I have the following data frame:
lm mean resids sd resids resid 1 resid 2 resid 3 intercept beta
1 0.000000e+00 6.2806844 -3.6261548 7.2523096 -3.6261548 103.62615 24.989340
2 -2.960595e-16 8.7515899 -5.0527328 10.1054656 -5.0527328 141.96786 -1.047323
3 -2.960595e-16 5.9138984 -3.4143908 6.8287817 -3.4143908 206.29046 -26.448694
4 3.700743e-17 0.5110845 0.2950748 -0.5901495 0.2950748 240.89801 -35.806642
5 7.401487e-16 6.6260504 3.8255520 -7.6511040 3.8255520 187.03479 -23.444762
6 5.921189e-16 8.7217431 5.0355007 -10.0710014 5.0355007 41.43239 3.138396
7 0.000000e+00 5.5269434 3.1909823 -6.3819645 3.1909823 -119.90628 27.817845
8 -1.480297e-16 1.0204260 -0.5891432 1.1782864 -0.5891432 -180.33773 35.623363
9 -5.921189e-16 6.9488186 -4.0119023 8.0238046 -4.0119023 -64.72245 21.820226
10 -8.881784e-16 8.6621512 -5.0010953 10.0021906 -5.0010953 191.65339 -5.218767
Each row represents an estimated linear model with window length 3. I used rollapply on a separate dataframe with the function lm(y~t) to extract the coefficients and intercepts into a new dataframe, which I have combined with the residuals from the same model and their corresponding means and residuals.
Since the window length is 3, it implies that there are 3 residuals as shown, per model, in resid 1, resid 2 and resid 3. The mean and sd of these are included accordingly.
I am seeking to predict the next observation, in essence, k+1, where k is the window length, using the intercept and beta.
Recall that lm1 takes observations 1,2,3 to estimate the intercept and the beta, and lm2 takes 2,3,4, lm3 takes 3,4,5, etc. The function for the prediction should be:
predict_lm1 = intercept_lm1 + beta_lm1*(k+1)
Where k+1 = 4. For lm2:
predict_lm2 = intercept_lm2 + beta_lm2*(k+1)
Where k+1 = 5.
Clearly, k increases by 1 every time I move down one row in the dataset. This is because the explanatory variable is time, t, which is a sequence increasing by one per observation.
Should I use a for loop, or an apply function here?
How can I make a function that iterates down the rows and calculates the predictions accordingly with the information found in that row?
Thanks.
EDIT:
I managed to find a possible solution by writing the following:
n=nrow(dataset)
for(i in n){
predictions = dataset$Intercept + dataset$beta*(k+1)
}
However, k does not increase by 1 per iteration. Thus, k+1 is always = 4.
How can I make sure k increases by 1 accordingly?
EDIT 2
I managed to add 1 to k by writing the following:
n=nrow(dataset)
for(i in n){
x = 0
x[i] = k + 1
preds = dataset$`(Intercept)` + dataset$t*(x[i])
}
However, the first prediction is overestimated. It should be 203, whereas it is estimated as 228, implying that it sets the explanatory variable as 1 too high.
Yet, the second prediction is correct. I am not sure what I am doing wrong. Any advice?
EDIT 3
I managed to find a solution as follows:
n=nrow(dataset)
for(i in n){
x = k + 1
preds = dataset$`(Intercept)` + dataset$t*(x)
x = x + 1
}
Your loop is not iterating:
dataset <- read.table(text="lm meanresids sdresids resid1 resid2 resid3 intercept beta
1 0.000000e+00 6.2806844 -3.6261548 7.2523096 -3.6261548 103.62615 24.989340
2 -2.960595e-16 8.7515899 -5.0527328 10.1054656 -5.0527328 141.96786 -1.047323
3 -2.960595e-16 5.9138984 -3.4143908 6.8287817 -3.4143908 206.29046 -26.448694
4 3.700743e-17 0.5110845 0.2950748 -0.5901495 0.2950748 240.89801 -35.806642
5 7.401487e-16 6.6260504 3.8255520 -7.6511040 3.8255520 187.03479 -23.444762
6 5.921189e-16 8.7217431 5.0355007 -10.0710014 5.0355007 41.43239 3.138396
7 0.000000e+00 5.5269434 3.1909823 -6.3819645 3.1909823 -119.90628 27.817845
8 -1.480297e-16 1.0204260 -0.5891432 1.1782864 -0.5891432 -180.33773 35.623363
9 -5.921189e-16 6.9488186 -4.0119023 8.0238046 -4.0119023 -64.72245 21.820226
10 -8.881784e-16 8.6621512 -5.0010953 10.0021906 -5.0010953 191.65339 -5.218767", header=T)
n <- nrow(dataset)
predictions <- data.frame()
for(i in 1:n){
k <- i ##not sure where k is coming from but put it here
predictions <- rbind(predictions, dataset$intercept[i] + dataset$beta[i]*(k+1))
}
predictions
Related
I run a CV Lasso with the cv.gamlr function in R. I can get the coefficients for the lambdas that correspond to the “1se” or “min” criterion.
set.seed(123)
lasso<-cv.gamlr(x = X, y = Y, family ='binomial')
coef(lasso,select = "1se")
coef(lasso,select = "min")
But what if I want to obtain the coefficients for a specific lambda, stored in the lasso$gamlr$lambda vector? Is it possible to obtain them?
For example, to get the coefficients for the first lambda in the model... Something like this:
lambda_100<- lasso$gamlr$lambda[100]
coef(lasso,select = lambda_100)
Of course, this sends the following error:
Error in match.arg(select) : 'arg' must be NULL or a character vector
Thanks :)
The coefficients are stored under lasso$gamlr$beta, in your example, you can access them like this:
library(gamlr)
x = matrix(runif(500),ncol=5)
y = rnorm(100)
cvfit <- cv.gamlr(x, y, gamma=1)
dim(cvfit$gamlr$beta)
[1] 5 100
length(cvfit$gamlr$lambda)
[1] 100
cvfit$gamlr$lambda[100]
seg100
0.00125315
cvfit$gamlr$beta[,drop=FALSE,100]
5 x 1 sparse Matrix of class "dgCMatrix"
seg100
1 0.12960060
2 -0.16406246
3 -0.46566731
4 0.08197053
5 -0.54170494
Or if you prefer it in a vector:
cvfit$gamlr$beta[,100]
1 2 3 4 5
0.12960060 -0.16406246 -0.46566731 0.08197053 -0.54170494
While working on an Rcpp program, I used the sample() function, which gave me the following error: "NAs not allowed in probability." I traced this issue to the fact that the probability vector I used had NA values in it. I have no idea how. Below is some R code that captures the errors:
n.0=20
n.1=20
n.reps=1
beta0.vals=rep(seq(-.3,.1,,n.0),n.reps)
beta1.vals=rep(seq(-7,0,,n.1),n.reps)
beta.grd=as.matrix(expand.grid(beta0.vals,beta1.vals))
n.rnd=200
beta.rnd.grd=cbind(runif(n.rnd,min(beta0.vals),max(beta0.vals)),runif(n.rnd,min(beta1.vals),max(beta1.vals)))
beta.grd=rbind(beta.grd,beta.rnd.grd)
N = 22670
count = 0
for(i in 1:dim(beta.grd)[1]){ # iterate through 600 possible beta values in beta grid
beta.ind = 0 # indicator for current pair of beta values
for(j in 1:N){ # iterate through all possible Nsums
logit = beta.grd[i,1]/N*(j - .1*N)^2 + beta.grd[i,2];
phi01 = exp(logit)/(1 + exp(logit))
if(is.na(phi01)){
count = count + 1
}
}
}
cat("Total number of invalid probabilities: ", count)
Here, $\beta_0 \in (-0.3, 0.1), \beta_1 \in (-7, 0), N = 22670, N_\text{sum} \in (1, N)$. Note that $N$ and $N_\text{sum}$ are integers, whereas the beta values may not be.
Since mathematically, $\phi_{01} \in (0,1)$, I'm assuming that NAs are arising because R is not liking extremely small values. I am receiving an overwhelming amount of NA values, too. More so than numbers. Why would I be getting NAs in this code?
Include print(logit) next to count = count + 1 and you will find lots of logit > 1000 values. exp(1000) == Inf so you divide Inf by Inf which will get you a NaN and NaN is NA:
> exp(500)
[1] 1.403592e+217
> Inf/Inf
[1] NaN
> is.na(NaN)
[1] TRUE
So your problems are not too small but to large numbers coming first out of the evaluation of exp(x) with x larger then roughly 700:
> exp(709)
[1] 8.218407e+307
> exp(710)
[1] Inf
Bernhard's answer correctly identifies the problem:
If logit is large, exp(logit) = Inf.
Here is a solution:
for(i in 1:dim(beta.grd)[1]){ # iterate through 600 possible beta values in beta grid
beta.ind = 0 # indicator for current pair of beta values
for(j in 1:N){ # iterate through all possible Nsums
logit = beta.grd[i,1]/N*(j - .1*N)^2 + beta.grd[i,2];
## This one isn't great because exp(logit) can be very large
# phi01 = exp(logit)/(1 + exp(logit))
## So, we say instead
## phi01 = 1 / ( 1 + exp(-logit) )
phi01 = plogis(logit)
if(is.na(phi01)){
count = count + 1
}
}
}
cat("Total number of invalid probabilities: ", count)
# Total number of invalid probabilities: 0
We can use the more stable 1 / (1 + exp(-logit)
(to convince yourself of this, multiply your expression with exp(-logit) / exp(-logit)),
and luckily either way, R has a builtin function plogis() that can calculate these probabilities quickly and accurately.
You can see from the help file (?plogis) that this function evaluates the expression I gave, but you can also double check to assure yourself
x = rnorm(1000)
y = 1 / (1 + exp(-x))
z = plogis(x)
all.equal(y, z)
[1] TRUE
I am trying to create multiple confusion matrices from one dataframe, with each matrix generated based off a different condition in the dataframe.
So for the dataframe below, I want a confusion matrix for when Value = 1, Value = 2, Value =3
observed predicted Value
1 1 1
0 1 1
1 0 2
0 0 2
1 1 3
0 0 3
and see the results like:
Value Sensitivity Specificity PPV NPV
1 .96 .71 .84 .95
2 .89 .63 .30 .45
3 .88 .95 .28 .80
This is what I tried with a reproducible example. I am trying to write a loop that looks at every row, determines if Age = 1, and then pulls the values from the predicted and observed columns to generate a confusion matrix. Then I manually pull out the values from the confusion matrix to write out sen, spec, ppv, and npv and tried to combine all the matrices together. And then the loop starts again with Age = 2.
data(scat)
df<-scat %>% transmute(observed=ifelse(Site=="YOLA","case", "control"), predicted=ifelse(Location=="edge","case", "control"),Age)
x<-1 #evaluate at ages 1 through 5
for (i in dim(df)[1]) { #for every row in df
while(x<6) { #loop stops at Age=5
if(x=df$Age) {
q<-confusionMatrix(data = df$predicted, reference = df$observed, positive = "case")
sensitivity = q$table[1,1]/(q$table[1,1]+q$table[2,1])
specificity = q$table[2,2]/(q$table[2,2]+q$table[1,2])
ppv = q$table[1,1]/(q$table[1,1]+q$table[1,2])
npv = q$table[2,2]/(q$table[2,2]+q$table[2,1])
matrix(c(sensitivity, specificity, ppv, npv),ncol=4,byrow=TRUE)
}
}
x <- x + 1 #confusion matrix at next Age value
}
final<- rbind(matrix) #combine all the matrices together
However, this loop is completely non-functional. I'm not sure where the error is.
Your code can be simplified and the desired output achieved like this:
library(caret)
library(dplyr)
data(scat)
df <- scat %>%
transmute(observed = factor(ifelse(Site == "YOLA","case", "control")),
predicted = factor(ifelse(Location == "edge","case", "control")),
Age)
final <- t(sapply(sort(unique(df$Age)), function(i) {
q <- confusionMatrix(data = df$predicted[df$Age == i],
reference = df$observed[df$Age == i],
positive = "case")$table
c(sensitivity = q[1, 1] / (q[1, 1] + q[2, 1]),
specificity = q[2, 2] / (q[2, 2] + q[1, 2]),
ppv = q[1, 1] / (q[1, 1] + q[1, 2]),
npv = q[2, 2] / (q[2, 2] + q[2, 1]))
}))
Resulting in
final
#> sensitivity specificity ppv npv
#> [1,] 0.0 0.5625000 0.00000000 0.8181818
#> [2,] 0.0 1.0000000 NaN 0.8000000
#> [3,] 0.2 0.5882353 0.06666667 0.8333333
#> [4,] 0.0 0.6923077 0.00000000 0.6923077
#> [5,] 0.5 0.6400000 0.25000000 0.8421053
However, it's nice to know why your own code didn't work, so here are a few issues that might be useful to consider:
You need factor columns rather than character columns for confusionMatrix
You were incrementing through the rows of df, but you need one iteration for each unique age, not each row in your data frame.
Your line to increment x happens outside of the while loop, so x never increments and the loop never terminates, so the console just hangs.
You are doing if(x = df$Age), but you need a == to test equality.
It doesn't make sense to compare x to df$Age anyway, because x is length 1 and df$Age is a long vector.
You have unnecessary repetition by doing q$table each time. You can just make q equal to q$table to make your code more readable and less error-prone.
You call matrix at the end of the loop, but you don't store it anywhere, so the whole loop doesn't actually do anything.
You are trying to rbind an object called matrix in the last line which doesn't exist
Your lack of spaces between math operators, commas and variables make the code less readable and harder to debug. I'm not just saying this as a stylistic point; it is a major source of errors I see frequently here on SO.
I have searched for an answer or a solution to this task with no success as of yet, so I do apologize if this is redundant.
I want to randomize the data between two columns. This is to simulate species misidentification in vegetation field data, so I want to assign some sort of probability of misidentification between the two columns as well. I would imagine that there is some way to do this using sample or the "permute" package.
I will select some readily available data for an example.
library (vegan)
data (dune)
If you type head (dune), then you can see that this is a data frame with sites as rows and species as columns. For convenience sake, we can presume some field tech has potential to misidentify Poa pratensis and Poa trivialis.
poa = data.frame(Poaprat=dune$Poaprat,Poatriv=dune$Poatriv)
head(poa)
Poaprat Poatriv
1 4 2
2 4 7
3 5 6
4 4 5
5 2 6
6 3 4
What would be the best way to randomize the values between these two columns (transferring between each other and/or adding to one when both are present). The resulting data may look like:
Poaprat Poatriv
1 6 0
2 4 7
3 5 6
4 5 4
5 0 7
6 4 3
P.S.
For the cringing ecologist out there: please realize, I have made this example in the interest of time and that I know relative cover values are not additive. I apologize for needing to do that.
*** Edit: For more clarity, the type of data being randomized would be percent cover estimates (so values between 0% and 100%). The data in this quick example are relative cover estimates, not counts.
You'll still need to replace the actual columns with the new ones and there may be a more elegant way to do this (it's late in EDT land) and you'll have to decide what else besides the normal distribution you'll want to use (i.e. how you'll replace sample()) but you get your swaps and adds with:
library(vegan)
library(purrr)
data(dune)
poa <- data.frame(
Poaprat=dune$Poaprat,
Poatriv=dune$Poatriv
)
map2_df(poa$Poaprat, poa$Poatriv, function(x, y) {
for (i in 1:length(x)) {
what <- sample(c("left", "right", "swap"), 1)
switch(
what,
left={
x[i] <- x[i] + y[i]
y[i] <- 0
},
right={
y[i] <- x[i] + y[i]
x[i] <- 0
},
swap={
tmp <- y[i]
y[i] <- x[i]
x[i] <- tmp
}
)
}
data.frame(Poaprat=x, Poatriv=y)
})
Here is my approach:
Let's define a function that will take a number of specimens (n) and a probability (p) that it could be labeled incorrectly. This function will sample a 1 with probability p and a 0 with 1-p. The sum of this random sampling will give how many of the n specimens were incorrect.
mislabel = function(x, p){
N_mis = sample(c(1,0), x, replace = T, prob = c(p, 1-p))
sum(N_mis)
}
Once defined the function, apply it to each column and store it into two new columns
p_miss = 0.3
poa$Poaprat_mislabeled = sapply(poa$Poaprat, mislabel, p_miss)
poa$Poatriv_mislabeled = sapply(poa$Poatriv, mislabel, p_miss)
The final number of specimens tagged for each species can be calculated by substracting the incorrect from same species and adding the incorrect from the other specimen.
poa$Poaprat_final = poa$Poaprat - poa$Poaprat_mislabeled + poa$Poatriv_mislabeled
poa$Poatriv_final = poa$Poatriv - poa$Poatriv_mislabeled + poa$Poaprat_mislabeled
Result:
> head(poa)
Poaprat Poatriv Poaprat_mislabeled Poatriv_mislabeled Poaprat_final Poatriv_final
1 4 2 0 0 4 2
2 4 7 1 2 5 6
3 5 6 0 3 8 3
4 4 5 1 2 5 4
5 2 6 0 3 5 3
6 3 4 1 2 4 3
Complete procedure:
mislabel = function(x, p){
N_mis = sample(c(1,0), x, replace = T, prob = c(p, 1-p))
sum(N_mis)
}
p_miss = 0.3
poa$Poaprat_mislabeled = sapply(poa$Poaprat, mislabel, p_miss)
poa$Poatriv_mislabeled = sapply(poa$Poatriv, mislabel, p_miss)
poa$Poaprat_final = poa$Poaprat - poa$Poaprat_mislabeled + poa$Poatriv_mislabeled
poa$Poatriv_final = poa$Poatriv - poa$Poatriv_mislabeled + poa$Poaprat_mislabeled
The p_miss variable is the probability of labeling incorrectly both species. You could also use a different value for each to simulate a non symmetrical chance that it may be easier to mislabel one of them compared to the other.
I just wanted to check in since accepting the answer from hrbrmstr. Given a little bit of time today, I went ahead and made a function that does this task with some degree of flexibility. It allows for inclusion of multiple species pairs, different probabilities between different species pairs (asymmetry in different direction), and includes explicitly the probability of the value staying the same.
misID = function(X, species,probs = c(0.1,0.1,0,0.8)){
library(purrr)
X2 = X
if (!is.matrix(species) == T){
as.matrix(species)
}
if (!is.matrix(probs) == T){
probs=matrix(probs,ncol=4,byrow=T)
}
if (nrow(probs) == 1){
probs = matrix(rep(probs[1,],nrow(species)),ncol=4,byrow=T)
}
for (i in 1:nrow(species)){
Spp = data.frame(X[species[i,1]],X[species[i,2]])
mis = map2_df(Spp[1],Spp[2],function(x,y) {
for(n in 1:length(x)) {
what = sample(c('left', 'right', 'swap','same'), size=1,prob=probs[i,])
switch(
what,
left = {
x[n] = x[n] + y[n]
y[n] = 0
},
right = {
y[n] = x[n] + y[n]
x[n] = 0
},
swap = {
tmp = y[n]
y[n] = x[n]
x[n] = tmp
},
same = {
x[n] = x[n]
y[n] = y[n]
}
)
}
misSpp = data.frame(x,y)
colnames(misSpp) =c(names(Spp[1]),names(Spp[2]))
return(misSpp)
})
X2[names(mis[1])] = mis[1]
X2[names(mis[2])] = mis[2]
}
return(X2)
}
There are probably a number of minor inefficiencies in here, but by and large it does what I need it to do. Sorry that there are no comments, but I did figure out how to handle getting the shuffled data into the data frame easily.
Thanks for pointing out the "purrr" package for me and also the switch function.
Example:
library(vegan)
library(labdsv)
data(dune)
#First convert relative abundances to my best guess at the % values in Van der Maarel (1979)
code = c(1,2,3,4,5,6,7,8,9)
value = c(0.1,1,2.5,4.25,5.5,20,40,60.5,90)
veg = vegtrans(dune,code,value)
specpairs = matrix(c("Poaprat","Poatriv","Trifprat","Trifrepe"),ncol=2,byrow=T) #create matrix of species pairs
probmat = matrix(c(0.3,0,0,0.7,0,0.5,0,0.5),ncol=4,byrow=T) #create matrix of misclassification probabilities
veg2 = misID(veg,specpairs,probs = probmat)
print(veg2)
I made a function g to calculate maximum likelihood following a multivariated normal distribution in order to classify the rows of a data frame test between 34 classes. So I need to use this function over a data frame and with several different set of parameters. My code works but it's to slow. I want to get it faster, maybe by removing the for loop and using some other apply's family function (which I don't have any experience).
g = function(x,p,mu,Sigma){
log(p) - log(det(Sigma)) - as.matrix(x-mu)%*%solve(Sigma)%*%t(as.matrix(x-mu))
}
mu = summaryBy(.~class,train,FUN=mean)
Sigma = by(train[,1:8],train$class,cov)
p = as.data.frame(table(train$class))
p$Freq = p$Freq/n
k = length(levels(train$class))
logver = NULL
for(j in 1:k){
logver = cbind(logver,apply(test,1,g,p=p$Freq[j],mu=mu[j,-1],Sigma=Sigma[[j]]))
}
preds = apply(logver,1,which.max)
The output logver must be a data frame with one row for each row of test and one col for each j, so in this case with 340000 rows and 34 cols.
p$Freq is a numeric vector 1x34. mu is a data frame 34x9 (fist col is a factor generated by summaryBy mean). Sigma is a list with 34 elements and each element is a 8x8 covariace matrix. test is a data frame 340000x8.
head(test)
band1 band2 band3 band4 band5 band6 band7 band9
1 2.0592 4.3630 6.6506 10.5952 18.4566 37.3683 36.9154 33.9467
2 2.5772 4.0766 6.0116 10.1476 18.8585 36.7654 36.2221 33.3717
3 2.8240 4.0766 6.4183 9.6813 18.5148 37.3113 35.7318 33.8367
4 2.8999 4.4317 7.4529 10.2842 18.4566 37.2513 37.3219 33.8367
5 2.8684 3.5324 7.5845 10.9021 19.2262 37.4758 36.0219 33.3472
6 2.4069 4.3256 6.0241 10.6668 20.0381 36.7203 36.4816 33.3472
head(train)
band1 band2 band3 band4 band5 band6 band7 band9 class
1 5.1224 8.1723 11.6837 15.6408 22.5884 33.9782 32.2985 32.1805 Green
2 5.4430 6.1158 9.8344 14.7719 23.2234 34.1247 32.0722 32.4367 Dry
3 4.5048 7.7364 11.9494 15.7740 22.6291 33.7642 32.4599 32.5217 Conifer
4 5.3120 6.9558 9.8344 14.3223 22.8088 34.0513 32.2985 32.4045 Snow
5 5.2907 6.6837 10.5367 16.0684 22.6291 33.7156 32.1650 32.4900 Ice
6 5.3120 6.8131 10.9727 15.7114 22.8088 34.6136 32.4772 32.4367 Soil
Thank you all