Dear R User,
I was wondering how I can unstack my data. For example I have following data set
data<-structure(list(siteS = structure(c(3L, 3L, 3L, 4L, 4L, 5L, 5L,
5L, 5L, 5L, 1L, 1L, 1L, 2L), .Label = c("11a", "12d",
"1a", "2v",
"6a"), class = "factor"), species = structure(c(1L, 2L, 3L,
1L,
4L, 1L, 4L, 5L, 2L, 3L, 1L, 6L, 3L, 1L), .Label = c("sa",
"sb",
"sc", "sd", "se", "sg"), class =
"factor"), abundance = c(31L,
55L, 62L, 42L, 40L, 30L, 84L, 10L, 23L, 74L, 11L, 51L, 37L, 15L
)), .Names = c("siteS", "species", "abundance"),
class = "data.frame", row.names = c(NA,
-14L))
I wanted to have this data into following format
newData<-structure(list(siteS = structure(c(3L, 4L, 5L, 1L, 2L), .Label =
c("11a",
"12d", "1a", "2v", "6a"), class =
"factor"), sa = c(31L, 42L,
30L, 11L, 15L), sb = c(55L, NA, 84L, NA, NA), sc = c(62L, NA,
10L, 37L, NA), sd = c(NA, 40L, 23L, NA, NA), se = c(NA, NA, 74L,
NA, NA), sg = c(NA, NA, NA, 51L, NA)), .Names = c("siteS",
"sa",
"sb", "sc", "sd", "se", "sg"),
class = "data.frame", row.names = c(NA,
-5L))
I tried several ways such as:> data.frame(unstack(data, species~siteS))
Error in data.frame(`11a` = c("sa", "sg", "sc"),
`12d` = "sa", `1a` = c("sa", :
arguments imply differing number of rows: 3, 1, 2, 5
[[alternative HTML version deleted]]
Hi Kristi
It doesn't seem that unstack can do it for you. Here is one way:
library(reshape)
data.molten <- melt(data, id = c("siteS", "species"))
cast(data.molten, siteS ~ species)
siteS sa sb sc sd se sg
1 11a 11 NA 37 NA NA 51
2 12d 15 NA NA NA NA NA
3 1a 31 55 62 NA NA NA
4 2v 42 NA NA 40 NA NA
5 6a 30 23 74 84 10 NA>
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> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at
r-project.org]
> On Behalf Of Kristi Glover
> Sent: 27. maj 2014 12:54
> To: R-help
> Subject: [R] How to unstack three columns into rows?
>
> Dear R User,
> I was wondering how I can unstack my data. For example I have following
> data set
>
> data<-structure(list(siteS = structure(c(3L, 3L, 3L, 4L, 4L, 5L, 5L,
> 5L, 5L, 5L, 1L, 1L, 1L, 2L), .Label = c("11a", "12d",
"1a", "2v",
> "6a"), class = "factor"), species = structure(c(1L, 2L,
3L, 1L,
> 4L, 1L, 4L, 5L, 2L, 3L, 1L, 6L, 3L, 1L), .Label = c("sa",
"sb",
> "sc", "sd", "se", "sg"), class =
"factor"), abundance = c(31L,
> 55L, 62L, 42L, 40L, 30L, 84L, 10L, 23L, 74L, 11L, 51L, 37L, 15L
> )), .Names = c("siteS", "species",
"abundance"), class = "data.frame",
> row.names = c(NA,
> -14L))
>
> I wanted to have this data into following format
>
> newData<-structure(list(siteS = structure(c(3L, 4L, 5L, 1L, 2L), .Label
= c("11a",
> "12d", "1a", "2v", "6a"), class =
"factor"), sa = c(31L, 42L,
> 30L, 11L, 15L), sb = c(55L, NA, 84L, NA, NA), sc = c(62L, NA,
> 10L, 37L, NA), sd = c(NA, 40L, 23L, NA, NA), se = c(NA, NA, 74L,
> NA, NA), sg = c(NA, NA, NA, 51L, NA)), .Names = c("siteS",
"sa",
> "sb", "sc", "sd", "se",
"sg"), class = "data.frame", row.names = c(NA,
> -5L))
>
> I tried several ways such as:
> > data.frame(unstack(data, species~siteS))
> Error in data.frame(`11a` = c("sa", "sg",
"sc"), `12d` = "sa", `1a` = c("sa", :
> arguments imply differing number of rows: 3, 1, 2, 5
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at 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.
Hi cast(data, siteS~species) gives you quite close result, only row ordering is different Regards Petr> -----Original Message----- > From: r-help-bounces at r-project.org [mailto:r-help-bounces at r- > project.org] On Behalf Of Kristi Glover > Sent: Tuesday, May 27, 2014 12:54 PM > To: R-help > Subject: [R] How to unstack three columns into rows? > > Dear R User, > I was wondering how I can unstack my data. For example I have following > data set > > data<-structure(list(siteS = structure(c(3L, 3L, 3L, 4L, 4L, 5L, 5L, > 5L, 5L, 5L, 1L, 1L, 1L, 2L), .Label = c("11a", "12d", "1a", "2v", > "6a"), class = "factor"), species = structure(c(1L, 2L, 3L, 1L, 4L, 1L, > 4L, 5L, 2L, 3L, 1L, 6L, 3L, 1L), .Label = c("sa", "sb", "sc", "sd", > "se", "sg"), class = "factor"), abundance = c(31L, 55L, 62L, 42L, 40L, > 30L, 84L, 10L, 23L, 74L, 11L, 51L, 37L, 15L )), .Names = c("siteS", > "species", "abundance"), class = "data.frame", row.names = c(NA, > -14L)) > > I wanted to have this data into following format > > newData<-structure(list(siteS = structure(c(3L, 4L, 5L, 1L, 2L), .Label > = c("11a", "12d", "1a", "2v", "6a"), class = "factor"), sa = c(31L, > 42L, 30L, 11L, 15L), sb = c(55L, NA, 84L, NA, NA), sc = c(62L, NA, 10L, > 37L, NA), sd = c(NA, 40L, 23L, NA, NA), se = c(NA, NA, 74L, NA, NA), sg > = c(NA, NA, NA, 51L, NA)), .Names = c("siteS", "sa", "sb", "sc", "sd", > "se", "sg"), class = "data.frame", row.names = c(NA, > -5L)) > > I tried several ways such as: > > data.frame(unstack(data, species~siteS)) > Error in data.frame(`11a` = c("sa", "sg", "sc"), `12d` = "sa", `1a` > c("sa", : > arguments imply differing number of rows: 3, 1, 2, 5 > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at 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.________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho kopie vyma?te ze sv?ho syst?mu. Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat. Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou modifikacemi ?i zpo?d?n?m p?enosu e-mailu. V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?: - vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en? smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu. - a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn? p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky ze strany p??jemce s dodatkem ?i odchylkou. - trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech. - odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn? zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje, p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen? zn?m?. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. 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. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient.
On May 27, 2014, at 3:54 AM, Kristi Glover wrote:> > data<-structure(list(siteS = structure(c(3L, 3L, 3L, 4L, 4L, 5L, 5L, > 5L, 5L, 5L, 1L, 1L, 1L, 2L), .Label = c("11a", "12d", "1a", "2v", > "6a"), class = "factor"), species = structure(c(1L, 2L, 3L, 1L, > 4L, 1L, 4L, 5L, 2L, 3L, 1L, 6L, 3L, 1L), .Label = c("sa", "sb", > "sc", "sd", "se", "sg"), class = "factor"), abundance = c(31L, > 55L, 62L, 42L, 40L, 30L, 84L, 10L, 23L, 74L, 11L, 51L, 37L, 15L > )), .Names = c("siteS", "species", "abundance"), class = "data.frame", row.names = c(NA, > -14L)) > > I wanted to have this data into following format > > newData<-structure(list(siteS = structure(c(3L, 4L, 5L, 1L, 2L), .Label = c("11a", > "12d", "1a", "2v", "6a"), class = "factor"), sa = c(31L, 42L, > 30L, 11L, 15L), sb = c(55L, NA, 84L, NA, NA), sc = c(62L, NA, > 10L, 37L, NA), sd = c(NA, 40L, 23L, NA, NA), se = c(NA, NA, 74L, > NA, NA), sg = c(NA, NA, NA, 51L, NA)), .Names = c("siteS", "sa", > "sb", "sc", "sd", "se", "sg"), class = "data.frame", row.names = c(NA, > -5L))> xtdat <- xtabs( abundance ~ siteS + species, data)# Zeros rather than NA> require(gtools) # for mixedsort()Loading required package: gtools> xtdat2 <- xtdat[ mixedsort(dimnames(xtdat)[[1]]) ,] > is.na(xtdat2) <- xtdat2==0 > xtdat2species siteS sa sb sc sd se sg 1a 31 55 62 NA NA NA 2v 42 NA NA 40 NA NA 6a 30 23 74 84 10 NA 11a 11 NA 37 NA NA 51 12d 15 NA NA NA NA NA -- David Winsemius Alameda, CA, USA