[R] how to do averaging of two tables (rows with columns)

[Rainer Schuermann]

Based upon my understanding of your problem, this should do the job:

> d3 <- dat1
> d3[2] <- ifelse( d3[2] == 1, dat2[1,2], NA )
> d3[3] <- ifelse( d3[3] == 1, dat2[2,2], NA )
> d3[4] <- "Nodata"
> d3[5] <- ifelse( d3[5] == 1, dat2[3,2], NA )
> d3$average <- rowMeans( d3[c(2,3,5)], na.rm = TRUE )

d3 is now the same as your dat3.

This is cumbersome, but maybe a starting point. 

Rgds,
Rainer


On Friday 11 May 2012 01:35:18 Kristi Glover wrote:
> 
> Hi R user,
> I saw some errors in the dat1.
> The correct dat1 is 
> 
> dat1  <- structure(list(X = structure(1:4, .Label = c("Plot1", 
> "Plot2", "plot3", "plot4"), class = "factor"), speciesX = c(1L, 0L, 0L, 1L), speciesY = 
> c(0L, 1L, 0L, 0L), speciesZ = c(1L, 1L, 0L, 1L), speciesXX = c(1L, 0L, 1L, 0L)), .Names = 
> c("X", "speciesX", "speciesY", "speciesZ", "speciesXX"), class = "data.frame", row.names = 
> c(NA, -4L)) 
> 
> 
> cheers,
> K
> 
> 
> > From: kristi.glover at hotmail.com
> > To: r-help at r-project.org
> > Date: Thu, 10 May 2012 19:03:38 -0300
> > Subject: [R] how to do averaging of  two tables (rows with columns)
> > 
> > 
> > 
> > Hi R user,
> >  I finally able to send you the table in 
> > readable format. I have seen that  some of you do send tables in email 
> > when asking questions, but why i could not send. Any way 
> > some of you helped me to send you the example table in a readable format. 
> > 
> > now,
> >  I want to concentrate on my problem. I am trying to get the information
> >  (dat 3)  from dat 1 and 2 in R. I have very big data but these data are
> >  just hypothetical data. my data structures are exactly same as dat 1 
> > and dat 2. I created dat 3 and dat4 manually to show what information I 
> > wanted to have.  
> > 
> > I am struggling to figure it out how I can do in R. I think it is not difficult.  I hope any one can help me. 
> > 
> > dat1 is the table of species occurrence (o means species absence, 1 means species presence).
> > 
> > dat1  <- structure(list(X = structure(1:4, .Label = c("Plot1", 
> > "Plot2", "plot3", "plot4"), class = "factor"), speciesX = c(1L, 0L, 1L, 0L), speciesY = 
> > c(0L, 1L, 0L, 0L), speciesZ = c(1L, 1L, 0L, 1L), speciesXX = c(0L, 0L, 1L, 0L)), .Names = 
> > c("X", "speciesX", "speciesY", "speciesZ", "speciesXX"), class = "data.frame", row.names = 
> > c(NA, -4L)) 
> > 
> > dat2 is the species tolerances value in each environmental variable
> > 
> > dat2  <- structure(list(X = structure(c(1L,
> > 
> > 3L, 2L), .Label = c("SpeciesX", "SpeciesXX", "SpeciesY"), class = "factor"), EnviA =
> > 
> > c(0.21, 0.1, 0.14), EnviB = c(0.4, 0.15, 0.16), EnviC = c(0.17, 0.18, 0.19)), .Names =
> > 
> > c("X", "EnviA", "EnviB", "EnviC"), class = "data.frame", row.names = c(NA, -3L))
> > 
> > 
> >  ## note (here in dat 2 there is no "species Z" you can see that )
> > Now, I want to get the average value of tolerances in each grid. like dat 3
> > 
> > the dat3 is based on the column EnviA.
> > 
> > dat3  <-structure(list(X = structure(1:4, .Label = c("plot1",
> > 
> >  "plot2", "plot3", "plot4"), class = "factor"), speciesX = c(0.21, NA, NA,
> > 
> >  0.21), speciesY = c(NA, 0.1, NA, NA), speciesZ = structure(c(1L, 1L, 1L,
> > 
> >  1L), .Label = "Nodata", class = "factor"), speciesXX = c(0.14, NA, 0.14,
> > 
> >  NA), average = c(0.175, 0.1, 0.14, 0.21)), .Names = c("X", "speciesX",
> > 
> >  "speciesY", "speciesZ", "speciesXX", "average"), class = "data.frame",
> > 
> >  row.names = c(NA, -4L))
> > 
> > 
> > dat4 is same thing as dat3 but here i used EnviB instead of EnviA.
> > 
> > dat4  <- structure(list(X = structure(1:4, .Label = c("plot1", "plot2", "plot3", "plot4"), class =
> > 
> > "factor"), speciesX = c(0.4, NA, NA, 0.4), speciesY = c(NA, 0.15, NA, NA), speciesZ =
> > 
> > structure(c(1L, 1L, 1L, 1L), .Label = "Nodata", class = "factor"), speciesXX = c(0.16, NA,
> > 
> > 0.16, NA), average = c(0.28, 0.15, 0.16, 0.4)), .Names = c("X", "speciesX", "speciesY",
> > 
> > "speciesZ", "speciesXX", "average"), class = "data.frame", row.names = c(NA, -4L))  
> > 
> > I hope you understand my problem and you can help me.
> > 
> > Thanks 
> > 
> > Kristi
> > 
> > 
> >  		 	   		   		 	   		   		 	   		  
> > 	[[alternative HTML version deleted]]
> > 
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> 
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.