !

 Sincerely
 Hannah



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Re: [R] Making a point-sampling matrix

[andrija djurovic]

Hi.
If I understood you correctly here is one approach to your solution:

 k - 5
 mat - diag(1,nrow=k,ncol=k)
 set.seed(10)
 samp - mat[sample(1:k,3,rep=FALSE),]
 samp
 [,1] [,2] [,3] [,4] [,5]
[1,]00100
[2,]01000
[3,]00010
 #check
 apply(samp,1,sum)
[1] 1 1 1
 apply(samp,2,sum)
[1] 0 1 1 1 0

I hope this helps
Andrija


On Sat, Feb 4, 2012 at 4:27 PM, Alaios ala...@yahoo.com wrote:
 Dear all,
 I would like to create a k,n matrix which that will include



 - k random elements with value 1
 -all rest zeros

 -one single 1 in each row

 - each column will have maximum one 1


 so far I have a so stupid function that with nested while tries to fit k 
 numbers randomly in a row by row fashion, checking also if the c column rule 
 is violated. In that case there is also another loop that will try to find a 
 new solution.

 That way it takes ages, as it is natural to complete a matrix of 300,512 as 
 the new ones conflict all the time.

 I am trying to find an easy way to make this algorithm way simpler that will 
 help me reduce the time needed.
 I was thinking then if R can help me by shuffling number from 1  to 300 
 randomly, so to avoid the first loop that does that goes row by row.. and 
 then find an easy way to check if the column rule is violated.

 Do you think that you can spend some time help me?

 Have a nice weekend

 Regards
 Alex

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Re: [R] mode of frequency distribution table

[andrija djurovic]

Hi. You can do something like this:
#find the most frequent values of x
 t - table(x)
 t[t==max(t)]
5
8
#sort table t based on frequencies
 t[order(as.numeric(t),decreasing = TRUE)]
x
  5   6   4  17   1   2  13  30 100 300
  8   5   4   2   1   1   1   1   1   1
#extract any range from sorted table
 t[order(as.numeric(t),decreasing = TRUE)][1:3]
x
5 6 4
8 5 4

I hope this helps.

Andrija


On Sun, Jan 8, 2012 at 1:48 PM, Mary Kindall mary.kind...@gmail.com wrote:
 In a frequency distribution table (bell shaped), how can we find the most
 frequent range?
 for example:

  x = c(1,2, 4,4,4,4, 5,5,5,6,6,5,5,5,5,5,6,6,6,13, 17,17,30,100,300)

 barplot(table(x))


 In the code above, which function do we use to find that the most
 frequent value range from 4 to 6.

 Thanks.



 --
 -
 Mary Kindall
 Yorktown Heights, NY
 USA

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Re: [R] Combining characters

[andrija djurovic]

Hi. You can use expand.grid here

expand.grid(x,y,z)

Andrija

On Wed, Jan 4, 2012 at 5:32 PM, jeremy jeremynamer...@gmail.com wrote:
 Hi all,

 I'm trying to combine exhaustively several character arrays in R like:
 x=c(one,two,three)
 y=c(yellow,blue,green)
 z=c(apple,cheese)

 in order to get concatenation of

 x[1] y[1] z[1]  (one yellow apple)
 x[1] y[1] z[2] (one yellow cheese)
 x[1] y[2] z[1](one blue apple)
 ...
 x[length(x)] y[length(y)] z[length(z)]  (three green cheese)

 Anyone has a solution ?
 Thank in advance

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Re: [R] Creating and assigning variable names in loop

[andrija djurovic]

Hi. You assign two times different values to variable label. Try this
and i hope you will notice a mistake:

i=1
label - paste(score, i, sep=_)
label
assign(label, x1+(x2*i) )
label

Instead of this you can do something like:
rm(list=ls())
n = 10
set.seed(1)
x1 = rnorm(n,0)
x2 = rnorm(n,0)
samp_data - data.frame(x1,x2)

for( i in 1:3) {
   samp_data - cbind(samp_data, x1+(x2*i))
   colnames(samp_data)[i+2] - paste(score, i, sep=_)
}
head(samp_data)

Andrija


On Wed, Dec 21, 2011 at 11:01 AM, aajit75 aaji...@yahoo.co.in wrote:
 Hello List

 I am trying to create and assign variable names in loop, but not able to get
 expected variable names.

 Here is the sample code

 n = 10
 set.seed(1)
 x1 = rnorm(n,0)
 x2 = rnorm(n,0)
 samp_data - data.frame(x1,x2)

 for( i in 1:3) {
    label - paste(score, i, sep=_)
    assign(label, value =x1+(x2*i) )
    samp_data - cbind(samp_data, get(label))
 }

 head(samp_data)
          x1                    x2          get(label)     get(label)
 get(label)
 1 -0.6264538  1.51178117  0.8853274  2.3971085  3.9088897
 2  0.1836433  0.38984324  0.5734866  0.9633298  1.3531730
 3 -0.8356286 -0.62124058 -1.4568692 -2.0781098 -2.6993504
 4  1.5952808 -2.21469989 -0.6194191 -2.8341190 -5.0488189
 5  0.3295078  1.12493092  1.4544387  2.5793696  3.7043005
 6 -0.8204684 -0.04493361 -0.8654020 -0.9103356 -0.9552692

 I am expecting new variables to be created in the samp_data are
 score_1   score_2   score_3, instead get(label)   get(label)   get(label)
 Where am I going wrong?

 Thanks in advance
 ~Ajit

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Re: [R] matching using which

[andrija djurovic]

Hi. Here is one approach:

if (length(b)0) data[,-b] else data

Andrija

On Thu, Dec 8, 2011 at 1:25 PM, Vikram Bahure economics.vik...@gmail.comwrote:

 Dear R users,

 I have a very simple query.

 I am using the following command, which should give me row no. for the
 matching colnames. It works well for matching the colnames but if there is
 no column matching it gives me outcome as integer(0) which I am not able to
 use in further calculation. It would be very helpful to have some insight.
 *
 *
 * b - which(colnames(data)%in%c(X.,X))*
 * b*
 *integer(0)*
 * b0*
 *logical(0)*
 *b- data[,-b]  # does not work properly*
 *
 *
 Regards
 Vikram

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[R] sum of deviations from the weighted mean

[andrija djurovic]

Hi all. I tried to calculate sum of deviations from the weighted mean and i
didn't get what i expected - 0. Here is an example:

 wt - c(10,25,38,22,5)
 x - 6:10
 wm - weighted.mean(x,wt)
 (x-wm)*wt
[1] -18.70 -21.75   4.94  24.86  10.65
 sum((x-wm)*wt)
[1] -1.24345e-14

With simple mean I got 0:
 sum(x-mean(x))
[1] 0

Could someone explain me why we didn't get 0, with weighted mean, as it is
expected?
Thanks in advance

Andrija

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Re: [R] sum of deviations from the weighted mean

[andrija djurovic]

Thank and sorry i should check this before asking the question.

Andrija

On Thu, Dec 8, 2011 at 3:20 PM, R. Michael Weylandt 
michael.weyla...@gmail.com michael.weyla...@gmail.com wrote:

 R FAQ 7.31 I'd imagine. (Floating point arithmetic and limitations thereof)

 Michael

 On Dec 8, 2011, at 9:15 AM, andrija djurovic djandr...@gmail.com wrote:

  Hi all. I tried to calculate sum of deviations from the weighted mean
 and i
  didn't get what i expected - 0. Here is an example:
 
  wt - c(10,25,38,22,5)
  x - 6:10
  wm - weighted.mean(x,wt)
  (x-wm)*wt
  [1] -18.70 -21.75   4.94  24.86  10.65
  sum((x-wm)*wt)
  [1] -1.24345e-14
 
  With simple mean I got 0:
  sum(x-mean(x))
  [1] 0
 
  Could someone explain me why we didn't get 0, with weighted mean, as it
 is
  expected?
  Thanks in advance
 
  Andrija
 
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Re: [R] Paste() - Get all possible combinations from multiple vectors

[andrija djurovic]

Hi. Maybe this will help you:

expand.grid(x,y,z)
apply(expand.grid(x,y,z),1, function (x) paste(x[1], x[2], x[3], sep=))

Andrija

On Tue, Dec 6, 2011 at 1:53 PM, Gaj Stan (BIGCAT) 
stan@maastrichtuniversity.nl wrote:

 Hello fellow R-users,

 Given are three vectors and the outcome would be all possible combinations
 in combination with the paste() function.
 For example:

 x - c(1:3)
 y - letters[1:3]
 z - LETTERS[1:3]

 My result would thus be 18 names based on all possible combinations
 between these vectors:
 1 a A 1 a B, 1 a C, 1 b A, 1 b B, 1 b C, 1 c A, 1 c B, 1
 c C, etc.

 To solve the issue above I would use:

 paste(rep(x, each=9), rep(y, each=6), z)

 This is a very straightforward example, but when I have vectors of
 different sizes, it will be much more difficult (for me) to create this. Is
 there an easier way to do this?

 Many thanks in advance,

  -- Stan

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[R] RODBC connect to Excel (64-bit Windows 7)

[andrija djurovic]

Hi to all.
 I have a problem to connect to an Excel database using RODBC.
Namely, I am using 64-bit R 2.14.0, under Windows 7 and I tried following:
library(RODBC)
 channel - odbcConnectExcel(results.xlsx)
Error in odbcConnectExcel(results.xlsx) :
  odbcConnectExcel is only usable with 32-bit Windows   # ok this is
clear why it doesn't work
 channel - odbcConnectExcel2007(results.xlsx) # this was
one of proposals from old R help posts, but it doesn't work for me
Warning messages:
1: In odbcDriverConnect(con, tabQuote = c([, ]), ...) :
  [RODBC] ERROR: state IM002, code 0, message [Microsoft][ODBC Driver
Manager] Data source name not found and no default driver specified
2: In odbcDriverConnect(con, tabQuote = c([, ]), ...) :
  ODBC connection failed

After that as it suggested in R data import/export manual I
installed AccessDatabaseEngine.exe. Also I used ODBC in
c:\Windows\SysWOW64\odbcad32 to create data source called result. I check
if it is created:
 odbcDataSources()
 dBASE Files
   Microsoft Access dBASE Driver (*.dbf, *.ndx, *.mdx)
  MS Access Database
  Microsoft Access Driver (*.mdb, *.accdb)
  Excel file
Microsoft Excel Driver (*.xls, *.xlsx, *.xlsm, *.xlsb)
* results*
Microsoft Excel Driver (*.xls, *.xlsx, *.xlsm, *.xlsb)

and tried again but it doesn't work:

 channel - odbcDriverConnect()
Warning messages:
1: In odbcDriverConnect() :
  [RODBC] ERROR: state IM014, code 0, message [Microsoft][ODBC Driver
Manager] The specified DSN contains an architecture mismatch between the
Driver and Application
2: In odbcDriverConnect() : ODBC connection failed


Could someone guide me what should I try next to fix a problem? Is this
some problem with Drivers or not?

Thanks in advance

Andrija

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Re: [R] RODBC connect to Excel (64-bit Windows 7)

[andrija djurovic]

Hi. Not specifically, but I used it before with 32-bit R under Windows XP
and it worked just fine so i though to keep using it.Anyway I will lookt at
XLConnect.
Thank you for suggestion.
Andrija

On Sun, Dec 4, 2011 at 7:41 PM, R. Michael Weylandt 
michael.weyla...@gmail.com michael.weyla...@gmail.com wrote:

 Do you need to use RODBC specifically? I've been using XLConnect quite a
 lot recently and have been quite pleased with it.

 Michael

 On Dec 4, 2011, at 9:40 AM, andrija djurovic djandr...@gmail.com wrote:

  Hi to all.
  I have a problem to connect to an Excel database using RODBC.
  Namely, I am using 64-bit R 2.14.0, under Windows 7 and I tried
 following:
  library(RODBC)
  channel - odbcConnectExcel(results.xlsx)
  Error in odbcConnectExcel(results.xlsx) :
   odbcConnectExcel is only usable with 32-bit Windows   # ok this is
  clear why it doesn't work
  channel - odbcConnectExcel2007(results.xlsx) # this was
  one of proposals from old R help posts, but it doesn't work for me
  Warning messages:
  1: In odbcDriverConnect(con, tabQuote = c([, ]), ...) :
   [RODBC] ERROR: state IM002, code 0, message [Microsoft][ODBC Driver
  Manager] Data source name not found and no default driver specified
  2: In odbcDriverConnect(con, tabQuote = c([, ]), ...) :
   ODBC connection failed
 
  After that as it suggested in R data import/export manual I
  installed AccessDatabaseEngine.exe. Also I used ODBC in
  c:\Windows\SysWOW64\odbcad32 to create data source called result. I check
  if it is created:
  odbcDataSources()
  dBASE Files
Microsoft Access dBASE Driver (*.dbf, *.ndx, *.mdx)
   MS Access Database
   Microsoft Access Driver (*.mdb, *.accdb)
   Excel file
  Microsoft Excel Driver (*.xls, *.xlsx, *.xlsm, *.xlsb)
 * results*
  Microsoft Excel Driver (*.xls, *.xlsx, *.xlsm, *.xlsb)
 
  and tried again but it doesn't work:
 
  channel - odbcDriverConnect()
  Warning messages:
  1: In odbcDriverConnect() :
   [RODBC] ERROR: state IM014, code 0, message [Microsoft][ODBC Driver
  Manager] The specified DSN contains an architecture mismatch between the
  Driver and Application
  2: In odbcDriverConnect() : ODBC connection failed
 
 
  Could someone guide me what should I try next to fix a problem? Is this
  some problem with Drivers or not?
 
  Thanks in advance
 
  Andrija
 
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Re: [R] equating approximate values

[andrija djurovic]

Hi. Maybe this can help you (you can try additionally to incorporate
threshold):

 set.seed(1)
 x-rnorm(10,10,1)
 values - sample(1:10,10)  #values that we are looking for
 mat - matrix(c(x,values),ncol=2)

 closest-function(x,values)#function is an example from The R book
(Crawley)
+ {
+ x[which(abs(x-values)==min(abs(x-values)))]
+ }

 apply(mat[,2,drop=FALSE],1,function(y) closest(mat[,1],y))
 [1] 10.183643  9.164371  9.164371  9.164371  9.164371  9.164371  9.164371
 9.164371
 [9]  9.164371  9.164371


On Sun, Dec 4, 2011 at 7:00 PM, vamshi999 vamshi...@gmail.com wrote:

 Hello List,

 I am having trouble finding the command for my problem.

 I have two arrays x and y. now i would like to compare the values of x and
 y
 and then get the index of x which is exactly or approximately equal(+/-
 some
 value ) to the values in y.
  x - runif(100,min=0,max=5)
  y - runif(10,min=0,max=5)


 the threshold value(+/-) value can vary. for this example lets take it to
 be
 .5

 I know the regular method of doing this by writing different if and for
 loops. But i have very big dataframe the computation time is very high for
 this method. can anyone please tell me if there any functions to do this.

 thank you for your help.

 --
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 http://r.789695.n4.nabble.com/equating-approximate-values-tp4157551p4157551.html
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Re: [R] Iteration in R

[andrija djurovic]

Hi. One approach is using replicate. See ?replicate:

replicate(3,rnorm(100,1,2))

Andrija

On Sat, Dec 3, 2011 at 7:10 PM, Martin Zonyrah martin20...@yahoo.comwrote:

 Hi,
 I need help. I am trying to iterate this command  x - rnorm(100, 1.0,
 2.0) one hundred times in R but I don't seem to have a clue.
 Can anyone help?
 Your help is very much appreciated.

 Martin

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Re: [R] Iteration in R

[andrija djurovic]

Hi Brad. Maybe something like this:

lapply(rep(1,6), function(x) rnorm(10,0,1))

Andrija

On Sat, Dec 3, 2011 at 8:21 PM, B77S bps0...@auburn.edu wrote:

 Hi Michael,
 How would you do this with lapply to return a list?
 I can't seem to get that to work (I haven't used these much and am trying
 to
 learn).
 Thanks
 Brad


 Michael Weylandt wrote
 
  ? replicate
 
  or a for loop
 
  or do all one hundred simulations at once
 
  x - matrix(rnorm(100^2, 1, 2), 100)
 
  It's going to depend on what you want to do with the numbers.
 
  Michael
 
  On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah lt;martin2005z@gt;
 wrote:
  Hi,
  I need help. I am trying to iterate this command  x - rnorm(100, 1.0,
  2.0) one hundred times in R but I don't seem to have a clue.
  Can anyone help?
  Your help is very much appreciated.
 
  Martin
 
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Re: [R] Transforming a string into a command

[andrija djurovic]

?eval
s- expression(log(a+b))
a-1
b-2
eval(s)

Andrija

On Sun, Nov 27, 2011 at 11:16 AM, Victor vdem...@gmail.com wrote:

 I would like to make a string executable, e.g,

 s- ln(a+b)
 a-1
 b-2

  execute string s to obtain ln(a+b) 

 How can I make it?

 Ciao fron Rome
 Vittorio

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Re: [R] Simplifying my code

[andrija djurovic]

Hi.
You haven't specified object to store results.

Try something like:

dat = c() #or dat = list() or matrix with specified ncol and nrow
for(i in 1:20){

dat[i]=complete(dat.mice,[i]

}

On Sun, Nov 27, 2011 at 4:02 PM, Christopher Desjardins desja...@umn.eduwrote:

 Hi,
 I have a pretty simple problem. Here is the code:

 dat1=complete(dat.mice,1)

 dat2=complete(dat.mice,2)

 dat3=complete(dat.mice,3)

 dat4=complete(dat.mice,4)

 dat5=complete(dat.mice,5)

 dat6=complete(dat.mice,6)

 dat7=complete(dat.mice,7)

 dat8=complete(dat.mice,8)

 dat9=complete(dat.mice,9)

 dat10=complete(dat.mice,10)

 dat11=complete(dat.mice,11)

 dat12=complete(dat.mice,12)

 dat13=complete(dat.mice,13)

 dat14=complete(dat.mice,14)

 dat15=complete(dat.mice,15)

 dat16=complete(dat.mice,16)

 dat17=complete(dat.mice,17)

 dat18=complete(dat.mice,18)

 dat19=complete(dat.mice,19)

 dat20=complete(dat.mice,20)


 I would like to simplify this into a for loop. I thought this would work:

 for(i in 1:20){

 dat[i]=complete(dat.mice,[i]

 }

 But it doesn't. Any tips?

 Chris

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Re: [R] Is there way to add a new row to a data frame in a specific location

[andrija djurovic]

Hi.
May be this:

df = data.frame( A=c('a','b','c'), B=c(1,2,3), C=c(10,20,30),
stringsAsFactors=FALSE)

newrow = c('X', 100, 200)

rbind(df,newrow)[c(1,4,2,3),]

Andrija


On Thu, Nov 24, 2011 at. 6:05 PM, Sammy Zee szee2...@gmail.com wrote:

 Is there easy way (without copying the existing rows to a temporary
 location and copying back) to add a new row to a specific index location in
 an existing data frame?

 Example

 df = data.frame( A= c('a','b','c'), B=c(1,2,3), C=(10,20,30))

 newrow = c('X', 100, 200)

 I want to add the newrow as the second row to the data frame df

 Please suggest a solution that is efficient for a data frame that can have
 millions of rows, and I want to add a new row in any given index location
 of the data frame.

 Thanks,
 Sammy

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Re: [R] Correlation Matrix in R

[andrija djurovic]

Hi,
one solution is to use sink. Check ?sink to see explanation and following
example.

sink(sink-examp.txt)
i - 1:10
outer(i, i, *)
sink()

Andrija

On Tue, Nov 1, 2011 at 10:43 AM, AlexC alexandre.chaus...@unil.ch wrote:

 Hello,

 Thank you for your replies.  I cannot run the function rcor.test even when
 having loaded package ltm.  Perhaps it has to do with the fact that I am
 using the latest version of R and this package wasn't created under that
 version

 The function corr.test in package psych works fine.  Is there anyway to
 export the results in a txt or csv file?  Since it isn't in a data frame
 format it cannot simply be exported using write.table

 Alexandre

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Re: [R] How to get Quartiles when data contains both numeric variables and factors

[andrija djurovic]

Hi,
you are almost there:

sapply(x, function(x) quantile(as.numeric(x), c(0.01, 0.99)))
  x1  x2   x3x4 x5 x6
1%  0.0351777 0.007628441 0.225533 0.4459064  1  1
99% 0.9938919 0.964901423 1.826894 3.6226944  3  2

Andrija

On Mon, Oct 31, 2011 at 2:09 PM, aajit75 aaji...@yahoo.co.in wrote:

 When data contains both factor and numeric variables, how to get quartiles
 for all numeric variables?
 n - 100
 x1 - runif(n)
 x2 - runif(n)
 x3 - x1 + x2 + runif(n)/10
 x4 - x1 + x2 + x3 + runif(n)/10
 x5 - factor(sample(c('a','b','c'),n,replace=TRUE))
 x6 - factor(1*(x5=='a' | x5=='c'))
 data1 - cbind(x1,x2,x3,x4,x5,x6)
 data - data.frame(data1)

 data - within(data,{x5 - factor(x5)})
 x - data

 qs - sapply(x, function(x) quantile(x, c(0.01, 0.99)))

 I get an error: Error in quantile.default(x, c(min_pct, max_pct)) : factors
 are not allowed

 Thanks for the help.


 --
 View this message in context:
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 Sent from the R help mailing list archive at Nabble.com.

 __
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Re: [R] How to get Quartiles when data contains both numeric variables and factors

[andrija djurovic]

Or this:
 str(x)
'data.frame':   100 obs. of  6 variables:
 $ x1: num  0.4548 0.0352 0.6353 0.6017 0.8588 ...
 $ x2: num  0.849 0.335 0.986 0.617 0.212 ...
 $ x3: num  1.35 0.46 1.67 1.23 1.14 ...
 $ x4: num  2.67 0.91 3.31 2.48 2.28 ...
 $ x5: Factor w/ 3 levels 1,2,3: 3 1 3 3 3 1 2 3 3 1 ...
 $ x6: num  2 2 2 2 2 2 1 2 2 2 ...

 sapply(x[,sapply(x,is.numeric)], function(x) quantile(as.numeric(x),
c(0.01, 0.99)))
   x1  x2   x3x4 x6
1%  0.0351777 0.007628441 0.225533 0.4459064  1
99% 0.9938919 0.964901423 1.826894 3.6226944  2

Hope this helps.

Andrija

On Mon, Oct 31, 2011 at 2:09 PM, aajit75 aaji...@yahoo.co.in wrote:

 When data contains both factor and numeric variables, how to get quartiles
 for all numeric variables?
 n - 100
 x1 - runif(n)
 x2 - runif(n)
 x3 - x1 + x2 + runif(n)/10
 x4 - x1 + x2 + x3 + runif(n)/10
 x5 - factor(sample(c('a','b','c'),n,replace=TRUE))
 x6 - factor(1*(x5=='a' | x5=='c'))
 data1 - cbind(x1,x2,x3,x4,x5,x6)
 data - data.frame(data1)

 data - within(data,{x5 - factor(x5)})
 x - data

 qs - sapply(x, function(x) quantile(x, c(0.01, 0.99)))

 I get an error: Error in quantile.default(x, c(min_pct, max_pct)) : factors
 are not allowed

 Thanks for the help.


 --
 View this message in context:
 http://r.789695.n4.nabble.com/How-to-get-Quartiles-when-data-contains-both-numeric-variables-and-factors-tp3955750p3955750.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 R-help@r-project.org mailing list
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Re: [R] why the a[-indx] does not work?

[andrija djurovic]

 as.logical(c(1,0,1,1))
[1]  TRUE FALSE  TRUE  TRUE

?as.logical



On Sun, Oct 30, 2011 at 8:50 PM, Alaios ala...@yahoo.com wrote:

 probably you mean




 For ‘[’-indexing only: ‘i’, ‘j’, ‘...’ can be logical
 vectors, indicating elements/slices to select.  Such vectors
 are recycled if necessary to match the corresponding extent.
 ‘i’, ‘j’, ‘...’ can also be negative integers, indicating
 elements/slices to leave out of the selection.



 How can i convert the positives to TRUE and zeros and FALSE?



 
 From: William Dunlap wdun...@tibco.com

 Sent: Sunday, October 30, 2011 9:17 PM
 Subject: RE: [R] why the a[-indx] does not work?

a[overLoadTesT==0]
   [1]  2  4  5  6  7  8  9 10
 Look into help('[') or help('Subscript') to see
 how integer and logical (Boolean) subscripts differ.

 Bill Dunlap
 Spotfire, TIBCO Software
 wdunlap tibco.com

  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On Behalf Of Alaios
  Sent: Sunday, October 30, 2011 11:52 AM
  To: R-help@r-project.org
  Subject: [R] why the a[-indx] does not work?
 
  Dear all,
 
  Could you please explain me why
 
   OverloadsTesT
  [1] 1 0 1 0 0 0 0 0 0 0
   a-matrix(data=seq(1,10),nrow=10)
   a
  [,1]
  [1,]1
  [2,]2
  [3,]3
  [4,]4
  [5,]5
  [6,]6
  [7,]7
  [8,]8
  [9,]9
  [10,]   10
   a[-OverloadsTesT]
  [1]  2  3  4  5  6  7  8  9 10
 
 
 
  the last line does not remove the first and third element and only does
 the first element.?
 
  What I want to do is for zeros to return the elements and for any
 positive value to remove it.
  What I am doing wrong?
 
  B.R
  Alex
  [[alternative HTML version deleted]]
 
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 http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
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[[alternative HTML version deleted]]

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Re: [R] why the a[-indx] does not work?

[andrija djurovic]

You are absolutely right it is better. This was just an example how one
can convert the positives to TRUE and zeros to FALSE as it was asked.

Andrija


On Sun, Oct 30, 2011 at 9:57 PM, William Dunlap wdun...@tibco.com wrote:

 I like to use numericVector != 0 instead of
 is.logical(numericVector) because the former
 more directly indicates what you want to happen
 instead of relying on knowledge that numeric 0
 maps to logical FALSE.

 Bill Dunlap
 Spotfire, TIBCO Software
 wdunlap tibco.com

  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On Behalf Of Alaios
  Sent: Sunday, October 30, 2011 1:40 PM
  To: andrija djurovic
  Cc: R-help@r-project.org
  Subject: Re: [R] why the a[-indx] does not work?
 
  I think this does the work
 
  return(m[!as.logical(data)])
 
 
  I am not sure though if this is the same with
 
  return(m[!as.logical(data)])
 
 
 
  
  From: andrija djurovic djandr...@gmail.com
 
  Cc: R-help@r-project.org R-help@r-project.org
  Sent: Sunday, October 30, 2011 9:58 PM
  Subject: Re: [R] why the a[-indx] does not work?
 
 
   as.logical(c(1,0,1,1))
  [1]  TRUE FALSE  TRUE  TRUE
 
  ?as.logical
 
 
 
 
 
 
  probably you mean
  
  
  
  
  For ‘[’-indexing only: ‘i’, ‘j’, ‘...’ can be logical
  vectors, indicating elements/slices to select.  Such vectors
  are recycled if necessary to match the corresponding extent.
  ‘i’, ‘j’, ‘...’ can also be negative integers, indicating
  elements/slices to leave out of the selection.
  
  
  
  How can i convert the positives to TRUE and zeros and FALSE?
  
  
  
  
  From: William Dunlap wdun...@tibco.com
  
  Sent: Sunday, October 30, 2011 9:17 PM
  Subject: RE: [R] why the a[-indx] does not work?
  
 a[overLoadTesT==0]
[1]  2  4  5  6  7  8  9 10
  Look into help('[') or help('Subscript') to see
  how integer and logical (Boolean) subscripts differ.
  
  Bill Dunlap
  Spotfire, TIBCO Software
  wdunlap tibco.com
  
   -Original Message-
   From: r-help-boun...@r-project.org [mailto:
 r-help-boun...@r-project.org] On Behalf Of Alaios
   Sent: Sunday, October 30, 2011 11:52 AM
   To: R-help@r-project.org
   Subject: [R] why the a[-indx] does not work?
  
   Dear all,
  
   Could you please explain me why
  
OverloadsTesT
   [1] 1 0 1 0 0 0 0 0 0 0
a-matrix(data=seq(1,10),nrow=10)
a
   [,1]
   [1,]1
   [2,]2
   [3,]3
   [4,]4
   [5,]5
   [6,]6
   [7,]7
   [8,]8
   [9,]9
   [10,]   10
a[-OverloadsTesT]
   [1]  2  3  4  5  6  7  8  9 10
  
  
  
   the last line does not remove the first and third element and only
 does the first element.?
  
   What I want to do is for zeros to return the elements and for any
 positive value to remove it.
   What I am doing wrong?
  
   B.R
   Alex
   [[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.
 [[alternative HTML version deleted]]
  
  
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Re: [R] generate randomly a value of a vector

[andrija djurovic]

Hi Boris.

Here is one approach:

N-100
a-rep(0,N)
a[sample(N,1)]-1
a
which(a==1)

Look ?sample, ?which.

Andrija

On Thu, Sep 8, 2011 at 10:42 AM, Boris Beranger borisberan...@gmail.comwrote:

 Hi everyone,

 I have a zero vector of length N and I would like to randomly allocate the
 value 1 to one of the values of this vector. I presume I have to use the
 uniform distribution but could someone tell me how I should process?

 Thanks in advance,
 Boris

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[R] subqueries in sqlQuery function (package RODBC)

[andrija djurovic]

Hi R users.

sorry for missing example and if question is to general but I am wondering
if it is possible to execute subqueries in function sqlQuery (package RODBC)
with opened connection with Excel or SQL server 2000. I couldn't find any
example of this.

And if it is possible what should be a correct syntax for this query:

SELECT ct,COUNT(*) as n
FROM (SELECT COUNT(*) AS ct FROM children
GROUP BY family_id) AS x
GROUP BY ct;

sqlQuery(connecton,  CORRECT SYNTAX )

(This query is an example from book Data Manipulation with R, Phil Spector,
page 47)

Thanks for any help

Andrija

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Re: [R] subqueries in sqlQuery function (package RODBC)

[andrija djurovic]

I thought subqueries in sense of commad inside the command (in my example
two select commands). It works as you proposed an I thought in this case
(subqueries) that I need different syntax for sqlQury function combining SQL
query and paste.

But now I have another problem and again sorry if it is to general and basic
but I just can't find the right option to set up.
Namely, when I import table from SQL server into R, columns that are defined
in SQL table as char (with leading zeros as 001, 002,...) are imported as
integers.

Could you, please,  guide me on some options that should be set up to solve
this problem?
On Wed, Aug 10, 2011 at 10:16 AM, Prof Brian Ripley
rip...@stats.ox.ac.ukwrote:

 In what sense is this a 'subquery'?  It is just an SQL command (write it on
 one line, no terminating ;, which is not part of the query).


 On Wed, 10 Aug 2011, andrija djurovic wrote:

   Hi R users.

 sorry for missing example and if question is to general but I am wondering
 if it is possible to execute subqueries in function sqlQuery (package
 RODBC)
 with opened connection with Excel or SQL server 2000. I couldn't find any
 example of this.

 And if it is possible what should be a correct syntax for this query:

 SELECT ct,COUNT(*) as n
 FROM (SELECT COUNT(*) AS ct FROM children
 GROUP BY family_id) AS x
 GROUP BY ct;

 sqlQuery(connecton,  CORRECT SYNTAX )

 (This query is an example from book Data Manipulation with R, Phil
 Spector,
 page 47)

 Thanks for any help

 Andrija

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 Professor of Applied Statistics,  
 http://www.stats.ox.ac.uk/~**ripley/http://www.stats.ox.ac.uk/~ripley/
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Re: [R] Simple loop

[andrija djurovic]

Hi.
There is no need to do this in a for loop.
Here is one approach:

x - read.table(textConnection(Site  Prof  H
1  1 24
1  1 16
1  1 67
1  2 23
1  2 56
1  2 45
2  1 67
2  1 46), header = TRUE)
closeAllConnections()
x
cbind(x,newCol=unlist(tapply(x[,3],paste(x[,1],x[,2],sep=),
function(x) x-min(x)))
   Site Prof  H newCol
11111 24  8
11211 16  0
11311 67 51
12112 23  0
12212 56 33
12312 45 22
21121 67 21
21221 46  0

Andrija


On Tue, May 3, 2011 at 5:44 PM, Woida71 w.gost...@ipp.bz.it wrote:

 Hello everybody,
 I am beginning with loops and functions and would be glad to have help in
 the following question:
 If i have a dataframe like this
 Site  Prof  H
 1  1 24
 1  1 16
 1  1 67
 1  2 23
 1  2 56
 1  2 45
 2  1 67
 2  1 46
 And I would like to create a new column that subtracts the minimum of H
 from
 H, but for S1 and P1
 only the minimum of the data points falling into this category should be
 taken.
 So for example the three first numbers of the new column write: 24-16,
 16-16, 67-16
 the following numbers refering to Site1 and Prof2 write: 23-23, 56-23,
 45-23.
 I think with two loops one refering to the Site, the other to the Prof, it
 should be possible to automatically
 create the new column.
 Thanks a lot for any help.

 --
 View this message in context:
 http://r.789695.n4.nabble.com/Simple-loop-tp3492819p3492819.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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 http://www.R-project.org/posting-guide.html
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Re: [R] Simple loop

[andrija djurovic]

William, you are right. Thanks for clarification.

Andrija

On Tue, May 3, 2011 at 9:04 PM, William Dunlap wdun...@tibco.com wrote:



 Bill Dunlap
 Spotfire, TIBCO Software
 wdunlap tibco.com

  -Original Message-
  From: r-help-boun...@r-project.org
  [mailto:r-help-boun...@r-project.org] On Behalf Of andrija djurovic
  Sent: Tuesday, May 03, 2011 11:28 AM
  To: Woida71
  Cc: r-help@r-project.org
  Subject: Re: [R] Simple loop
 
  Hi.
  There is no need to do this in a for loop.
  Here is one approach:
 
  x - read.table(textConnection(Site  Prof  H
  1  1 24
  1  1 16
  1  1 67
  1  2 23
  1  2 56
  1  2 45
  2  1 67
  2  1 46), header = TRUE)
  closeAllConnections()
  x
  cbind(x,newCol=unlist(tapply(x[,3],paste(x[,1],x[,2],sep=),
  function(x) x-min(x)))
 Site Prof  H newCol
  11111 24  8
  11211 16  0
  11311 67 51
  12112 23  0
  12212 56 33
  12312 45 22
  21121 67 21
  21221 46  0

 That works when Site and Prof are ordered as shown, but if
 they are not sorted cbind(...,tapply) won't line up the the
 new entries with the old rows properly.  Try doing it on
 x[8:1,] to see this.

 ave() can deal that problem:
   cbind(x, newCol2 = with(x, ave(H, Site, Prof,
 FUN=function(y)y-min(y
Site Prof  H newCol2
  111 24   8
  211 16   0
  311 67  51
  412 23   0
  512 56  33
  612 45  22
  721 67  21
  821 46   0
  Warning message:
  In min(y) : no non-missing arguments to min; returning Inf
 The warning is unfortunate: ave() calls FUN even for when
 there is no data for a particular group (Site=2, Prof=2 in this
 case).

 Bill Dunlap
 Spotfire, TIBCO Software
 wdunlap tibco.com


 
  Andrija
 
 
  On Tue, May 3, 2011 at 5:44 PM, Woida71 w.gost...@ipp.bz.it wrote:
 
   Hello everybody,
   I am beginning with loops and functions and would be glad
  to have help in
   the following question:
   If i have a dataframe like this
   Site  Prof  H
   1  1 24
   1  1 16
   1  1 67
   1  2 23
   1  2 56
   1  2 45
   2  1 67
   2  1 46
   And I would like to create a new column that subtracts the
  minimum of H
   from
   H, but for S1 and P1
   only the minimum of the data points falling into this
  category should be
   taken.
   So for example the three first numbers of the new column
  write: 24-16,
   16-16, 67-16
   the following numbers refering to Site1 and Prof2 write:
  23-23, 56-23,
   45-23.
   I think with two loops one refering to the Site, the other
  to the Prof, it
   should be possible to automatically
   create the new column.
   Thanks a lot for any help.
  
   --
   View this message in context:
   http://r.789695.n4.nabble.com/Simple-loop-tp3492819p3492819.html
   Sent from the R help mailing list archive at Nabble.com.
  
   __
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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
   and provide commented, minimal, self-contained, reproducible code.
  
 
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Re: [R] adding a name to cross tab margins

[andrija djurovic]

Hi.

Have a look also at ?addmargins.

addmargins(table(fS,fF),c(1,2),FUN=list(total.row=sum,total.col=sum))

Andrija
On Fri, Apr 15, 2011 at 3:29 PM, Dmitry Berman ravenb...@gmail.com wrote:

 Listers,

 I have created a cross-tab matrix using the following code:

 S -

 c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,1,2,3,1,1,2,3,1,2,2)
 F -

 c(1,2,3,1,2,3,1,1,1,1,2,3,1,1,3,2,2,2,1,2,3,1,1,1,1,2,3,3,1,3,1,3,1,1,2,3,1,1,3,2,3,2,1,1,1,2,3,1,1,2)
 table(S,F)
 fF -factor(F, labels = c(Dem,Ind,Rep))
 fS -factor(S, labels = c(Dem,Ind,Rep))
 PiD - table(fS,fF)
 PiD - cbind(PiD, margin.table(PiD,1))
 PiD - rbind(PiD, margin.table(PiD,2))

 Now I would like to add the Column Name and Row Name Total to the last
 row
 and last column (where the marginal totals now hang out). When I use the
 command

 colnames(PiD[,4]) - c(total) I get the error:

 Error in `colnames-`(`*tmp*`, value = WTF) :
  attempt to set colnames on object with less than two dimensions


 Since I am still new to R, I can't really understand the error... Can
 someone tell me what I am doing wrong here?

 Thanks

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Re: [R] adding a name to cross tab margins

[andrija djurovic]

Sorry, I've just reversed the names of dimensions :)

addmargins(table(fS,fF),c(1,2),FUN=list(total.col=sum,total.row=sum))

Andrija




On Fri, Apr 15, 2011 at 3:50 PM, andrija djurovic djandr...@gmail.comwrote:

 Hi.

 Have a look also at ?addmargins.

 addmargins(table(fS,fF),c(1,2),FUN=list(total.row=sum,total.col=sum))

 Andrija
  On Fri, Apr 15, 2011 at 3:29 PM, Dmitry Berman ravenb...@gmail.comwrote:

 Listers,

 I have created a cross-tab matrix using the following code:

 S -

 c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,1,2,3,1,1,2,3,1,2,2)
 F -

 c(1,2,3,1,2,3,1,1,1,1,2,3,1,1,3,2,2,2,1,2,3,1,1,1,1,2,3,3,1,3,1,3,1,1,2,3,1,1,3,2,3,2,1,1,1,2,3,1,1,2)
 table(S,F)
 fF -factor(F, labels = c(Dem,Ind,Rep))
 fS -factor(S, labels = c(Dem,Ind,Rep))
 PiD - table(fS,fF)
 PiD - cbind(PiD, margin.table(PiD,1))
 PiD - rbind(PiD, margin.table(PiD,2))

 Now I would like to add the Column Name and Row Name Total to the last
 row
 and last column (where the marginal totals now hang out). When I use the
 command

 colnames(PiD[,4]) - c(total) I get the error:

 Error in `colnames-`(`*tmp*`, value = WTF) :
  attempt to set colnames on object with less than two dimensions


 Since I am still new to R, I can't really understand the error... Can
 someone tell me what I am doing wrong here?

 Thanks

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Re: [R] Converting a categorical variable to multiple dichotemous variables

[andrija djurovic]

hi:
here is one solution:

cat-as.factor(c(1,1,3,2,4))
model.matrix(~cat-1,cat)

cbind(cat,model.matrix(~cat-1,cat))
Andrija

On Tue, Apr 12, 2011 at 2:17 PM, Shane Phillips sphill...@lexington1.netwrote:

 I have a categorical variable in a dataframe similar to the following...

 cat
 1
 1
 3
 2
 4

 I need to convert it to 4 dichotemous variables for each observations
 like...

 cat1cat2cat3cat4
 1   0   0   0
 1   0   0   0
 0   0   1   0
 0   1   0   0
 0   0   0   1


 Thanks in advance!

 Shane

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Re: [R] Creating a symmetric contingency table from two vectors with different length of levels in R

[andrija djurovic]

Hi:

Here is one solution:

a-factor(c(1,2,4,5,6))
b-factor(c(2,2,4,5,5))
b1-factor(b,levels=c(levels(b),levels(a)[levels(a)%in%levels(b)==FALSE]))
table(a,b1)

but be aware that levels of b is a subset of levels of a.

Andrija

On Wed, Apr 6, 2011 at 10:39 AM, suparna mitra 
mi...@informatik.uni-tuebingen.de wrote:

 Hello,
 How can I create a symmetric contingency table from two categorical vectors
 having different length of levels?
 For example one vector has 98 levels
 TotalData1$Taxa.1
  [1] Aconoidasida Actinobacteria (class)
 Actinopterygii   Alphaproteobacteria
  [5] AmoebozoaAmphibia
 Anthozoa Aquificae (class)
 and so on .
 98 Levels: Aconoidasida Actinobacteria (class) 

  and the other vector has 105 levels
 TotalData1$Taxa.2
[1] FlavobacteriaProteobacteria
 Bacteroidetes/Chlorobi group Bacteria
[5] EpsilonproteobacteriaEpsilonproteobacteria
  Epsilonproteobacteria
 and so on  ..
 105 Levels: Acidobacteria Aconoidasida Actinobacteria (class) 

 Now I want to create a symmetric contingency table.
 Any quick idea will be really helpful.
 Best regards,
 Mitra

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Re: [R] Value between which elements of a vector?

[andrija djurovic]

Hi:

try this and have a look at ?cut just to change the lables:

data.frame(
weeks=sprintf('%s %d','week',1:30),
specialweeks=cut(1:30,c(0,2,5,12,18,19,20),right = FALSE))

Andrija

On Tue, Apr 5, 2011 at 1:20 PM, beatleb rhelpfo...@gmail.com wrote:

 Dear R-useRs,

 I am looking for a why to perform the following:

 specialweeks-c(0,2,5,12,18,19,20)
 weeks-c(1:30)

 Now I would like that for every week it is even between which elements of
 vector special weeks it is.
 For weeks after 20, the value NA or 20, or even 20-30is fine.

 Thus for
 week 1: 0-2
 week 2: 2-5
 week 3: 2-5
 week 4: 2-5
 week 5: 5-12
 ect

 It is not relevant if those intervals are captured in a matrix, in a vector
 or whatever.

 I hope that you can help me!

 With best regards,

 Brenda Grondman



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Re: [R] Help in splitting ists into sub-lists

[andrija djurovic]

Hi:

here is one solution. Not so elegant but maybe it will give you some ideas:

mylist1-rep(mylist,c(2,2))
a-matrix(c(index1,!index1,index2,!index2),ncol=4)
mylist2-list()
for (i in 1:4)
{
mylist2[[i]]-mylist1[[i]][,a[,i]]
}

mylist2

Andrija

On Sun, Apr 3, 2011 at 6:56 PM, Axel Urbiz axel.ur...@gmail.com wrote:

 Dear List,

 Let's say I have a list whose components are 2 matrices (as exemplified in
 the mylist object below). I'd like to create a list with components being
 4 matrices based on an logical index vector. is there a way to simplify
 what
 I'm doing to obtain the results in mylist2? I'd like something that would
 work on an arbitrary number of elements in mylist.

 mylist - list(matrix(1:9,3,3), matrix(10:18,3,3))
 index1 - c(TRUE,FALSE,TRUE)
 index2- c(FALSE,TRUE,TRUE)
 mylist2 - list(mylist[[1]][,index1],
  mylist[[1]][,index1==FALSE],
  mylist[[2]][,index2],
  mylist[[2]][,index2==FALSE])

 Thanks in advance,
 Axel.

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Re: [R] Matrix manipulation

[andrija djurovic]

HI,
here is another solution:

int - sample(1:20,10)
int
 [1] 10  4  5  2 14 17  9 11 16 13

mat-matrix(11:30,ncol=4)
 mat
 [,1] [,2] [,3] [,4]
[1,]   11   16   21   26
[2,]   12   17   22   27
[3,]   13   18   23   28
[4,]   14   19   24   29
[5,]   15   20   25   30

 mat[apply(mat,1, function(x) any(int==x[1])),]
   [,1] [,2] [,3] [,4]
[1,]   11   16   21   26
[2,]   13   18   23   28
[3,]   14   19   24   29

Andrija


On Sat, Apr 2, 2011 at 7:08 AM, Joseph N. Paulson
josephpaul...@gmail.comwrote:

 Hi all!

 I have a vector, let's say for example int - sample(1:20,10);
 for now:

 now I have a matrix...
 M = m x n
 where the first column is a feature column and most likely shares at
 least
 one of the int (interesting) numbers.

 I want to extract the rows where int[] = M[,1]

 I thought:
 rownames(int)-int;
 rownames(M)-M[,1];

 M[rownames(int),] would work, but it doesn't... (I assume because I have
 rownames(int) that are not found in M[,1]. Neither does,
 rownames(M)==rownames(int)...

 Any help would be greatly appreciated!

 Thank you!

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[R] export list to csv

[andrija djurovic]

Hi everybody.

I have list like this:

l-list(data.frame(q1=c(1,2,check),q2=c(3,check,5)),
data.frame(q1=c(check,1),q2=c(4,5)))
names(l)-c(A,B)
rownames(l[[1]])-c(aa,bb,cc)
rownames(l[[2]])-c(aa,bb)

Every object has the same number of columns but different number of rows.
Does anyone know if it is possible to export such kind of list, into one csv
file, and  keeping all the names?

Thanks in advance

Andrija

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Re: [R] export list to csv

[andrija djurovic]

Soryy, I didn't explain well what I want. I would like to have a table in
csv on txt file like this:

$A
  q1q2
aa 1 3
bb 2 check
cc check 5
$B
  q1 q2
aa check  4
bb 1  5
The same as write.csv of any data frame.




On Wed, Mar 16, 2011 at 4:03 PM, Henrique Dallazuanna www...@gmail.comwrote:

 Use dput:

 dput(l, file = l_str.txt)

 Then, to read again:

 l - dget(file = 'l_str.txt')

 On Wed, Mar 16, 2011 at 11:55 AM, andrija djurovic djandr...@gmail.com
 wrote:
  Hi everybody.
 
  I have list like this:
 
  l-list(data.frame(q1=c(1,2,check),q2=c(3,check,5)),
  data.frame(q1=c(check,1),q2=c(4,5)))
  names(l)-c(A,B)
  rownames(l[[1]])-c(aa,bb,cc)
  rownames(l[[2]])-c(aa,bb)
 
  Every object has the same number of columns but different number of rows.
  Does anyone know if it is possible to export such kind of list, into one
 csv
  file, and  keeping all the names?
 
  Thanks in advance
 
  Andrija
 
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 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O


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Re: [R] export list to csv

[andrija djurovic]

Thanks everyone for different solutions.
Every solution works very well. For my purpose, this function sink does what
I was looking for.

Andrija

On Wed, Mar 16, 2011 at 4:23 PM, Roman Luštrik roman.lust...@gmail.comwrote:

 How about?

 sink(andrija.csv)
 l
 sink()

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 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] data.frame transformation

[andrija djurovic]

Thank you Bill for this additional solution.

Andrija

On Tue, Mar 15, 2011 at 12:16 AM, bill.venab...@csiro.au wrote:

 It is possible to do it with numeric comparisons, as well, but to make life
 comfortable you need to turn off the warning system temporarily.

 df - data.frame(q1 = c(0,0,33.33,check),
 q2 = c(0,33.33,check,9.156),
 q3 = c(check,check,25,100),
 q4 = c(7.123,35,100,check))

 conv - function(x, cutoff) {
oldOpt - options(warn = -1)
on.exit(options(oldOpt))
x - as.factor(x)
lev - as.numeric(levels(x))
levels(x)[!is.na(lev)  lev  cutoff] - .
x
 }

 Check:
  (df1 - data.frame(lapply(df, conv, cutoff = 10)))
 q1q2q3q4
 1 . . check .
 2 . 33.33 check35
 3 33.33 check25   100
 4 check .   100 check
 

 Bill Venables.

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On Behalf Of David Winsemius
 Sent: Tuesday, 15 March 2011 6:29 AM
 To: andrija djurovic
 Cc: r-help@r-project.org
 Subject: Re: [R] data.frame transformation


 On Mar 14, 2011, at 3:51 PM, andrija djurovic wrote:

  I would like to hide cells with values less the 10%, so . or just
   doesn't make me any difference. Also I used apply combined with
  as.character:
 
  apply(df, 2, function(x)  ifelse(as.character(x)  10,.,x))
 
  This is, probably not a good solution, but it works except that I
  lose  row names and because of that I was wondering if there is some
  other way to do this.
 
  Anyway thank you both i will try to do this before combining numbers
  and strings.

 I saw your later assertion that it didn't work which surprised me. My
 version of your data followed my advice not to use factors and your
 effort did succeed when the columns were character rather than factor.
 I put back the row numbers by coercing back to a data.frame. `apply`
 returns a matrix.

   df-data.frame(q1=c(0,0,33.33,check),q2=c(0,33.33, check,9.156),
 + q3=c(check,check,25,100),q4=c(7.123,35,100,check),
 stringsAsFactors=FALSE)
   as.data.frame(apply(df, 2, function(x)  ifelse(as.character(x) 
 10,.,x)))
  q1q2q3q4
 1 . . check 7.123
 2 . 33.33 check35
 3 33.33 .25   100
 4 check 9.156   100 check

 There is a danger of using character collation in that if there are
 any leading characters in those strings that are below 1 such as a
 blank or any other punctuation, they will get dotted.

   ,  1
 [1] TRUE
   .  1
 [1] TRUE
   -  1
 [1] TRUE

 And 1.check would also get dotted

   1.check  10
 [1] TRUE

 
  Andrija
 
  On Mon, Mar 14, 2011 at 8:11 PM, David Winsemius dwinsem...@comcast.net
   wrote:
 
  On Mar 14, 2011, at 2:52 PM, andrija djurovic wrote:
 
  Hi R users,
 
  I have following data frame
 
  df-data.frame(q1=c(0,0,33.33,check),q2=c(0,33.33,check,9.156),
  q3=c(check,check,25,100),q4=c(7.123,35,100,check))
 
  and i would like to replace every element that is less then 10
  with . (dot)
  in order to obtain this:
 
 q1q2q3q4
  1 . . check .
  2 . 33.33 check35
  3 33.33 check25   100
  4 check .   100 check
 
  I had a lot of difficulties because each variable is factor.
 
  Right, so comparisons with  will throw an error.  I would
  sidestep the factor problem with stringsAsFactors=FALSE in the
  data.frame call. You might want to reconsider the . as a missing
  value. If you are coming from a SAS background, you should try to
  get comfortable with NA or NA_character as a value.
 
 
  df-data.frame(q1=c(0,0,33.33,check),q2=c(0,33.33,check,9.156),
   q3=c(check,check,25,100),q4=c(7.123,35,100,check),
  stringsAsFactors=FALSE)
 
  is.na(df) - t(apply(df, 1, function(x)  as.numeric(x)  10))
 
  Warning messages:
  1: In FUN(newX[, i], ...) : NAs introduced by coercion
  2: In FUN(newX[, i], ...) : NAs introduced by coercion
  3: In FUN(newX[, i], ...) : NAs introduced by coercion
  4: In FUN(newX[, i], ...) : NAs introduced by coercion
   df
  q1q2q3q4
  1  NA  NA check  NA
  2  NA 33.33 check35
 
  3 33.33 check25   100
  4 check  NA   100 check
 
 
  Could someone help me with this?
 
  Thanks in advance for any help.
 
  Andrija
 
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  David Winsemius, MD
  West Hartford, CT
 
 

 David Winsemius, MD
 West Hartford, CT

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[[alternative

Re: [R] applying to dataframe rows

[andrija djurovic]

Hi,

maybe this:

df-data.frame(a=c(1,2,3,Inf,4,Inf),b=c(Inf,2,3,4,5,8))

df[apply(df,1, function(x) !any(x==Inf)),]
df[apply(df,1, function(x) any(x==Inf)),]

Andrija


On Tue, Mar 15, 2011 at 10:44 PM, Alexy Khrabrov delivera...@gmail.comwrote:

 How do I apply a function to every row of a dataframe most naturally?
  Specifically, I'd like to filter out any row which contains an Inf in any
 column.  Since all columns are numeric, I guess max should work on a row...

 -- Alexy
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