So I have been working through a population ecology exercise using the popbio package in R-Studio that focuses on using Leslie Matrix's. I have successfully created a Leslie matrix with the proper dimensions using the Fecundity (mx) and Annual Survival values (sx) that I have calculated with my life table. I then am trying to use the pop.projection function in the popbio package to multiply my Leslie matrix (les.mat) by a starting population vector (N0) followed by the number of time intervals (4 years). It is my understanding that you should be able to take a Leslie matrix and multiply by a population vector to calculate a population size after a set number of time intervals. Have I done something wrong here, when I try to run my pop.projection line of code I get the following error message in R:
"> projA <- pop.projection(les.mat,N0,10)
Error in A %*% n : non-conformable arguments"
Could the problem be an issue with my pop.projection function? I am thinking it may be an issue with by N0 argument (population vector), when I look at my N0 values it seems like it has been saved in R as a "Numeric Type", should I be converting it into its own matrix, or as it's own vector somehow to get my pop.projection line of code to run? Any advice would be greatly appreciated, the short code I have been using will be linked below!
Sx <- c(0.8,0.8,0.7969,0.6078,0.3226,0)
mx <- c(0,0,0.6,1.09,0.2,0)
Fx <- mx # fecundity values
S <- Sx # dropping the first value
F <- Fx
les.mat <- matrix(rep(0,36),nrow=6)
les.mat[1,] <- F
les.mat
for(i in 1:5){
les.mat[(i+1),i] <- S[i]
}
les.mat
N0 <- c(100,80,64,51,31,10,0)
projA <- pop.projection(les.mat,N0,10)
The function uses matrix multiplication on the first and second arguments so they must match. The les.mat matrix is 6x6, but N0 is length 7. Try
projA <- pop.projection(les.mat, N0[-7], 10) # Delete last value
or
projA <- pop.projection(les.mat, N0[-1], 10) # Delete first value
Related
I am trying to calculate the area under the curve for every 10ms of a short piece of EEG wave. To first practice this I made a small dataset to run the auc (from package {flux}) function on.
x <- seq(1:10)
y <- c(0:4,5:1)
df <- data.frame(x,y)
attach(df)
plot(x,y)
for (i in 1:10){
x1 <- c(i,(i+1))
y1 <- c(subset(y, x == i),subset(y, x == (i+1)))
auc(x1,y1,thresh = 0)
rm(y1,x1,i)
}
The loop should try to subset two data points from each variable and then run a auc over those data points. However, when running the loop, I get this error:
Error in seq.default(x[1], x[2], length.out = dens) : 'to' must be a finite number
When I run the subset and auc code outside of the loop, it works no problem. Can anyone tell me what's going wrong in the loop?
Thanks for updating the question. It's not because of the control statement (for loop), the error gets thrown precisely when i=10 -- because the length of your x-coords and y-coords vectors are different. Specifically c(10,11) vs c(1). But you have no point at x=11 !
just stop the loop early, at the appropriate time
In R, I am trying to build the matrix 'pi' through an inductive process, I've tried the following code:
lambda<-0.2 #Tuning Parameter
T<-521
pi<-matrix(0,ncol=3,nrow=T-1) #Empty matrix for portfolio pi, 3 columns for 3 stocks, T-1 time periods
pi[1,]<-1/3 #starting weights: Equal Weighted Portfolio
V<-matrix(0,ncol=1,nrow=T-1) #Empty vector for portfolio relative value process
V[1] <- 1 #Starting wealth = $1
FE<-matrix(0,ncol=1,nrow=T-1) #Emptry vector for free energy at each time point
K<-matrix(0,ncol=1,nrow=T-1) #Empty vector for kappa (convex parameter)
#will only be T-2 values for K=kappa, ignore the first to keep index the same as other variables
for( i in (1:T-1)){
V[i+1]<-sum(pi[i,]*(1+r[i,]))/Vmu[i+1]
FE[i]<-log(V[i+1]) - log(V[i]) - sum(pi[i,]*(mu[i+1,]/mu[i,]))
K[i+1]<-min( 1 , lambda*FE[i]/abs( sum(1+(log(pi[i,]/mu[i+1]))) ) )
pi[i+1,]<-pi[i,] + K[i+1]*(mu[i+1,]-pi[i,])
}
the code isn't working, and I am just wondering if the for loop works in the following steps:
find V[2] then FE [2] then K[2] then pi[2] then repeat for 3,4,...
(which is what I want)
or is it doing
2. find V[2], V[3],...........V[T] then FE[2],FE[3],....,FE[T],......etc
(which is not what I want.)
If it is doing it the second way, how can I adjust it so that it follows the first method?
i'm comparing different measures of distance and similarity for vector profiles (Subtest results) in R, most of them are easy to compute and/or exist in dist().
Unfortunately, one that might be interesting and is to difficult for me to calculate myself is Cattel's Rp. I can not find it in R.
Does anybody know if this exists already?
Or can you help me to write a function?
The formula (Cattell 1994) of Rp is this:
(2k-d^2)/(2k + d^2)
where:
k is the median for chi square on a sample of size n;
d is the sum of the (weighted=m) difference between the two profiles,
sth like: sum(m(x(i)-y(i)));
one thing i don't know is, how to get the chi square median in there
Thank you
What i get without defining the k is:
Rp.Cattell <- function(x,y){z <- (2k-(sum(x-y))^2)/(2k+(sum(x-y))^2);return(z)}
Vector examples are:
x <- c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758)
y <- c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925)
They are measures by the same device, but related to different bodyparts. They don't need to be standartised or weighted, i would say.
This page gives a general formula for k, and then gives a more thorough method using SAS/IML which pretty much gives the same results. So I used the general formula, added calculation of degrees of freedom, which leads to this:
Rp.Cattell <- function(x,y) {
dof <- (2-1) * (length(y)-1)
k <- (1-2/(9*dof))^3
z <- (2*k-sum(sum(x-y))^2)/(2*k+sum(sum(x-y))^2)
return(z)
}
x <- c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758)
y <- c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925)
Rp.Cattell(x, y)
# [1] -0.9012083
Does this figure appear to make sense?
Trying to verify the function, I found out now that the median of chisquare is the chisquare value for 50% probability - relating to random. So the function should be:
Rp.Cattell <- function(x,y){
dof <- (2-1) * (length(y)-1)
k <- qchisq(.50, df=dof)
z <- (2k-(sum(x-y))^2)/(2k+(sum(x-y))^2);
return(z)}
It is necessary though to standardize the Values before, so the results are distributed correctly.
So:
library ("stringr")
# they are centered already
x <- as.vector(scale(c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758),center=F, scale=T))
y <- as.vector(scale(c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925),center=F, scale=T))
Rp.Cattell(x, y) -0.584423
This sounds reasonable now - or not?
I consider calculation of z is incorrect.
You need to calculate the sum of the squared differences. Not the square of the sum of differences. Besides product operator is missing in 2k.
It should be
z <- (2*k-sum((x-y)^2))/(2*k+sum((x-y)^2))
Do you agree?
I need your helps to explain how I can obtain the same result as this function does:
gini(x, weights=rep(1,length=length(x)))
http://cran.r-project.org/web/packages/reldist/reldist.pdf --> page 2. Gini
Let's say, we need to measure the inocme of the population N. To do that, we can divide the population N into K subgroups. And in each subgroup kth, we will take nk individual and ask for their income. As the result, we will get the "individual's income" and each individual will have particular "sample weight" to represent for their contribution to the population N. Here is example that I simply get from previous link and the dataset is from NLS
rm(list=ls())
cat("\014")
library(reldist)
data(nls);data
help(nls)
# Convert the wage growth from (log. dollar) to (dollar)
y <- exp(recent$chpermwage);y
# Compute the unweighted estimate
gini_y <- gini(y)
# Compute the weighted estimate
gini_yw <- gini(y,w=recent$wgt)
> --- Here is the result----
> gini_y = 0.3418394
> gini_yw = 0.3483615
I know how to compute the Gini without WEIGHTS by my own code. Therefore, I would like to keep the command gini(y) in my code, without any doubts. The only thing I concerned is that the way gini(y,w) operate to obtain the result 0.3483615. I tried to do another calculation as follow to see whether I can come up with the same result as gini_yw. Here is another code that I based on CDF, Section 9.5, from this book: ‘‘Relative
Distribution Methods in the Social Sciences’’ by Mark S. Handcock,
#-------------------------
# test how gini computes with the sample weights
z <- exp(recent$chpermwage) * recent$wgt
gini_z <- gini(z)
# Result gini_z = 0.3924161
As you see, my calculation gini_z is different from command gini(y, weights). If someone of you know how to build correct computation to obtain exactly
gini_yw = 0.3483615, please give me your advices.
Thanks a lot friends.
function (x, weights = rep(1, length = length(x)))
{
ox <- order(x)
x <- x[ox]
weights <- weights[ox]/sum(weights)
p <- cumsum(weights)
nu <- cumsum(weights * x)
n <- length(nu)
nu <- nu/nu[n]
sum(nu[-1] * p[-n]) - sum(nu[-n] * p[-1])
}
This is the source code for the function gini which can be seen by entering gini into the console. No parentheses or anything else.
EDIT:
This can be done for any function or object really.
This is bit late, but one may be interested in concentration/diversity measures contained in the [SciencesPo][1] package.
I am attempting to create a weights object in R with the mat2listw function. I have a very large spatial weights matrix (roughly 22,000x22,000)
that was created in Excel and read into R, and I'm now trying to implement:
library(spdep)
SW=mat2listw(matrix)
I am getting the following error:
Error in if (any(x<0)) stop ("values in x cannot be negative"): missing
value where TRUE/FALSE needed.
What's going wrong here? My current matrix is all 0's and 1's, with no
missing values and no negative elements. What am I missing?
I'd appreciate any advice. Thanks in advance for your help!
Here is a simple test to your previous comment:
library(spdep)
m1 <-matrix(rbinom(100, 1, 0.5), ncol =10, nrow = 10) #create a random 10 * 10 matrix
m2 <- m1 # create a duplicate of the first matrix
m2[5,4] <- NA # assign an NA value in the second matrix
SW <- mat2listw(m1) # create weight list matrix
SW2 <- mat2listw(m2) # create weight list matrix
The first matrix one does not fail, but the second matrix does. The real question is now why your weight matrix is created containing NAs. Have you considered creating spatial weight matrix in r? Using dnearneigh or other function.