[R] melt function chooses wrong id variable with large datasets
PIKAL Petr
petr.pikal at precheza.cz
Fri Apr 17 11:07:13 CEST 2015
Yes
It could be. But anyway, if you wanted to melt your frame and be sure to have norm column added to months column you shall use
melt(dataset, id.vars=NULL, na.rm=TRUE)
construction.
Without it and considering data you sent I get
> head(dd.m0)
norm variable value
1 45.8713463281901 januari 38.1
2 24.047250681782984 januari 32.4
3 3.7533684144746324 januari 34.5
4 38.594241119279324 januari 20.7
5 26.391897460120358 januari 21.5
6 61.746470001194638 januari 23.1
...
without this id.vars=NULL parameter.
Cheers
Petr
PS. Keep your post to Rhelp - others may learn from it and anothers can provide you with more elaborated explanation.
From: Joachim Audenaert [mailto:Joachim.Audenaert at pcsierteelt.be]
Sent: Friday, April 17, 2015 8:23 AM
To: PIKAL Petr
Subject: RE: [R] melt function chooses wrong id variable with large datasets
Hello,
I upgraded R tot version 3.1.3 and now everything in the script works perfectly. Could the troubles be due to the fact that I was running an older version?
Met vriendelijke groeten - With kind regards,
Joachim Audenaert
onderzoeker gewasbescherming - crop protection researcher
PCS | proefcentrum voor sierteelt - ornamental plant research
________________________________
Schaessestraat 18, 9070 Destelbergen, België
T: +32 (0)9 353 94 71 | F: +32 (0)9 353 94 95
E: joachim.audenaert at pcsierteelt.be<mailto:joachim.audenaert at pcsierteelt.be> | W: www.pcsierteelt.be<http://www.pcsierteelt.be/>
From: PIKAL Petr <petr.pikal at precheza.cz<mailto:petr.pikal at precheza.cz>>
To: Joachim Audenaert <Joachim.Audenaert at pcsierteelt.be<mailto:Joachim.Audenaert at pcsierteelt.be>>
Cc: "r-help at r-project.org<mailto:r-help at r-project.org>" <r-help at r-project.org<mailto:r-help at r-project.org>>
Date: 16/04/2015 13:41
Subject: RE: [R] melt function chooses wrong id variable with large datasets
________________________________
Hi
With this dataset I get
> dd.m0<-melt(dataset, na.rm=T)
Using norm as id variables
> head(dd.m0)
norm variable value
1 45.8713463281901 januari 38.1
2 24.047250681782984 januari 32.4
3 3.7533684144746324 januari 34.5
4 38.594241119279324 januari 20.7
5 26.391897460120358 januari 21.5
6 61.746470001194638 januari 23.1
>
or
dd.m<-melt(dataset, id.vars=NULL, na.rm=T)
> head(dd.m)
variable value
1 januari 38.1
2 januari 32.4
3 januari 34.5
4 januari 20.7
5 januari 21.5
6 januari 23.1
> tail(dd.m)
variable value
255 norm 4.856812959269508
256 norm 5.3982910143166514
257 norm 46.553976273304215
258 norm 17.566272518985429
259 norm 20.552451905814117
260 norm 61.894775704479279
The latter will put norm to the same column as months. Is it intended?
Maybe you want
> dd.m1<-melt(dataset[,-13], na.rm=T)
No id variables; using all as measure variables
> head(dd.m1)
variable value
1 januari 38.1
2 januari 32.4
3 januari 34.5
4 januari 20.7
5 januari 21.5
6 januari 23.1
> tail(dd.m1)
variable value
235 december 20.7
236 december 30.9
237 december 36.2
238 december 21.0
239 december 20.2
240 december 21.3
Cheers
Petr
From: Joachim Audenaert [mailto:Joachim.Audenaert at pcsierteelt.be]
Sent: Thursday, April 16, 2015 1:13 PM
To: PIKAL Petr
Cc: r-help at r-project.org<mailto:r-help at r-project.org>
Subject: RE: [R] melt function chooses wrong id variable with large datasets
Hello,
This is a part of my dataset:
structure(list(januari = c(38.1, 32.4, 34.5, 20.7, 21.5, 23.1,
29.7, 36.6, 36.1, 20.6, 20.4, 30.1, 38.7, 41.4, 37, 36, 37, 38,
23, 26.7), februari = c(31.5, 36.2, 38.2, 26.4, 20.9, 21.5, 30.2,
33.4, 32.6, 22.2, 21.7, 30, 35.7, 32.8, 39.3, 25.5, 23, 19.9,
21.3, 20.8), maart = c(34.2, 27, 24.2, 19.9, 19.7, 21.5, 30.6,
30, 19, 19.6, 20.6, 23.6, 17.9, 17.3, 21.4, 24.1, 20.9, 30.1,
32.6, 21.3), april = c(26.3, 29.6, 30.3, 23.6, 28.4, 20.7, 24.1,
27.3, 23.2, 18.3, 24.6, 27.4, 20.4, 18.1, 25.2, 19.8, 21, 23.7,
19.6, 18.1), mei = c(23.7, 24, 17.2, 23.2, 25.2, 17.2, 16, 15.6,
13.4, 16, 16.8, 14.6, 19.4, 21, 19.5, 18.5, 13.3, 13.7, 14.3,
14.1), juni = c(17.7, 14.2, 16.6, 15.7, 13.7, 14.7, 13.1, 12.9,
15.4, 11.9, 15.2, 15.3, 16.5, 16.1, 11.7, 11.2, 11.5, 10.8, 16.1,
14.8), juli = c(15.7, 14.5, 10.8, 10.5, 13.4, 12.2, 13.2, 13,
12.4, 13.1, 9.8, 10.5, 13.4, 11, 13.1, 15, 16.7, 16.1, 18.2,
15.7), augustus = c(12.9, 12.8, 15.2, 14.5, 17.2, 14.5, 14.4,
11, 13.1, 13.6, 14.6, 12.7, 13.6, 12.7, 15.5, 17.4, 15.2, 14.2,
17.7, 19.2), september = c(15.6, 15.5, 15.9, 15.1, 16, 19.4,
21.5, 23.7, 18.7, 23.8, 18, 16.2, 18.5, 20.6, 18.3, 22.5, 26.9,
19.4, 15.9, 20.5), oktober = c(21.4, 20.8, 14, 17, 23, 26.4,
19.6, 22.7, 26.9, 14.7, 15.2, 19.8, 26.9, 20.2, 14.3, 14.8, 18.5,
21.7, 21.4, 21.8), november = c(24.7, 26.2, 29, 21.6, 17.1, 16.9,
19.1, 24.7, 25.4, 19.8, 18.2, 16.3, 17, 17.7, 15.5, 14.7, 15.8,
19.9, 20.4, 23.3), december = c(19.8, 27, 21, 33, 22.6, 28.3,
21.1, 19, 17.3, 27, 30.2, 24.8, 17.9, 17.9, 20.7, 30.9, 36.2,
21, 20.2, 21.3), norm = c("45.8713463281901", "24.047250681782984",
"3.7533684144746324", "38.594241119279324", "26.391897460120358",
"61.746470001194638", "6.8321020448487992", "11.933109250115226",
"51.951891096493924", "37.424611852237945", "5.1587836676942374",
"36.552835044409434", "31.781209673851027", "29.09146215582853",
"4.856812959269508", "5.3982910143166514", "46.553976273304215",
"17.566272518985429", "20.552451905814117", "61.894775704479279"
)), .Names = c("januari", "februari", "maart", "april", "mei",
"juni", "juli", "augustus", "september", "oktober", "november",
"december", "norm"), row.names = c(NA, 20L), class = "data.frame")
I transform my dataset with the following script:
y <- melt(dataset,na.rm=TRUE)
variable <- y[,1]
value <- y[,2]
and can then perform a levene test as follows:
LEVENE <- leveneTest(value~variable,y)
When the dataset is small, lets say less than 100 values per column everything works great. I get the message:
No id variables; using all as measure variables
When the dataset is much bigger I get the following message
Using norm as id variables, why does this function pick norm as id variable? and how can I tell R that each column title is my variable
Met vriendelijke groeten - With kind regards,
Joachim Audenaert
onderzoeker gewasbescherming - crop protection researcher
PCS | proefcentrum voor sierteelt - ornamental plant research
________________________________
Schaessestraat 18, 9070 Destelbergen, België
T: +32 (0)9 353 94 71 | F: +32 (0)9 353 94 95
E: joachim.audenaert at pcsierteelt.be<mailto:joachim.audenaert at pcsierteelt.be> | W: www.pcsierteelt.be<http://www.pcsierteelt.be/>
From: PIKAL Petr <petr.pikal at precheza.cz<mailto:petr.pikal at precheza.cz>>
To: Joachim Audenaert <Joachim.Audenaert at pcsierteelt.be<mailto:Joachim.Audenaert at pcsierteelt.be>>, "r-help at r-project.org<mailto:r-help at r-project.org>" <r-help at r-project.org<mailto:r-help at r-project.org>>
Date: 16/04/2015 12:13
Subject: RE: [R] melt function chooses wrong id variable with large datasets
________________________________
Hi
There is something weird with your data and melt function.
AFAIK melt does not use first row as id.variables.
What is result of
str(dataset)
Instead of
melt(dataset,id.vars=dataset[1,], na.rm=TRUE)
melt expects something like
melt(dataset, id.vars=c("norm, "jaar"), na.rm=TRUE)
If you want more specific answer you shall show us part of your data, preferably copy output of
dput(dataset[1:20,])
into your mail.
Cheers
Petr
> -----Original Message-----
> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Joachim
> Audenaert
> Sent: Thursday, April 16, 2015 11:37 AM
> To: r-help at r-project.org<mailto:r-help at r-project.org>
> Subject: [R] melt function chooses wrong id variable with large
> datasets
>
> Hello all,
>
> I'm using a large dataset consisting of 2 groups of data, 2 columns in
> excel with a header (group name) and 15 000 rows of data. I would like
> like to compare this data, so I transform my dataset with the melt
> function to get 1 column of data and 1 column of ID variables, then I
> can apply different statistical tests. With small datasets this works
> great, the melt function automatically chooses the name in row 1 as ID
> variable and melts the data, thus giving me a matrix with all ID
> variables in column one and the data accordingly in column 2.
> With this big dataset however it chooses the whole first column as ID
> variables in stead of the first row. Is there a reason why this happens
> and how can I make sure the first row is chosen as ID variabele and the
> lower rows as data?
>
> If I specify that I want the first row to be the id variable I also get
> error.
>
> melt(dataset,id.vars=dataset[1,], na.rm=TRUE)
>
> Error: id variables not found in data: norm, jaar
>
> Are there alternative ways to create a good reshaped dataset?
>
> Met vriendelijke groeten - With kind regards,
>
> Joachim Audenaert
> onderzoeker gewasbescherming - crop protection researcher
>
> PCS | proefcentrum voor sierteelt - ornamental plant research
>
> Schaessestraat 18, 9070 Destelbergen, Belgi
> T: +32 (0)9 353 94 71 | F: +32 (0)9 353 94 95
> E: joachim.audenaert at pcsierteelt.be<mailto:joachim.audenaert at pcsierteelt.be> | W: www.pcsierteelt.be
>
> Heb je je individuele begeleiding bemesting (CVBB) al aangevraagd? |
> Het PCS op LinkedIn Disclaimer | Please consider the environment before
> printing. Think green, keep it on the screen!
> [[alternative HTML version deleted]]
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Tento e-mail a jakékoliv k nìmu pøipojené dokumenty jsou dùvìrné a jsou urèeny pouze jeho adresátùm.
Jestli¾e jste obdr¾el(a) tento e-mail omylem, informujte laskavì neprodlenì jeho odesílatele. Obsah tohoto emailu i s pøílohami a jeho kopie vyma¾te ze svého systému.
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- a obsahuje-li nabídku, je adresát oprávnìn nabídku bezodkladnì pøijmout; Odesílatel tohoto e-mailu (nabídky) vyluèuje pøijetí nabídky ze strany pøíjemce s dodatkem èi odchylkou.
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Tento e-mail a jakékoliv k nìmu pøipojené dokumenty jsou dùvìrné a jsou urèeny pouze jeho adresátùm.
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V pøípadì, ¾e je tento e-mail souèástí obchodního jednání:
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- a obsahuje-li nabídku, je adresát oprávnìn nabídku bezodkladnì pøijmout; Odesílatel tohoto e-mailu (nabídky) vyluèuje pøijetí nabídky ze strany pøíjemce s dodatkem èi odchylkou.
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- odesílatel tohoto emailu informuje, ¾e není oprávnìn uzavírat za spoleènost ¾ádné smlouvy s výjimkou pøípadù, kdy k tomu byl písemnì zmocnìn nebo písemnì povìøen a takové povìøení nebo plná moc byly adresátovi tohoto emailu pøípadnì osobì, kterou adresát zastupuje, pøedlo¾eny nebo jejich existence je adresátovi èi osobì jím zastoupené známá.
This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients.
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The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email.
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