similar to: search species with all absence in a presence-absence matrix

Displaying 20 results from an estimated 300 matches similar to: "search species with all absence in a presence-absence matrix"

2013 Sep 27
3
Compare species presence and absence between sites
Dear List, I want to compare the presence and absence of bird species based on the sites in a matrix. The matrix has 5 rows for Island A, B, C, D, and E. It has 100 columns for bird species D001-D100. In each cell of the matrix, the presence-absence of bird species will be recorded as 1 or 0. (For example, if species D001 is found on Island D, the matrix cell of species D001 and Island D
2007 Feb 08
5
remove component from list or data frame
Sorry to ask such a simple question, but I can't find the answer after extensive searching the docs and the web. How do you remove a component from a list? For example say you have: lst<-c(5,6,7,8,9) How do you remove, for example, the third component in the list? lst[[3]]]<-NULL generates an error: "Error: more elements supplied than there are to replace" Also,
2013 Apr 10
6
means in tables
Hi. I have 2 tables, with same dimensions (8000 x 5). Something like: tab1: V1 V2 V3 V4 V5 14.23 1.71 2.43 15.6 127 13.20 1.78 2.14 11.2 100 13.16 2.36 2.67 18.6 101 14.37 1.95 2.50 16.8 113 13.24 2.59 2.87 21.0 118 tab2: V1 V2 V3 V4 V5 1.23 1.1 2.3 1.6 17 1.20 1.8 2.4 1.2 10 1.16 2.6 2.7 1.6 11 1.37 1.5 2.0 1.8 13 1.24 2.9 2.7 2.0 18 I need generate a table of averages, the
2013 Apr 13
2
Comparison of Date format
Hi, ?In the example you provided, it looks like the dates in Date2 happens first.? So, I changed it a bit.? DataA<- read.table(text=" ID,Status,Date1,Date2 ??? ??? ?????? 1,A,3-Feb-01,15-May-01 ??? ??? 1,B,15-May-01,16-May-01 ??? ??? 1,A,16-May-01,3-Sep-01 ??? ??? ??? ??? ??? 1,B,3-Sep-01,13-Sep-01 ??? ??? ??? ??? ??? 1,C,13-Sep-01,26-Feb-04 ??? ??? ??? ??? ???
2008 Oct 01
3
lapply where each list object has multiple parts
Hi. I have a list where each object in the list has multiple parts. I'd like to take the mean of just one part of each object. Is it possible to do this with lapply? If not, can you recommend another function? Thanks. eric > x1 <- c(0,1,2,3) > x2 <- c(7,8) > x3 <- c(2,6,6,8) > x4 <- c(4,8) > > Lst1 <- list(label1 = x1,label2 = x2) > Lst2 <-
2013 Sep 05
2
binary symmetric matrix combination
Hi, May be this helps: m1<- as.matrix(read.table(text=" y1 g24 y1 0 1 g24 1 0 ",sep="",header=TRUE)) m2<-as.matrix(read.table(text="y1 c1 c2 l17 ?y1 0 1 1 1 ?c1 1 0 1 1 ?c2 1 1 0 1 ?l17 1 1 1 0",sep="",header=TRUE)) m3<- as.matrix(read.table(text="y1 h4??? s2???? s30 ?y1 0 1 1 1 ?h4 1 0 1 1 ?s2 1 1 0 1 ?s30 1 1 1
2007 Jul 18
1
Neuman-Keuls
hello, I have programmed this function to calculate the Neuman-Keuls test but I have a problem the function return an empty list and I don't know why. summary(fm1) E <- sqrt((summary(fm1)[[1]]["Residuals","Mean Sq"])/length(LR)) lst <- list() lst1 <- list() lst2 <- list() NK <- function (x) { if (length(x) == 2) { Tstudent <- t.test(subset(exple,
2013 Nov 08
2
making chains from pairs
Hello, having a data frame like test with pairs of characters I would like to create chains. For instance from the pairs A/B and B/I you get the vector A B I. It is like jumping from one pair to the next related pair. So for my example test you should get: A B F G H I C F I K D L M N O P > test V1 V2 1 A B 2 A F 3 A G 4 A H 5 B F 6 B I 7 C F 8 C I 9 C K 10 D L
2013 Apr 18
6
count each answer category in each column
Hey, Is it possible that R can calculate each options under each column and return a summary table? Suppose I have a table like this: Gender Age Rate Female 0-10 Good Male 0-10 Good Female 11-20 Bad Male 11-20 Bad Male >20 N/A I want to have a summary table including the information that how many answers in each category, sth like this: X
2013 Aug 25
1
Capturing the whole output using R
Hi, May be this helps: #Creating some dummy data.? set.seed(24) lst1<-lapply(1:8,function(x) ts(sample(1:25,20,replace=TRUE))) set.seed(49) lst2<-lapply(1:8,function(x) ts(sample(1:45,20,replace=TRUE))) ??Find_Max_CCF() #No vignettes or demos or help files found with alias or concept or #title matching ?Find_Max_CCF? using regular expression matching. Found a function with the same
2013 May 02
3
R issue with unequal large data frames with multiple columns
I'm a bit of an amateur R programmer. I can do simple R scenarios but my handle on complex grammatical issues isn't steady. I have 12 CSV files that I've read into dataframes. Each has 8 columns and over 2000000 rows. Each dataframe has data associated by time component and a date component in the format of: X.DATE and then X.TIME X.DATE is in the format of MMDDYYYY and X.TIME is
2006 Jul 26
1
.Call question
Writing R Ext says to treat R objects that are arguments to .Call as read only (i.e. don't modify). I have a long list of lists that and I want to avoid the overhead of a copy in my C code. I would just like to modify some of the elements of list by replacing them with elements of exactly the same size/type. below is an example of the essence of the problem. This seems to work. Is this
2013 Jun 07
4
matched samples, dataframe, panel data
I R-helpers #I have a data panel of thousands of firms, by year and industry and #one dummy variable that separates the firms in two categories: 1 if the firm have an auditor; 0 if not #and another variable the represents the firm dimension (total assets in thousand of euros) #I need to create two separated samples with the same number os firms where #one firm in the first have a corresponding
2009 Sep 29
1
Comparing vectors from lists
Hi guys, I still did not solve my problem properly! I have to compare the values of two lists of 250 numbers as a result of using the ?by function! List1 of 250  $ 0   : num [1:28] 22 11 31...  $ 1   : num [1:15] 12 14 9 ... .. .. ..  - attr(*, "dim")= int 250  - attr(*, "dimnames")=List of 1 List2 of 250  $ 0   : num [1:24] 20 12 22...  $ 1   : num [1:17] 11 12 19 ... .. ..
2014 Aug 25
5
problema con campos que tienen formato fecha
Hola Javier, Muchas gracias por responder tan rápido! Yo trabajo en Mac OS X 10.9.4. Versión 0.98.953 de RStudio. Versión 3.0.2 (2013-09-25) de R. ## Este es el script que estoy trabajando. Se trata de una rutina para automatizar el cálculo de la duración del evento. setwd("/Users/angelacamargosanabria/Documents/ANGELITA/1-DOC/1-TESIS/4-PAPERS/1-Mamiferos/DATOS/Bases") BASE <-
2002 Nov 28
4
Mime-Version: 1.0
I am using expression() to incorporate text into graphics. To create a superscript, I use the '^' character. Can someone please tell me the character to use to create a subscript? Thank you ----------------------------------------------------- Christine Donnelly Statistical Consulting Unit The Graduate School John Dedman Mathematical Sciences Building (Bldg 27) Australian National
2013 Mar 10
2
list + lapply insead of matrix + apply
I need to develop a simple list manipulation. Although it seems easier to do it in matrix form, but I need it in list form. I have a matrix x <- matrix(c(12.1, 3.44, 0.1, 3, 12, 33.1, 1.1, 23), nrow=2) for list form example, the conversion is x.list <- lapply(seq_len(nrow(x)), function(i) x[i,]) ### list version calcnorm=function(a, b){ diff <- mapply("-", a, b)
2013 Apr 24
2
Distance matrices Combinations
Dear UseRs, MY PROBLEM IS A SMALL PIECE OF A REAL BIG AND A COMPLICATED PROBLEM. IF I DELIBERATE IN A VERY SIMPLE WAY THEN ALL I WANT IS TO PUT ALL THE POSSIBLE COMBINATIONS OF 75 DISTANCE MATRICES (BY TAKING 4 MATRICES, MORE COMMONLY 75C4), in the following equation. t<-as.matrix((MAT1)^2+(MAT2)^2+(MAT3)^2+(MAT4)^2+,upper=T,diag=T)) Then "1215450" values of "t"(one for
2012 Nov 09
1
Mean of matrices entries in list of lists
Hey there, I've got a list of lists with matrices: A list with 13 entries (representing years), each of them another list with 12 matrices (representing one month). In each matrix there are as many rows as there are hours in the different months and 2 columns, since two meteorological parameters are measured. What I want to do is to calculate the hourly mean values for each month over the
2013 Aug 30
4
mean
When I try to apply mean to a list, I get the answer : argument is not numeric or logical: returning NA Could you help me? (I am a very beginner) -- View this message in context: http://r.789695.n4.nabble.com/mean-tp4674999.html Sent from the R help mailing list archive at Nabble.com.