I want to obtain a dataframe with simulated values which have a specific correlation to each other.
I need to use this function, but in the returned output there are also negative values, which do not have meaning for my purposes:
COR <- function (n, xmean, xsd, ymean, ysd, correlation) {
x <- rnorm(n)
y <- rnorm(n)
z <- correlation * scale(x)[,1] + sqrt(1 - correlation^2) *
scale(resid(lm(y ~ x)))[,1]
xresult <- xmean + xsd * scale(x)[,1]
yresult <- ymean + ysd * z
data.frame(x=xresult,y=yresult)
}
Please note that my question starts from this previous post (currently closed):
another similar discussion
Is there a method able to exclude from the final output all the rows which have at least one negative value? (in another terms, x and y must be always positives).
I spent many hours without any concrete result.....
Filtering rows which have at least one negative value can be done with the apply function, e.g.
df <- simcor(100, 1, 1, 1, 1, 0.8)
filter <- apply(df, 1, function(x) sum(x < 0) > 0)
df <- df[!filter,]
plot(df)
First, I create a dataframe df from your funcion. Then, I apply the function sum(x < 0) > 0 rowwise to the dataframe (the second argument of apply, 1 indicates to go along the first dimension of the dataframe or array). This will create a logical vector that is TRUE for every row with at least one negative value. Subsetting the dataframe with the inverse of that (!filter) leaves you with all rows that have no negative values.
UPDATE:
Seems like the package VineCopula offers functions to create distributions with a given correlation. However, I did not dive into the math as deep so I was not able to fully grasp how copulas (i.e. multivariate probability distributions) work. Using this package, you can at least create e.g. two gaussian distributions.
library(VineCopula)
BC <- BiCop(family = 1, par = 0.9)
sim <- BiCopSim(N = 1000, obj = BC)
cor(sim[,1], sim[,2])
plot(sim)
You might be able to then scale the resulting matrix to achieve a certain standard derivation.
Suppose I have two continuous vectors such like:
set.seed(123)
df <- data.frame(x = rnorm(100),
y = rnorm(100,3,5))
with(df, cor(x,y))
My question is how to find a percentile of x so that to maximize the absolute correlation of x and y such that:
perc <- quantile(df$x, 0.3)
df1 <- subset(df, x > perc)
with(df1, cor(x,y))
Namely how to find perc?
This problem is ill defined. Take your example data set and the function you want to find the maximum of (copied from #coffeinjunky):
set.seed(123)
df <- data.frame(x = rnorm(100),
y = rnorm(100,3,5))
findperc <- function(prop, dat) {
perc <- quantile(dat$x, prop)
with(subset(dat, dat$x > perc), abs(cor(x,y)))
}
Now plot the result of findperc for percentiles between 0 and 1.
x <- seq(0,1,0.01)
plot(x,sapply(x,findperc,df),type="l")
The circled point indicates that found by optimize as in #coffeinjunky's answer. This is clearly only a local maximum. The applicability of the warning from #Thierry, "You need to rethink the question. As soon a x and y contain only 2 element the correlation will be either 1 or -1", should be apparent on the right hand side of the plot.
In general, the fact that you are getting moderate to high correlations when starting with independently generated random variables should warn you that your results are spurious and method suspect.
Well, why not take your question literally, and just search for it? For instance, try:
findperc <- function(prop, dat) {
perc <- quantile(dat$x, prop)
with(subset(dat, dat$x > perc), abs(cor(x,y)))
}
optimize(findperc, lower=0, upper=1, maximum=T, dat=df)
This defines a function that computes the absolute correlation between your vectors based on the corresponding percentile (which here is a single value), just as in your example code. And then I feed this function to a linear optimizer which searches for the input that produces the maximum value for the output.
Edit: Thanks to #A. Webb's answer I learned that optimize uses a gradient search as opposed to a grid search. I thought that this was the main difference between optim and optimize, a clearly wrong assumption I should have checked myself. However, just to provide a solution using grid search that will get you closer to the global maximum, one could use the following:
x <- seq(0,0.97,0.01)
x[which.max(sapply(x, findperc, dat=df))]
Note that I have cut x here at 97%. This ensures that at least 3 people are left in the sample (given a sample size of 100).
I want to calculate the differential response of y to x (continuous) depending on the categorical variable z.
In the standard lm setup:
lm(y~ x:z)
However, I want to do this while allowing for Impulse Indicator Saturation (IIS) in the 'gets' package. However, the following syntax produces an error:
isat(y, mxreg=x:z, iis=TRUE)
The error message is of the form:
"Error in solve.qr(out, tol = tol, LAPACK = LAPACK) :
singular matrix 'a' in 'solve"
1: In x:z :
numerical expression has 96 elements: only the first used
2: In x:z :
numerical expression has 96 elements: only the first used"
How should I modify the syntax?
Thank you!
At the moment, alas, isat doesn't provide the same functionality as lm on categorical/character variables, nor on using * and :. We hope to address that in a future release.
In the meantime you'll have to create distinct variables in your dataset representing the interaction. I guess something like the following...
library(gets)
N <- 100
x <- rnorm(N)
z <- c(rep("A",N/4),rep("B",N/4),rep("C",N/4),rep("D",N/4))
e <- rnorm(N)
y <- 0.5*x*as.numeric(z=="A") + 1.5*x*as.numeric(z=="B") - 0.75*x*as.numeric(z=="C") + 5*x*as.numeric(z=="D") + e
lm.reg <- lm(y ~ x:z)
arx.reg.0 <- arx(y,mxreg=x:z)
data <- data.frame(y,x,z,stringsAsFactors=F)
for(i in z[duplicated(z)==F]) {
data[[paste("Zx",i,sep=".")]] <- data$x * as.numeric(data$z==i)
}
arx.reg.1 <- arx(data$y,mxreg=data[,c("x","Zx.A","Zx.B","Zx.C")])
isat.1 <- isat(data$y,mc=TRUE,mxreg=data[,c("x","Zx.A","Zx.B","Zx.C")],max.block.size=20)
Note that as you'll be creating dummies for each category, there's a chance those dummies will cause singularity of your matrix of explanatory variables (if, as in my example, isat automatically uses 4 blocks). Using the argument max.block.size enables you to avoid this problem.
Let me know if I haven't addressed your particular point.
I am experiencing difficulties estimating a BMA-model via glib(), due to multicollinearity issues, even though I have clearly specified which columns to use. Please find the details below.
The data I'll be using for the estimation via Bayesian Model Averaging:
Cij <- c(357848,766940,610542,482940,527326,574398,146342,139950,227229,67948,
352118,884021,933894,1183289,445745,320996,527804,266172,425046,
290507,1001799,926219,1016654,750816,146923,495992,280405,
310608,1108250,776189,1562400,272482,352053,206286,
443160,693190,991983,769488,504851,470639,
396132,937085,847498,805037,705960,
440832,847631,1131398,1063269,
359480,1061648,1443370,
376686,986608,
344014)
n <- length(Cij);
TT <- trunc(sqrt(2*n))
i <- rep(1:TT,TT:1); #row numbers: year of origin
j <- sequence(TT:1) #col numbers: year of development
k <- i+j-1 #diagonal numbers: year of payment
#Since k=i+j-1, we have to leave out another dummy in order to avoid multicollinearity
k <- ifelse(k == 2, 1, k)
I want to evaluate the effect of i and j both via levels and factors, but of course not in the same model. Since I can decide to include i and j as factors, levels, or not include them at all and for k either to include as level, or exclude, there are a total of 18 (3x3x2) models. This brings us to the following data frame:
X <- data.frame(Cij,i.factor=as.factor(i),j.factor=as.factor(j),k,i,j)
X <- model.matrix(Cij ~ -1 + i.factor + j.factor + k + i + j,X)
X <- as.data.frame(X[,-1])
Next, via the following declaration I specify which variables to consider in each of the 18 models. According to me, no linear dependence exists in these specifications.
model.set <- rbind(
c(rep(0,9),rep(0,9),0,0,0),
c(rep(0,9),rep(0,9),0,1,0),
c(rep(0,9),rep(0,9),0,0,1),
c(rep(0,9),rep(0,9),1,0,0),
c(rep(1,9),rep(0,9),0,0,0),
c(rep(0,9),rep(1,9),0,0,0),
c(rep(0,9),rep(0,9),0,1,1),
c(rep(0,9),rep(0,9),1,1,0),
c(rep(0,9),rep(1,9),0,1,0),
c(rep(0,9),rep(0,9),1,0,1),
c(rep(1,9),rep(0,9),0,0,1),
c(rep(1,9),rep(0,9),1,0,0),
c(rep(0,9),rep(1,9),1,0,0),
c(rep(1,9),rep(1,9),0,0,0),
c(rep(0,9),rep(0,9),1,1,1),
c(rep(0,9),rep(1,9),1,1,0),
c(rep(1,9),rep(0,9),1,0,1),
c(rep(1,9),rep(1,9),1,0,0))
Then I call the glib() function, telling it to select the specified columns from X according to model.set.
library(BMA)
model.glib <- glib(X,Cij,error="poisson", link="log",models=model.set)
which results in the error
Error in glim(x, y, n, error = error, link = link, scale = scale) : X matrix is not full rank
The function first checks whether the matrix is f.c.r, before it evaluates which columns to select from X via model.set. How do I circumvent this, or is there any other way to include all 18 models in the glib() function?
Thank you in advance.
In trying to figure out which one is better to use I have come across two issues.
1) The W statistic given by wilcox.test is different from that of coin::wilcox_test. Here's my output:
wilcox_test:
Exact Wilcoxon Mann-Whitney Rank Sum Test
data: data$variableX by data$group (yes, no)
Z = -0.7636, p-value = 0.4489
alternative hypothesis: true mu is not equal to 0
wilcox.test:
Wilcoxon rank sum test with continuity correction
data: data$variable by data$group
W = 677.5, p-value = 0.448
alternative hypothesis: true location shift is not equal to 0
I'm aware that there's actually two values for W and that the smaller one is usually reported. When wilcox.test is used with comma instead of "~" I can get the other value, but this comes up as W = 834.5. From what I understand, coin::statistic() can return three different statistics using ("linear", "standarized", and "test") where "linear" is the normal W and "standardized" is just the W converted to a z-score. None of these match up to the W I get from wilcox.test though (linear = 1055.5, standardized = 0.7636288, test = -0.7636288). Any ideas what's going on?
2) I like the options in wilcox_test for "distribution" and "ties.method", but it seems that you can not apply a continuity correction like in wilcox.test. Am I right?
I encountered the same issue when trying to apply Wendt formula to compute effect sizes using the coin package, and obtained aberrant r values due to the fact that the linear statistic outputted by wilcox_test() is unadjusted.
A great explanation is already given here, and therefore I will simply address how to obtain adjusted U statistics with the wilcox_test() function. Let's use a the following data frame:
d <- data.frame( x = c(rnorm(n = 60, mean = 10, sd = 5), rnorm(n = 30, mean = 16, sd = 5)),
g = c(rep("a",times = 60), rep("b",times = 30)) )
We can perform identical tests with wilcox.test() and wilcox_test():
w1 <- wilcox.test( formula = x ~ g, data = d )
w2 <- wilcox_test( formula = x ~ g, data = d )
Which will output two distinct statistics:
> w1$statistic
W
321
> w2#statistic#linearstatistic
[1] 2151
The values are indeed totally different (albeit the tests are equivalent).
To obtain the U statistics identical to that of wilcox.test(), you need to subtract wilcox_test()'s output statistic by the minimal value that the sum of the ranks of the reference sample can take, which is n_1(n_1+1)/2.
Both commands take the first level in the factor of your grouping variable g as reference (which will by default be alphabetically ordered).
Then you can compute the smallest sum of the ranks possible for the reference sample:
n1 <- table(w2#statistic#x)[1]
And
w2#statistic#linearstatistic- n1*(n1+1)/2 == w1$statistic
should return TRUE
Voilà.
It seems to be one is performing Mann-Whitney's U and the other Wilcoxon rank test, which is defined in many different ways in literature. They are pretty much equivalent, just look at the p-value. If you want continuity correction in wilcox.test just use argument correct=T.
Check https://stats.stackexchange.com/questions/79843/is-the-w-statistic-outputted-by-wilcox-test-in-r-the-same-as-the-u-statistic