R predict() function returning too many values - r

I am using the r predict function, and it is returning more values than I expected it too. I created a linear model for the data to predict MDC from PKWH, MDT, and MDT2, then I created new data for input values into the predict function. The original data for utility has 24 values for each column of MDC, PKWH, MDT, and MDT2.
fit2 <- lm(MDC ~ MDT + MDT2 + PKWH*(1 + MDT + MDT2), data =
utility)
predict <- predict(fit2, data = data.frame(PKWH = 9, MDT = 75, MDT2
= 5625))
I expected the predict() function to produce 1 predicted value for the inputs of PKWH = 9 | MDT = 75 | MDT2 = 5625, but it gave me these 24 values.
1 2 3 4 5 6 7
56.67781 51.66653 45.05200 42.12583 38.98647 38.80904 42.60033
8 9 10 11 12 13 14
46.86545 49.51928 54.15163 61.54441 68.00122 49.17722 45.27917
15 16 17 18 19 20 21
42.88154 40.93468 38.39330 37.80963 39.47550 41.58780 42.94447
22 23 24
46.25884 49.27053 53.98732
Also, when I plug the new input values to calculate the predicted value using the coefficients from the linear model, I get 55.42165 which is not found on the list of the 24 values from the predict() function.


first, I wouldn't name your result predict - you want to save that for the function. You need
predicted_data <- predict(fit2, newdata = data.frame(PKWH = 9, MDT = 75, MDT2
= 5625))
It's not throwing an error because predict has a catch-all (...) at the end where input to data is heading, but it's giving you the predictions for the data you fit the model with.

Related

How can I use SOM algorithm for classification prediction

I would like to see If SOM algorithm can be used for classification prediction.
I used to code below but I see that the classification results are far from being right. For example, In the test dataset, I get a lot more than just the 3 values that I have in the training target variable. How can I create a prediction model that will be in alignment to the training target variable?
library(kohonen)
library(HDclassif)
data(wine)
set.seed(7)
training <- sample(nrow(wine), 120)
Xtraining <- scale(wine[training, ])
Xtest <- scale(wine[-training, ],
center = attr(Xtraining, "scaled:center"),
scale = attr(Xtraining, "scaled:scale"))
som.wine <- som(Xtraining, grid = somgrid(5, 5, "hexagonal"))
som.prediction$pred <- predict(som.wine, newdata = Xtest,
trainX = Xtraining,
trainY = factor(Xtraining$class))
And the result:
$unit.classif
[1] 7 7 1 7 1 11 6 2 2 7 7 12 11 11 12 2 7 7 7 1 2 7 2 16 20 24 25 16 13 17 23 22
[33] 24 18 8 22 17 16 22 18 22 22 18 23 22 18 18 13 10 14 15 4 4 14 14 15 15 4

This might help:
SOM is an unsupervised classification algorithm, so you shouldn't expect it to be trained on a dataset that contains a classifier label (if you do that it will need this information to work, and will be useless with unlabelled datasets)
The idea is that it will kind of "convert" an input numeric vector to a network unit number (try to run your code again with a 1 per 3 grid and you'll have the output you expected)
You'll then need to convert those network units numbers back into the categories you are looking for (that is the key part missing in your code)
Reproducible example below will output a classical classification error. It includes one implementation option for the "convert back" part missing in your original post.
Though, for this particular dataset, the model overfitts pretty quickly: 3 units give the best results.
#Set and scale a training set (-1 to drop the classes)
data(wine)
set.seed(7)
training <- sample(nrow(wine), 120)
Xtraining <- scale(wine[training, -1])
#Scale a test set (-1 to drop the classes)
Xtest <- scale(wine[-training, -1],
center = attr(Xtraining, "scaled:center"),
scale = attr(Xtraining, "scaled:scale"))
#Set 2D grid resolution
#WARNING: it overfits pretty quickly
#Errors are 36% for 1 unit, 63% for 2, 93% for 3, 89% for 4
som_grid <- somgrid(xdim = 1, ydim=3, topo="hexagonal")
#Create a trained model
som_model <- som(Xtraining, som_grid)
#Make a prediction on test data
som.prediction <- predict(som_model, newdata = Xtest)
#Put together original classes and SOM classifications
error.df <- data.frame(real = wine[-training, 1],
predicted = som.prediction$unit.classif)
#Return the category number that has the strongest association with the unit
#number (0 stands for ambiguous)
switch <- sapply(unique(som_model$unit.classif), function(x, df){
cat <- as.numeric(names(which.max(table(
error.df[error.df$predicted==x,1]))))
if(length(cat)<1){
cat <- 0
}
return(c(x, cat))
}, df = data.frame(real = wine[training, 1], predicted = som_model$unit.classif))
#Translate units numbers into classes
error.df$corrected <- apply(error.df, MARGIN = 1, function(x, switch){
cat <- switch[2, which(switch[1,] == x["predicted"])]
if(length(cat)<1){
cat <- 0
}
return(cat)
}, switch = switch)
#Compute a classification error
sum(error.df$corrected == error.df$real)/length(error.df$real)

Feeding new data into predict() for multiple regression? [duplicate]

This question already has an answer here:
r predict function returning too many values [closed]
(1 answer)
Closed 6 years ago.
Assume I have have fit a regression model with multiple predictor variables in R, like in the following toy example:
n <- 20
x <- rnorm(n)
y <- rnorm(n)
z <- x + y + rnorm(n)
m <- lm(z ~ x + y + I(y^2))
Now I have new date, consisting of x and y values, and I want to predict the corresponding z values:
x.new <- rnorm(5)
y.new <- rnorm(5)
Question: How should I best call predict to apply the fitted model to the new data?
Here are a few things I tried, which do not work:
Attempt 1. Trying to use the x.new and y.new as the columns of a new data frame:
> predict(m, data=data.frame(x=x.new, y=y.new))
1 2 3 4 5 6 7
-0.0157090 1.1667958 -1.3797101 0.1185750 0.7786496 1.7666232 -0.6692865
8 9 10 11 12 13 14
1.9720532 0.3514206 1.1677019 0.6441418 -2.3010431 -0.3228424 -0.2181511
15 16 17 18 19 20
-0.8883275 0.4549592 -1.0377040 0.1750522 -2.4542843 1.2250101
This gave 20 values instead of 5, so cannot be right.
Attempt 2: Maybe predict got confused because the y^2 values were not supplied? Try to use model.frame to provide data in the correct form.
> predict(m, model.frame(~ x.new + y.new + I(y.new^2)))
1 2 3 4 5 6 7
-0.0157090 1.1667958 -1.3797101 0.1185750 0.7786496 1.7666232 -0.6692865
8 9 10 11 12 13 14
1.9720532 0.3514206 1.1677019 0.6441418 -2.3010431 -0.3228424 -0.2181511
15 16 17 18 19 20
-0.8883275 0.4549592 -1.0377040 0.1750522 -2.4542843 1.2250101
Warning message:
'newdata' had 5 rows but variables found have 20 rows
Again, this results in 20 values (plus a warning), so cannot be right.

The parameter is newdata (not data) when telling predict what to predict for.
predict(m, newdata = data.frame(x = x.new, y = y.new))

mcmcglmm loop to create many chains

Following up from this question (see for reproducible data frame) I want to run MCMCGLMM n times, where n is the number of randomisations. I have tried to construct a loop which runs all the chains, and saves them (to retrieve the posterior distributions of the randomised variable later) but I am encountering problems.
This is what the data frame looks like (when n = 5, hence R1-R5), A = response variable, L and V are random effect variables, B is a fixed effect, R1-R5 are random assignments of L with structure of V maintained:
ID L B V A R1 R2 R3 R4 R5
1 1_1_1 1 1 1 11.1 6 19 21 1 31
2 1_1_1 1 1 1 6.9 6 19 21 1 31
3 1_1_4 1 1 4 7.7 2 24 8 22 22
4 1_1_4 1 1 4 10.5 2 24 8 22 22
5 1_1_5 1 1 5 8.5 11 27 14 17 22
6 1_1_7 1 1 7 11.2 5 24 13 18 25
I can create the names I want to assign to my chains, and the names of the variable that changes with each run of the MCMC chain (R1-Rn):
n = 5
Rs = as.vector(rep(NA,n))
for(i in 1:n){
Rs[i] = paste("R",i, sep = "")
}
Rs
Output:
> Rs
[1] "R1" "R2" "R3" "R4" "R5"
I then tried this loop to produce 5 chains:
for(i in 1:n){
chains[i] = MCMCglmm(A ~1 + B,
random = as.formula(paste0("~" ,Rs[i], " + Vial")),
rcov = ~units,
nitt = 500,
thin = 2,
burnin = 50,
prior = prior2,
family = "gaussian",
start = list(QUASI = FALSE),
data = df)
}
Thanks Roland for helping to get the random effect to call properly, previously I was getting an error Error in buildZ(rmodel.terms[r] ... object Rs[i] not found- fixed by as.formula
But this stores all of the data in chains and seemingly only the $Sol components, but I need to be able to access the values within the VCV, specifically the posterior distributions of the R variables (e.g. summary(chainR1$VCV))
In summary: It seems I am making a mistake in how I assign the chain names, does anyone have a suggestion of how to do this, and save the posterior distributions or even the whole chain?

Using assign was a key point:
n = 10 #Number of chains to run
chainVCVdf = matrix(rep(NA, times = ((nitt-burnin)/thin)*n), ncol = n)
colnames(chainVCVdf)=c(rep("X", times = n))
for(i in 1:n){
assign("chainX",paste0("chain",Rs[i]))
chainX = MCMCglmm(A ~1 + B,
random = as.formula(paste0("~" ,Rs[i], " + V")),
rcov = ~units,
nitt = nitt,
thin = thin,
burnin = burnin,
prior = prior1,
family = "gaussian",
start = list(QUASI = FALSE),
data = df)
assign("chainVCV", chainX$VCV[,1])
chainVCVdf[,i]=(chainVCV)
colnames(chainVCVdf)[i] = colnames(chainX$VCV)[1]
}
It then became possible to build a matrix of the VCV component that I am interested in (namely the randomised L assignment in columns R1-Rn)

It seems as though you want to run a number of different MCMCglmm formulas in a loop. #Roland has helped you found the solution to this (although I personally would create the formulas prior to the loop). #Roland also points out that in order to save the results of each model, you should save them in a list - rather than a chain as you are currently doing. You could also save each model as an .RData file, as seen in the end of the question. To formalize an answer to this question I would perform this in the following way:
Rs = paste0("~R", 1:5, " + V") ## Create all model formulae
chainNames = paste0("chainR", 1:5) ## Names for each model
chains = list() ## Initialize list
## Loop over models
for(i in 1:length(Rs)){
chains[[i]] = MCMCglmm(A ~1 + B,
random = formula(Rs[i]),
rcov = ~units,
nitt = 500,
thin = 2,
burnin = 50,
prior = prior2,
family = "gaussian",
start = list(QUASI = FALSE),
data = df)
}
names(chains) = chainNames ## Name each model
save(chains, "chainsR1-R5.Rdata") ## Save all model output
A side note, paste0 is the same as paste, but with the argument sep="" by default

R - Probabilistic Neural Network - Iris dataset

I have the R iris dataset which I am using for a PNN. The 3 species have been recoded from level 0 to 3 as follows: 0 is setosa, 1 is versicolor, 2 is virginica. Training set is 75%
Q1. I don't understand the function pred_pnn, if anyone is good in R perhaps you can explain how it works
Q2. The output of the test set or prediction is shown below, I don't understand the output because it is supposed to be something close to either 0,1,2
data = read.csc("c:/iris-recoded.csv" , header = T)
size = nrow(data)
length = ncol(data)
index <- 1:size
positions <- sample(index, trunc(size * 0.75))
training <- data[positions,]
testing <- data[-positions,1:length-1]
result = data[-positions,]
result$actual = result[,length]
result$predict = -1
nn1 <- smooth(learn(training), sigma = 0.9)
pred_pnn <- function(x, nn){
xlst <- split(x, 1:nrow(x))
pred <- foreach(i = xlst, .combine = rbind) %dopar% {
data.frame(prob = guess(nn, as.matrix(i))$probabilities[1], row.names =NULL)
}
}
print(pred_pnn(testing, nn1))
prob
1 1.850818e-03
2 9.820653e-03
3 6.798603e-04
4 7.421435e-03
5 2.168817e-03
6 3.277354e-03
7 6.541173e-03
8 1.725332e-04
9 2.081845e-03
10 2.491388e-02
11 7.679823e-03
12 1.291811e-03
13 2.197234e-06
14 1.316366e-03
15 1.421219e-05
16 4.639239e-05
17 3.671907e-04
18 1.460001e-04
19 4.382849e-05
20 2.387543e-05
21 1.011196e-05
22 2.719982e-04
23 4.445472e-04
24 1.281762e-04
25 5.931106e-09
26 9.741870e-08
27 9.236434e-09
28 8.384690e-08
29 3.311667e-07
30 6.045306e-11
31 2.949265e-08
32 2.070014e-10
33 8.043735e-06
34 2.136666e-08
35 5.604398e-08
36 2.455841e-07
37 3.445977e-07
38 7.314647e-07

I'm assuming you're using the pnn package. Documentation for ?guess would lead us to believe that it does similar to what predict does for other models. In other words, it predicts to which class the observation belongs to. Everything else in there for bookkeeping. Why you get only the probabilities? Because the person who wrote the function made it that way by extracting guess(x)$probabilities and returning only that. If you look at the raw output, you would also get predicted class tucked in away in $category list element.

Logistic Regression : not actual out put with predict function

I am new to R, when I am going to estimate a logistic model using glm() it's not predicting the response, but gives a not actual output on calling predict function like 1 for every input at my predict function.
Code:
ex2data1R <- read.csv("/media/ex2data1R.txt")
x <-ex2data1R$x
y <-ex2data1R$y
z <-ex2data1R$z
logisticmodel <- glm(z~x+y,family=binomial(link = "logit"),data=ex2data1R)
newdata = data.frame(x=c(10),y=(10))
predict(logisticmodel, newdata, type="response")
Output:
> predict(logisticmodel, newdata, type="response")
1
1.181875e-11
Data(ex2data1R.txt) :
"x","y","z"
34.62365962451697,78.0246928153624,0
30.28671076822607,43.89499752400101,0
35.84740876993872,72.90219802708364,0
60.18259938620976,86.30855209546826,1
79.0327360507101,75.3443764369103,1
45.08327747668339,56.3163717815305,0
61.10666453684766,96.51142588489624,1
75.02474556738889,46.55401354116538,1
76.09878670226257,87.42056971926803,1
84.43281996120035,43.53339331072109,1
95.86155507093572,38.22527805795094,0
75.01365838958247,30.60326323428011,0
82.30705337399482,76.48196330235604,1
69.36458875970939,97.71869196188608,1
39.53833914367223,76.03681085115882,0
53.9710521485623,89.20735013750205,1
69.07014406283025,52.74046973016765,1
67.94685547711617,46.67857410673128,0
70.66150955499435,92.92713789364831,1
76.97878372747498,47.57596364975532,1
67.37202754570876,42.83843832029179,0
89.67677575072079,65.79936592745237,1
50.534788289883,48.85581152764205,0
34.21206097786789,44.20952859866288,0
77.9240914545704,68.9723599933059,1
62.27101367004632,69.95445795447587,1
80.1901807509566,44.82162893218353,1
93.114388797442,38.80067033713209,0
61.83020602312595,50.25610789244621,0
38.78580379679423,64.99568095539578,0
61.379289447425,72.80788731317097,1
85.40451939411645,57.05198397627122,1
52.10797973193984,63.12762376881715,0
52.04540476831827,69.43286012045222,1
40.23689373545111,71.16774802184875,0
54.63510555424817,52.21388588061123,0
33.91550010906887,98.86943574220611,0
64.17698887494485,80.90806058670817,1
74.78925295941542,41.57341522824434,0
34.1836400264419,75.2377203360134,0
83.90239366249155,56.30804621605327,1
51.54772026906181,46.85629026349976,0
94.44336776917852,65.56892160559052,1
82.36875375713919,40.61825515970618,0
51.04775177128865,45.82270145776001,0
62.22267576120188,52.06099194836679,0
77.19303492601364,70.45820000180959,1
97.77159928000232,86.7278223300282,1
62.07306379667647,96.76882412413983,1
91.56497449807442,88.69629254546599,1
79.94481794066932,74.16311935043758,1
99.2725269292572,60.99903099844988,1
90.54671411399852,43.39060180650027,1
34.52451385320009,60.39634245837173,0
50.2864961189907,49.80453881323059,0
49.58667721632031,59.80895099453265,0
97.64563396007767,68.86157272420604,1
32.57720016809309,95.59854761387875,0
74.24869136721598,69.82457122657193,1
71.79646205863379,78.45356224515052,1
75.3956114656803,85.75993667331619,1
35.28611281526193,47.02051394723416,0
56.25381749711624,39.26147251058019,0
30.05882244669796,49.59297386723685,0
44.66826172480893,66.45008614558913,0
66.56089447242954,41.09209807936973,0
40.45755098375164,97.53518548909936,1
49.07256321908844,51.88321182073966,0
80.27957401466998,92.11606081344084,1
66.74671856944039,60.99139402740988,1
32.72283304060323,43.30717306430063,0
64.0393204150601,78.03168802018232,1
72.34649422579923,96.22759296761404,1
60.45788573918959,73.09499809758037,1
58.84095621726802,75.85844831279042,1
99.82785779692128,72.36925193383885,1
47.26426910848174,88.47586499559782,1
50.45815980285988,75.80985952982456,1
60.45555629271532,42.50840943572217,0
82.22666157785568,42.71987853716458,0
88.9138964166533,69.80378889835472,1
94.83450672430196,45.69430680250754,1
67.31925746917527,66.58935317747915,1
57.23870631569862,59.51428198012956,1
80.36675600171273,90.96014789746954,1
68.46852178591112,85.59430710452014,1
42.0754545384731,78.84478600148043,0
75.47770200533905,90.42453899753964,1
78.63542434898018,96.64742716885644,1
52.34800398794107,60.76950525602592,0
94.09433112516793,77.15910509073893,1
90.44855097096364,87.50879176484702,1
55.48216114069585,35.57070347228866,0
74.49269241843041,84.84513684930135,1
89.84580670720979,45.35828361091658,1
83.48916274498238,48.38028579728175,1
42.2617008099817,87.10385094025457,1
99.31500880510394,68.77540947206617,1
55.34001756003703,64.9319380069486,1
74.77589300092767,89.52981289513276,1
Let me know am I doing something wrong?

I'm not seeing any problem. Here are predictions for x,y = 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100:
newdata = data.frame(x=seq(30, 100, 5) ,y=seq(30, 100, 5))
predict(logisticmodel, newdata, type="response")
1 2 3 4 5 6
2.423648e-06 1.861140e-05 1.429031e-04 1.096336e-03 8.357794e-03 6.078786e-02
7 8 9 10 11 12
3.320041e-01 7.923883e-01 9.670066e-01 9.955766e-01 9.994218e-01 9.999247e-01
13 14 15
9.999902e-01 9.999987e-01 9.999998e-01
You were predicting x=10, y=10 which is way outside the range of your x, y values (30 - 100), but the prediction was zero which fits these results. When x and y are low (30 - 55), the prediction for z is zero. when x and y are high (75 - 100), the prediction is one (or nearly one). It may be easier to interpret the results if you round them to a few decimals:
round(predict(logisticmodel, newdata, type="response") , 5)
1 2 3 4 5 6 7 8 9 10
0.00000 0.00002 0.00014 0.00110 0.00836 0.06079 0.33200 0.79239 0.96701 0.99558
11 12 13 14 15
0.99942 0.99992 0.99999 1.00000 1.00000
Here is a simple way to predict a category and compare the results with your data:
predict <- ifelse(predict(logisticmodel, type="response")>.5, 1, 0)
xtabs(~predict+ex2data1R$z)
ex2data1R$z
predict 0 1
0 34 5
1 6 55
We used predict() on your original data and then created a rule that picks 1 if the probability is greater than .5 and 0 if it is not. Then we use xtabs() to compare the predictions to the data. When z is 0, we correctly predict zero 34 times and incorrectly predict one 6 times. When z is 1 we correctly predict one 55 times and incorrectly predict zero 5 times. We are correct 89% of the time (34+55)/100*100. You could explore the accuracy of prediction if you use .45 or .55 as the cutoff instead of .5.

In my opinion all is correct, as you can read from R manual:
newdata - optionally, a data frame in which to look for variables with
which to predict. If omitted, the fitted linear predictors are used.
If you have data frame with 1 record it will produce prediction only for that one.
For more details see R manual/glm/predict
or just in R console, after loading library glm put:
?glm

You can also use the following command to make the confusion matrix:
predict <- ifelse(predict(logisticmodel, type="response")>.5, 1, 0)
table(predict,ex2data1R$z)

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