Re: [R] about data manipulation

[Jim Lemon]

Hi lily,
If you want to use aggregate, supply the name of the function:

aggregate(flow~year, data=df, "sum")

You can also use "by" like this

by(df$flow,df$year,FUN=sum)

I assume that you don't have to worry about missing months in a year.

Jim
:


On Thu, Dec 1, 2016 at 3:06 PM, lily li  wrote:
> Hi R users,
>
> I'm trying to manipulate dataset, but met some difficulties.
>
> df
> year   month   flow
> 2006 33.5
> 2006 43.8
> 2006 521
> 2006 632
> 2007 34.1
> 2007 44.4
> ...
>
> I want to calculate total flow for each year, and use the code below:
> aggregate(flow~year, data=df, sum)
> But it gave the error message:
> Error in get(as.character(FUN), mode = "function", envir = envir) :
>   object 'FUN' of mode 'function' was not found
>
> What is the problem and how to solve it? Thanks for your help.
>
> [[alternative HTML version deleted]]
>
> __
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> and provide commented, minimal, self-contained, reproducible code.

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Re: [R] about data manipulation

[Berend Hasselman]

> On 1 Dec 2016, at 05:06, lily li  wrote:
> 
> Hi R users,
> 
> I'm trying to manipulate dataset, but met some difficulties.
> 
> df
> year   month   flow
> 2006 33.5
> 2006 43.8
> 2006 521
> 2006 632
> 2007 34.1
> 2007 44.4
> ...
> 
> I want to calculate total flow for each year, and use the code below:
> aggregate(flow~year, data=df, sum)
> But it gave the error message:
> Error in get(as.character(FUN), mode = "function", envir = envir) :
>  object 'FUN' of mode 'function' was not found
> 
> What is the problem and how to solve it? Thanks for your help.
> 

Not enough information.
If I try this

df <- read.table(text="year   month   flow
2006 33.5
2006 43.8
2006 521
2006 632
2007 34.1
2007 44.4
", header=TRUE)

df

aggregate(flow~year, data=df, sum)

I get a correct answer. So you are likely doing something weird and not showing 
us all.


>   [[alternative HTML version deleted]]
> 

Plan text mail. Has been asked many, many times before.


Berend Hasselman


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Re: [R] Command for simple effects following 2 way anova with interaction

[Richard M. Heiberger]

## Use the split argument to summary.aov

## This tests the levels of tension within each level of wool using a common
## Residuals sum of squares.

sapply(warpbreaks, levels)

model2 <- aov(breaks ~ wool/tension, data = warpbreaks)
colnames(model.matrix(model2))
## [1] "(Intercept)""woolB"  "woolA:tensionM" "woolB:tensionM"
## [5] "woolA:tensionH" "woolB:tensionH"
sapply(warpbreaks[2:3], levels)
summary(model2,
split=list("wool:tension"=list("woolA/tension"=c(1,3),
   "woolB/tension"=c(2,4


## The less good choice has the same numerator sums of squares but
## different Residuals sums of squares, with fewer df for each,
## and comes to different conclusions.
model3A <- aov(breaks ~ tension,
   data = warpbreaks[warpbreaks$wool=="A",])
summary(model3A)

model3B <- aov(breaks ~ tension,
   data = warpbreaks[warpbreaks$wool=="B",])
summary(model3B)



## The boxplots show that model2 better describes the data.
## There is a significant difference for woolA and not for woolB.
library(lattice)
bwplot(breaks ~ tension | wool, data=warpbreaks)

On Wed, Nov 30, 2016 at 10:55 PM, Ashim Kapoor  wrote:
> Dear All,
>
> Suppose I do :-
>
> head(warpbreaks)
> model1<- aov(breaks ~ wool*tension,data = warpbreaks)
> summary(model1)
>
> There is significant interaction. So I need to test for simple effects of
> wool at each level of  tension and vice versa. I can do a subset and then
> do a one way anova for each level of tension and wool. My query is that is
> there any command programmed which can do this automatically?
>
> Best Regards,
> Ashim
>
> [[alternative HTML version deleted]]
>
> __
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> https://stat.ethz.ch/mailman/listinfo/r-help
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> and provide commented, minimal, self-contained, reproducible code.

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[R] about data manipulation

[lily li]

Hi R users,

I'm trying to manipulate dataset, but met some difficulties.

df
year   month   flow
2006 33.5
2006 43.8
2006 521
2006 632
2007 34.1
2007 44.4
...

I want to calculate total flow for each year, and use the code below:
aggregate(flow~year, data=df, sum)
But it gave the error message:
Error in get(as.character(FUN), mode = "function", envir = envir) :
  object 'FUN' of mode 'function' was not found

What is the problem and how to solve it? Thanks for your help.

[[alternative HTML version deleted]]

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

[R] Command for simple effects following 2 way anova with interaction

[Ashim Kapoor]

Dear All,

Suppose I do :-

head(warpbreaks)
model1<- aov(breaks ~ wool*tension,data = warpbreaks)
summary(model1)

There is significant interaction. So I need to test for simple effects of
wool at each level of  tension and vice versa. I can do a subset and then
do a one way anova for each level of tension and wool. My query is that is
there any command programmed which can do this automatically?

Best Regards,
Ashim

[[alternative HTML version deleted]]

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[R] Complex Survey MSE for prediction with ML

[Ganz, Carl]

Hello,

I have been toying with the survey package's withReplicates function, which 
lets users easily extend the survey package to support any weighted statistic. 
There are a number of ML algorithms in various packages that accept weights, 
and it is fairly easy to use them with withReplicates. Below is a naïve example:

library(survey)
library(rpart)
library(gbm)

data(api)

# create survey object
dstrat<-svydesign(id=~1,strata=~stype, weights=~pw, data=apistrat, fpc=~fpc)

rstrat<-as.svrepdesign(dstrat)

# try rpart
predr <- as.data.frame(withReplicates(rstrat, function(w, data) {
  predict(rpart(api00~ell+meals+mobility,data=data,weights=w))
}))

# try gbm
predg <- as.data.frame(withReplicates(rstrat, function(w, data) {
  predict(gbm(api00~ell+meals+mobility,data=data,weights=w,
  n.trees=100))
}))

# try regular svyglm
preds <- as.data.frame(predict(svyglm(api00~ell+meals+mobility,rstrat)))

head(data.frame(predr,predg,preds))

With rpart, the standard errors are absurdly large, and clearly incorrect. With 
gbm, the results seem reasonable. 

I see in this extremely old post that you can't use quantile regression with 
withReplicates for some survey designs and expect to get reasonable results: 
https://stat.ethz.ch/pipermail/r-help/2008-August/171620.html

Quantiles and survey stats are messy business so that issue may be unique to 
quantile regressions, but based on that post it would seem that the function, 
and survey design need to have certain properties for withReplicates to 
generate valid SEs. This is not documented with withReplicates though. 

So my question is, what properties does an ML algorithm/survey design need for 
withReplicates to generate valid SEs?

Kind Regards,
Carl Ganz

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Re: [R] transpose rows and columns for large data

[John Dougherty]

On Tue, 29 Nov 2016 17:22:47 + (UTC)
Elham - via R-help  wrote:

> Is there another way (I prefer a way with Excel)?

Search on "friends don't let friends use excel for statistics."
Spreadsheets are an inherently perilous way to do statistics and Excel
specifically is notoriously poor. In fact the reason I originally began
using dedicated statistical packages (STATA first and now R) is that
spreadsheets (Excel in my case) can throw subtle errors that can create
problems. immediately, or even worse - later.  I had Excel return a
negative variance. Since variance is a squared value, unless you are
dealing with some very exotic numbers including imaginary values, a
negative variance is an absurd result.  Further investigation revealed
that other stat routines provided with Excel at the time were also
throwing errors that could look reasonable and thus be missed.  It was
also simply using erroneously constructed methods and providing
outright wrong results to things like Chi-square calculations.  

-- 

John

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Re: [R] stringi behaves differently in 2 similar situations

[Sarah Goslee]

A dot is treated differently if it has a number on no, one, or both sides.

> stri_extract_all_words("me.com", simplify = TRUE)
 [,1]
[1,] "me.com"
> stri_extract_all_words("me1.com", simplify = TRUE)
 [,1]  [,2]
[1,] "me1" "com"
> stri_extract_all_words("me1.2com", simplify = TRUE)
 [,1]
[1,] "me1.2com"

?stri_extract_all_words

sent me to

?"stringi-search-boundaries"

which suggests that you should spend some time with the user guide:

 _Boundary Analysis_ - ICU User Guide, http://userguide.icu-project.org/boundaryanalysis>


Depending on your objective, you might be better off with strsplit()
separating on whitespace.

Sarah

On Wed, Nov 30, 2016 at 3:51 PM, Dimitri Liakhovitski
 wrote:
> Hello!
>
> library(stringi)
>
> stri_extract_all_words("me.com", simplify = TRUE) # returns with a dot
> stri_extract_all_words("watch32.com", simplify = TRUE)  # removes the dot
>
> Why is the dot removed only in the second case?
> How is it possible to ask it NOT to remove the dot in the second case?
>
> Thanks a lot!
>

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

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[R] stringi behaves differently in 2 similar situations

[Dimitri Liakhovitski]

Hello!

library(stringi)

stri_extract_all_words("me.com", simplify = TRUE) # returns with a dot
stri_extract_all_words("watch32.com", simplify = TRUE)  # removes the dot

Why is the dot removed only in the second case?
How is it possible to ask it NOT to remove the dot in the second case?

Thanks a lot!


-- 
Dimitri Liakhovitski

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[R] Question regarding Naive Bayes

[PHILIP GLADWIN]

Hello,


 
I am working with the naïve bayes function inlibrary(e1071).


 
The function calls are:

transactions.train.nb = naiveBayes(as.factor(DealerID) ~

   as.factor(Manufacturer) 

    + as.factor(RangeDesc)

    +as.factor(BodyType)  

    +as.factor(FuelType) 

    +as.factor(PaintColour)

    +as.factor(TransmissionType) 

    +as.factor(Mileage)

    +as.factor(Registration),

 data=transactions.train, 

 na.action=na.omit)


 
where transactions.train is a dataframe with dimension 2032rows by 14 columns.


 
and


 
transactions.test.nb = predict(transactions.train.nb,transactions.test[,-1], 
type='raw')


 
An example of the result are

View(transactions.test.nb)


 
Reduced results shown:

    188 225 229 
270 273

                    
 

1  0.000984  0.000492  0.000492 
 0.000492  0.001476

2  0.000984  0.000492  0.000492 
 0.000492  0.001476

3  0.000984  0.000492  0.000492 
 0.000492  0.001476

4  0.000984  0.000492  0.000492 
 0.000492  0.001476

5  0.000984  0.000492  0.000492 
 0.000492  0.001476


 
I was struggling to understand why the returnedprobabilities are the same for 
each column as I was hoping for them to bedifferent.

Dealer ID should have a different probability to row 1 than row 2.Each row does 
sum to 1.


 
Transactions.train represents 67% of the full set of data.

I’ve tried introducing laplace smoothing, and experimentedwith increasing and 
decreasing the number of parameters used to generate thetraining naivebayes 
object

But as of yet I can’t figure it out.  Could anybody help?


 
Kind regards,

Phil,


[[alternative HTML version deleted]]

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[R] Meta-Analysis using RVE (robumeta)

[Janina Steinert]

Hello,

I'm runnig a meta-analysis, correcting for clustering of standard errors
with the robumetacommand. R gives the following warning:

If df < 4, do not trust the results

In some cases my df is in fact <4. What does this tell me then? Does it
simply mean that the number of included studies is likely too small (often
aroung 5-6, but with more than 10 single effect sizes)?

Is there anything else I could do for these cases?
Thank you

[[alternative HTML version deleted]]

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[R] Improve performance of RConnection and Rserve

[Ranjana Girish]

Hi all,

If we execute an R script directly in R prompt in Linux environment, it
takes around 30 mins to complete.

The same R script if we execute from java using RConnection, eval and
Rserve for each line it takes thrice the time.

Basically the performance is very bad. And in most of the occasion it never
ends up in result.

Can anyone please tell how to improve the performance using Rconnection.

Thanks in advance

Ranjana

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Re: [R] Breaking down a list into a table

[Jeff Newmiller]

The tidyr::separate function modifies data frames.  You did not give it a data 
frame. Re-read your preferred R introductory material on what a data frame is 
(R comes with "Introduction to R").

You also did not give a reproducible example... e.g. does not execute as-is in 
a clean R work space. 

library(tidyr)
input <- data.frame( kop=rep( "w;d;e;f", 5 ) )
output <- separate( input, kop, into=c( "C1", "C2", "C3", "C4" ), sep=";" )

You really should read the documentation for any contributed package you use. 
(This presumes that you keep track of the fact that you are using them... which 
is a good idea as there are thousands of them). I recommend

vignette("tidy-data")

and

?separate

in this case. 
-- 
Sent from my phone. Please excuse my brevity.

On November 30, 2016 7:18:19 AM PST, Ferri Leberl  wrote:
>
>Thank to Ulrik for the hint.
>However, I don't comprehend the function until now:
>
>For example I made up an array "input":
>
>input
> kop  
>[1,] "w;d;e;f"
>[2,] "w;d;e;f"
>[3,] "w;d;e;f"
>[4,] "w;d;e;f"
>[5,] "w;d;e;f"
>
>and tried to break it into four cols with commmand:
>
>output<-separate(into,kop,into=c("a","b","c","d"),sep=";")
>
>R returned:
>
>Fehler in UseMethod("separate_") : 
>nicht anwendbare Methode für 'separate_' auf Objekt der Klasse
>"c('matrix', 'character')" angewendet
>
>Could you please explain me my mistake?
>
>Thank you in advance!
>
>Yours, Ferri
>
>
>
>
>
> 
>
>Gesendet: Dienstag, 22. November 2016 um 14:57 Uhr
>Von: "Ulrik Stervbo" 
>An: "Ferri Leberl" , "r-helpr-project.org"
>
>Betreff: Re: [R] Breaking down a list into a table
>
>Hi Ferri,
> It sounds like the function 'separate' from the tidyr package is what
>you look for,
> 
>HTH
>Ulrik 
>
>On Tue, 22 Nov 2016 at 14:49 Ferri Leberl
> wrote:
>
>Dear All,
>I asked for support to deal with a hirarchy within a character
>separated list.
>I solved the problem crudely but effectively by
>
>- Choosing for a TSV as input, where in columns that may contain
>several (or as well no) items the items are separated via semicolon
>- adding semicolons to the first row to grant that the first row has
>the maximum number of semicolons of this column
>- grasping the column(x<-myarray[,y], where y is some integer value)
>and saving it as a TSV (with only one column)
>- importing it again, defining it semicolumn-separated, with fill
>option
>
>To all those who feel pain reading this: Is there a shortcut?
>Thank you in advance.
>Yours, Ferri
>
>__
>R-help@r-project.org[mailto:R-help@r-project.org] mailing list -- To
>UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help[https://stat.ethz.ch/mailman/listinfo/r-help]
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html[http://www.R-project.org/posting-guide.html]
>and provide commented, minimal, self-contained, reproducible code.
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
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>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Breaking down a list into a table

[Bert Gunter]

Well, if i understand correctly, there's really no need to use special
packages for such simple tasks. The strsplit() function in base R does
nicely (see ?strplit):

Here is a reproducible example, --which you did not provide; please do
so in future:

> testdat <- c("abc;def;hij","qq;sdff;kuqw","oprrt;lbas;rw")

> testdat
[1] "abc;def;hij"   "qq;sdff;kuqw"  "oprrt;lbas;rw"

## note that this is a vector, not an array/data frame with 1 column.
You can subscript it in the call below -- e.g. testdat[,1] -- if it is
an array.

> out <- strsplit(testdat,split= ";")
> out
[[1]]
[1] "abc" "def" "hij"

[[2]]
[1] "qq"   "sdff" "kuqw"

[[3]]
[1] "oprrt" "lbas"  "rw"

## Note that the output of strsplit() is a list because it is designed
to handle the more general case of unequal numbers of pieces in each
row. To reform it to a matrix (i.e. 2d array) use rbind in a do.call
(?do.call) construct:


> do.call(rbind,out)
 [,1][,2]   [,3]
[1,] "abc"   "def"  "hij"
[2,] "qq""sdff" "kuqw"
[3,] "oprrt" "lbas" "rw"

All of this presupposes familiarity with basic R data structures and
constructs. If you do not have such familiarity yet, then I think you
would do well to go through some R tutorials to gain it (if you wish
to continue to use R). Packages are nice, with many helpful features,
but you still need to know the basics of the language -- and R *is* a
language.

As far as your error message is concerned, separate() is apparently an
S3 generic function (more reading!) but has no method for matrices. If
you had done the call as:

output<-separate(into,input[,1] ,into=c("a","b","c","d"),sep=";")

## giving the data as a vector rather than an array of 1 column
it probably would have worked (but you need to check, as I don't use
the tidyr package).

Cheers,

-- Bert

Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Wed, Nov 30, 2016 at 7:18 AM, Ferri Leberl  wrote:
>
> Thank to Ulrik for the hint.
> However, I don't comprehend the function until now:
>
> For example I made up an array "input":
>
> input
>  kop
> [1,] "w;d;e;f"
> [2,] "w;d;e;f"
> [3,] "w;d;e;f"
> [4,] "w;d;e;f"
> [5,] "w;d;e;f"
>
> and tried to break it into four cols with commmand:
>
> output<-separate(into,kop,into=c("a","b","c","d"),sep=";")
>
> R returned:
>
> Fehler in UseMethod("separate_") :
>   nicht anwendbare Methode für 'separate_' auf Objekt der Klasse "c('matrix', 
> 'character')" angewendet
>
> Could you please explain me my mistake?
>
> Thank you in advance!
>
> Yours, Ferri
>
>
>
>
>
>
>
> Gesendet: Dienstag, 22. November 2016 um 14:57 Uhr
> Von: "Ulrik Stervbo" 
> An: "Ferri Leberl" , "r-helpr-project.org" 
> 
> Betreff: Re: [R] Breaking down a list into a table
>
> Hi Ferri,
>  It sounds like the function 'separate' from the tidyr package is what you 
> look for,
>
> HTH
> Ulrik
>
> On Tue, 22 Nov 2016 at 14:49 Ferri Leberl 
>  wrote:
>
> Dear All,
> I asked for support to deal with a hirarchy within a character separated list.
> I solved the problem crudely but effectively by
>
> - Choosing for a TSV as input, where in columns that may contain several (or 
> as well no) items the items are separated via semicolon
> - adding semicolons to the first row to grant that the first row has the 
> maximum number of semicolons of this column
> - grasping the column(x<-myarray[,y], where y is some integer value) and 
> saving it as a TSV (with only one column)
> - importing it again, defining it semicolumn-separated, with fill option
>
> To all those who feel pain reading this: Is there a shortcut?
> Thank you in advance.
> Yours, Ferri
>
> __
> R-help@r-project.org[mailto:R-help@r-project.org] mailing list -- To 
> UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help[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] Breaking down a list into a table

[Ferri Leberl]

Thank to Ulrik for the hint.
However, I don't comprehend the function until now:

For example I made up an array "input":

input
 kop  
[1,] "w;d;e;f"
[2,] "w;d;e;f"
[3,] "w;d;e;f"
[4,] "w;d;e;f"
[5,] "w;d;e;f"

and tried to break it into four cols with commmand:

output<-separate(into,kop,into=c("a","b","c","d"),sep=";")

R returned:

Fehler in UseMethod("separate_") : 
  nicht anwendbare Methode für 'separate_' auf Objekt der Klasse "c('matrix', 
'character')" angewendet

Could you please explain me my mistake?

Thank you in advance!

Yours, Ferri





 

Gesendet: Dienstag, 22. November 2016 um 14:57 Uhr
Von: "Ulrik Stervbo" 
An: "Ferri Leberl" , "r-helpr-project.org" 

Betreff: Re: [R] Breaking down a list into a table

Hi Ferri,
 It sounds like the function 'separate' from the tidyr package is what you look 
for,
 
HTH
Ulrik 

On Tue, 22 Nov 2016 at 14:49 Ferri Leberl 
 wrote:

Dear All,
I asked for support to deal with a hirarchy within a character separated list.
I solved the problem crudely but effectively by

- Choosing for a TSV as input, where in columns that may contain several (or as 
well no) items the items are separated via semicolon
- adding semicolons to the first row to grant that the first row has the 
maximum number of semicolons of this column
- grasping the column(x<-myarray[,y], where y is some integer value) and saving 
it as a TSV (with only one column)
- importing it again, defining it semicolumn-separated, with fill option

To all those who feel pain reading this: Is there a shortcut?
Thank you in advance.
Yours, Ferri

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