Hi everyone I have a dataframe "data" wich is the result of join multiple csv (400 rows and 600cols every csv). The "data" dataframe has n rows and m columns (200000 rows and 600 cols) , and I have add a new colum, "csvdata", in which I specify the number of csv at wich those data belong. So, the dataframe "data" looks like: x1 x2 x3 .... xn csvdata 21 23 32 .... 12 1 27 21 39 .... 14 1 24 22 30 .... 11 1 .............................................. 21 24 32 .... 19 2 27 21 39 .... 14 2 .............................................. 27 22 30 .... 11 n I want to store into a matrix the mean values of different substes of data of every csv, for example: region1,1 (rows 1:20,columns 1:20) for every "csvdata" value region 2,1 (rows 21:40,columns 1:20) para every "csvdata" value .... And so on for hole data.frame. I have tryed: area1<-tapply(as.matrix(data[1:20,1]),datos$csvdata,mean,na.rm=T) area2<-tapply(as.matrix(data[1:20,1]),datos$csvdata,mean,na.rm=T) But this error is the output I obtain: Error in tapply(data[1:30, ], datos$nueva, mean, na.rm = T) : arguments must have same length I?m sure that it is not very complex to do it, but I have no idea of how to do it. Thanks for all. [[alternative HTML version deleted]]
Hi See in line> -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Jes?s > Para Fern?ndez > Sent: Wednesday, November 18, 2015 11:20 AM > To: r-help at r-project.org > Subject: [R] Get means of matrix > > Hi everyone > > I have a dataframe "data" wich is the result of join multiple csv (400 > rows and 600cols every csv). The "data" dataframe has n rows and m > columns (200000 rows and 600 cols) , and I have add a new colum, > "csvdata", in which I specify the number of csv at wich those data > belong. > > So, the dataframe "data" looks like: > > x1 x2 x3 .... xn csvdata > 21 23 32 .... 12 1 > 27 21 39 .... 14 1 > 24 22 30 .... 11 1 > .............................................. > 21 24 32 .... 19 2 > 27 21 39 .... 14 2 > .............................................. > 27 22 30 .... 11 n > > > > I want to store into a matrix the mean values of different substes of > data of every csv, for example: > > region1,1 (rows 1:20,columns 1:20) for every "csvdata" value region 2,1 > (rows 21:40,columns 1:20) para every "csvdata" value .... > > And so on for hole data.frame. > > I have tryed: > > area1<-tapply(as.matrix(data[1:20,1]),datos$csvdata,mean,na.rm=T) > area2<-tapply(as.matrix(data[1:20,1]),datos$csvdata,mean,na.rm=T) > > But this error is the output I obtain: > > Error in tapply(data[1:30, ], datos$nueva, mean, na.rm = T) : > arguments must have same lengthMost probably length(datos$csvdata) is not 20 as you specified by data[1:20,1]. If you have already number of csv file in your data$csvdata you can do area <- aggregate(data[,1:n], list(data$csvdata), mean, na.rm=TRUE) If you want to aggregate each 20 rows you can elaborate index ind <- rep(1:20, each=20) area <- aggregate(data[,1:n], list(ind), mean, na.rm=TRUE) Cheers Petr> > I m sure that it is not very complex to do it, but I have no idea of > how to do it. > > Thanks for all. > > > [[alternative HTML version deleted]]________________________________ 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.
Hi Jesus, While I do not have your data and cannot test this, the problem may be that you are using two different names for the data frame. Is this more or less what you want? datos<-data.frame(x1=sample(20:40,40,TRUE),x2=sample(20:40,40,TRUE), x3=sample(20:40,40,TRUE),csvdata=rep(1:2,each=20)) sapply(datos[datos$csvdata==1,],mean,na.rm=TRUE) sapply(datos[datos$csvdata==2,],mean,na.rm=TRUE) Jim On Wed, Nov 18, 2015 at 9:19 PM, Jes?s Para Fern?ndez < j.para.fernandez at hotmail.com> wrote:> Hi everyone > > I have a dataframe "data" wich is the result of join multiple csv (400 > rows and 600cols every csv). The "data" dataframe has n rows and m columns > (200000 rows and 600 cols) , and I have add a new colum, "csvdata", in > which I specify the number of csv at wich those data belong. > > So, the dataframe "data" looks like: > > x1 x2 x3 .... xn csvdata > 21 23 32 .... 12 1 > 27 21 39 .... 14 1 > 24 22 30 .... 11 1 > .............................................. > 21 24 32 .... 19 2 > 27 21 39 .... 14 2 > .............................................. > 27 22 30 .... 11 n > > > > I want to store into a matrix the mean values of different substes of data > of every csv, for example: > > region1,1 (rows 1:20,columns 1:20) for every "csvdata" value > region 2,1 (rows 21:40,columns 1:20) para every "csvdata" value > .... > > And so on for hole data.frame. > > I have tryed: > > area1<-tapply(as.matrix(data[1:20,1]),datos$csvdata,mean,na.rm=T) > area2<-tapply(as.matrix(data[1:20,1]),datos$csvdata,mean,na.rm=T) > > But this error is the output I obtain: > > Error in tapply(data[1:30, ], datos$nueva, mean, na.rm = T) : > arguments must have same length > > I?m sure that it is not very complex to do it, but I have no idea of how > to do it. > > Thanks for all. > > > [[alternative HTML version deleted]] > > > ______________________________________________ > R-help at 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. >[[alternative HTML version deleted]]
Hi:
Here's another way to look at the problem. Instead of manually adding
a new column after k datasets have been read in, read your individual
data files into a list, as long as they all have the same variable
names and the same class (in this case, data.frame). Then create a
vector of names for the list components and use 'apply family' logic
to get the column means, returning the combined results to a data
frame or matrix. Here's a toy example to illustrate the point.
Firstly, three data frames are created and saved to external files:
# Create some artificial data and ship to external files
d1 <- data.frame(x1 = rpois(10, 20), x2 = rpois(10, 23), x3 = rpois(10, 25))
d2 <- data.frame(x1 = rpois(10, 20), x2 = rpois(10, 23), x3 = rpois(10, 25))
d3 <- data.frame(x1 = rpois(10, 20), x2 = rpois(10, 23), x3 = rpois(10, 25))
write.csv(d1, file = "d1.csv", row.names = TRUE, quote = FALSE)
write.csv(d2, file = "d2.csv", row.names = TRUE, quote = FALSE)
write.csv(d3, file = "d3.csv", row.names = TRUE, quote = FALSE)
###
# Now, read them back in and store them in a list object
# Vector of file names to process
files <- paste0("d", 1:3, ".csv")
# Create the list of data frames and assign names to list components
L <- lapply(files, function(x) read.csv(x, header = TRUE))
names(L) <- paste0("d", 1:3)
# Compute column means from each list component and row bind them
# Method 1: base R
do.call(rbind, lapply(L, colMeans))
# Method 2: plyr package
library(plyr)
ldply(L, colMeans)
Dennis
On Wed, Nov 18, 2015 at 2:19 AM, Jes?s Para Fern?ndez
<j.para.fernandez at hotmail.com> wrote:> Hi everyone
>
> I have a dataframe "data" wich is the result of join multiple csv
(400 rows and 600cols every csv). The "data" dataframe has n rows and
m columns (200000 rows and 600 cols) , and I have add a new colum,
"csvdata", in which I specify the number of csv at wich those data
belong.
>
> So, the dataframe "data" looks like:
>
> x1 x2 x3 .... xn csvdata
> 21 23 32 .... 12 1
> 27 21 39 .... 14 1
> 24 22 30 .... 11 1
> ..............................................
> 21 24 32 .... 19 2
> 27 21 39 .... 14 2
> ..............................................
> 27 22 30 .... 11 n
>
>
>
> I want to store into a matrix the mean values of different substes of data
of every csv, for example:
>
> region1,1 (rows 1:20,columns 1:20) for every "csvdata" value
> region 2,1 (rows 21:40,columns 1:20) para every "csvdata" value
> ....
>
> And so on for hole data.frame.
>
> I have tryed:
>
> area1<-tapply(as.matrix(data[1:20,1]),datos$csvdata,mean,na.rm=T)
> area2<-tapply(as.matrix(data[1:20,1]),datos$csvdata,mean,na.rm=T)
>
> But this error is the output I obtain:
>
> Error in tapply(data[1:30, ], datos$nueva, mean, na.rm = T) :
> arguments must have same length
>
> I?m sure that it is not very complex to do it, but I have no idea of how to
do it.
>
> Thanks for all.
>
>
> [[alternative HTML version deleted]]
>
>
> ______________________________________________
> R-help at 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.