I attempted to prove to myself that predict() will not give incorrect predictions, when labels and levels (the underlying integer for the factor level) of newdata do not match that of the train data.
I think I did prove that, and I'm sharing that code below, but I'd just like to ask what exactly R is doing when predicting for newdata. I know it is not appending newdata to training data, does it translate the factor labels of newdata into the corresponding representation of train data before predicting?
options(stringsAsFactors = TRUE)
dat <- data.frame(x = rep(c("cat", "dog", "bird", "horse"), 100), y = rgamma(100, shape=3, scale = 300))
model <- glm(y~., family = Gamma(link = "log"), data = dat)
coefficients(model)
# (Intercept) xcat xdog xhorse
# 6.5816536 0.2924488 0.3586094 0.2740487
newdata1 <- data.frame(x = "cat")
newdata2 <- data.frame(x = "bird")
newdata3 <- data.frame(x = "dog")
predict.glm(object = model, newdata = newdata1, type = "response")
# 1
# 966.907
exp(6.5816536 + 0.2924488) #intercept + cat coef
# [1] 966.9071
predict.glm(object = model, newdata = newdata2, type = "response")
# 1
# 721.7318
exp(6.5816536)
# [1] 721.7318
predict.glm(object = model, newdata = newdata3, type = "response")
# 1
# 1033.042
exp(6.5816536 + 0.3586094)
# [1] 1033.042
unclass(dat$x)
# [1] 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3
# [87] 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4
# [173] 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3
# [259] 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4
# [345] 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4 2 3 1 4
# attr(,"levels")
# [1] "bird" "cat" "dog" "horse"
unclass(newdata1$x)
# [1] 1
# attr(,"levels")
# [1] "cat"
unclass(newdata2$x)
# [1] 1
# attr(,"levels")
# [1] "bird"
Model object has an xlevels recording factor levels used for model estimation. For your example, we have:
model$xlevels
#$x
#[1] "bird" "cat" "dog" "horse"
When your new data is presented in prediction, factor levels will be matched. For example, your newdata1 will be matched to "cat" levels, and this is the second level in xlevels. Thus, predict will have no difficulty finding the correct coefficients for that level.
Related
Let's say you run the example for a latent class model from ?gmnl:
library(mlogit)
library(gmnl)
## Examples using the Electricity data set from the mlogit package
data("Electricity", package = "mlogit")
Electr <- mlogit.data(Electricity, id.var = "id", choice = "choice",
varying = 3:26, shape = "wide", sep = "")
## Estimate a LC model with 2 classes
Elec.lc <- gmnl(choice ~ pf + cl + loc + wk + tod + seas| 0 | 0 | 0 | 1,
data = Electr,
subset = 1:3000,
model = 'lc',
panel = TRUE,
Q = 2)
summary(Elec.lc)
You get a fitted model with coefficient estimates for two classes (class 1 & 2). Is there a way to extract (or predict) for each observation, what the most likely class is that this observation belongs to?
After several helpful comments and lots of digging, it seems that there is an undocumented feature that allows you to get predicted class probabilities, which are stored in Wnq. You get one entry per observation and the number of columns matches the number of latent classes (Q = 2 from above), and entries sum to 1.
## Get class probabilities
head(Elec.lc$Wnq)
init
[1,] 0.5547805 0.4452195
[2,] 0.5547805 0.4452195
[3,] 0.5547805 0.4452195
[4,] 0.5547805 0.4452195
[5,] 0.5547805 0.4452195
[6,] 0.5547805 0.4452195
The fitted model contains a matrix called prob.alt which gives the probability of each choice, so you can do:
predictions <- apply(Elec.cor$prob.alt,1, which.max)
predictions
#> [1] 1 1 2 3 1 4 4 3 3 3 2 1 2 2 3 1 1 1 2 3 4 4 4 1 1 4 1 1 4 4 4 2 4 3 1 2 4
#> [38] 4 4 1 1 4 1 1 4 4 4 2 1 1 2 3 4 4 4 2 4 3 4 2 1 4 2 2 2 2 4 2 1 3 4 3 4 4
#> [75] 4 1 4 2 3 2 2 1 3 3 4 3 4 1 1 4 2 1 4 4 2 2 2 2 2 2 1 4 2 2 2 2 1 2 2 4 3
#> [112] 1 1 1 2 3 4 4 4 2 4 3 4 1 1 4 2 1 4 4 2 2 1 4 2 2 2 2 1 2 1 2 4 3 2 2 2 2
#> [149] 1 4 2 2 2 1 2 1 4 3 2 2 2 1 2 1 1 4 2 1 4 2 2 2 2 1 2 1 1 4 3 2 2 2 2 1 4
#> [186] 2 2 2 2 4 2 1 4 3 2 2 2 2 2 1 1 4 2 1 4 4 3 2 2 4 4 1 3 4 1 2 4 3 1 1 1 2
#> [223] 3 4 4 4 1 2 4 2 3 4 4 1 3 4 2 3 3 2 4 1 1 4 4 4 2 1 3 1 2 1 1 2 3 1 4 4 2
#> [260] 4 3 2 1 2 4 2 3 3 4 1 3 4 2 3 3 4 4 4 4 4 1 3 2 3 1 3 3 1 4 2 1 4 4 2 2 1
#> [297] 3 1 1 4 2 4 1 2 4 1 1 4 4 4 2 1 1 2 3 4 4 4 2 4 3 4 1 1 1 2 3 1 4 4 3 4 3
#> [334] 2 1 1 4 1 1 4 4 2 2 1 3 1 3 1 4 2 2 2 2 1 2 1 3 4 3 2 2 2 2 1 4 3 2 2 2 1
#> [371] 2 4 4 1 3 4 2 3 3 2 1 3 3 3 3 4 1 1 4 1 1 4 4 2 2 2 4 2 3 4 4 4 1 4 2 3 2
#> [408] 1 4 3 2 2 2 1 2 1 1 4 3 1 1 2 3 4 4 4 3 3 3 2 1 2 4 3 4 4 4 3 4 3 4 3 4 1
#> [445] 1 4 1 1 4 4 4 2 1 4 2 2 2 2 1 2 1 3 4 3 1 4 2 2 2 2 1 2 4 2 4 3 3 3 4 1 1
#> [482] 4 2 1 4 4 2 2 2 2 3 1 1 1 2 3 4 4 4 2 2 4 2 3 4 4 4 3 4 2 3 2 2 4 2 3 4 4
#> [519] 1 1 4 2 3 2 2 4 1 1 4 4 4 2 2 3 1 3 2 1 2 2 1 4 4 2 2 2 4 2 1 4 3 2 2 2 4
#> [556] 2 1 1 4 2 1 4 2 2 2 2 1 2 1 2 4 3 1 1 2 3 4 4 4 2 4 3 4 2 4 4 4 3 4 2 3 3
#> [593] 3 1 3 3 1 1 2 3 1 4 4 3 4 3 2 1 2 2 2 2 1 4 3 2 2 2 2 2 2 4 2 3 3 4 1 3 4
#> [630] 2 3 3 2 3 1 1 4 4 4 2 2 3 1 3 1 1 2 3 1 4 4 3 3 3 4 1 4 4 4 3 4 1 4 3 1 1
#> [667] 3 3 2 2 3 1 1 1 2 3 1 4 4 2 1 4 2 2 2 2 1 2 1 1 4 2 1 1 2 3 4 4 4 2 4 3 4
#> [704] 1 2 2 2 2 1 4 2 2 2 2 4 2 2 2 2 2 1 4 3 2 2 2 4 2 1 4 2 2 2 2 4 2 1 3 4 3
#> [741] 1 4 3 2 2 2 2 2 1 1
If we compare these predictions to the actual choice, we see that the prediction is correct about 50% of the time (the values in the diagonal are correct):
table(predictions, Electricity$choice[1:750])
#>
#> predictions 1 2 3 4
#> 1 78 35 28 32
#> 2 40 129 40 33
#> 3 16 27 57 24
#> 4 27 36 38 110
Created on 2022-08-06 by the reprex package (v2.0.1)
I have a feeling that this object Wnq is not class membership probabilities though.
Even in your example above, when calling Elec.lc$Wnq, you seem to have obtained a list of probabilities of class membership for your individuals, but critically they are all equal across individuals.
When looking for this I also found myself with the same problem. I think Elec.lc$Wnq is just the mean of class membership probabilities.
I have not looked throughly in the gmnl code, but I think the object Qir is what you should look for ?
I have been doing some hierarchical clusterings in R. Its worked out fine up til now, producing hclust objects left and center, but suddenly not anymore. Now it will only produce lists when performing:
mydata.clusters <- hclust(dist(mydata[, 1:8]))
mydata.clustercut <- cutree(mydata.clusters, 4)
and when trying to:
table(mydata.clustercut, mydata$customer_lifetime)
it doesnt produce a table, but an endless print of the values (Im guessing from the list).
The cutree function provide the grouping to which each observation belong to. For example:
iris.clust <- hclust(dist(iris[,1:4]))
iris.clustcut <- cutree(iris.clust, 4)
iris.clustcut
# [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2
# [52] 2 2 3 2 3 2 3 2 3 3 3 3 2 3 2 3 3 2 3 2 3 2 2 2 2 2 2 2 3 3 3 3 2 3 2 2 2 3 3 3 2 3 3 3 3 3 2 3 3 2 2
# [103] 4 2 2 4 3 4 2 4 2 2 2 2 2 2 2 4 4 2 2 2 4 2 2 4 2 2 2 4 4 4 2 2 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2
Additional comparison can then be done by using this as a grouping variable for the observed data:
new.iris <- data.frame(iris, gp=iris.clustcut)
# example to visualise quickly the Species membership of each group
library(ggplot2)
ggplot(new.iris, aes(gp, fill=Species)) +
geom_bar()
everyone! I've been asked to create an K-means algorithm on R, but I don't really know the language, so I've found some example code on the internet, and decided to use. I've looked into it, learned the functions that are being used in it, and corrected it a bit, because it didn't work very well. Here's the code:
# Creating a sample of data
y=rnorm(500,1.65)
x=rnorm(500,1.15)
x=cbind(x,y)
centers <- x[sample(nrow(x),5),]
# A function for calculating the distance between centers and the rest of the dots
euclid <- function(points1, points2) {
distanceMatrix <- matrix(NA, nrow=dim(points1)[1], ncol=dim(points2)[1])
for(i in 1:nrow(points2)) {
distanceMatrix[,i] <- sqrt(rowSums(t(t(points1)-points2[i,])^2))
}
distanceMatrix
}
# A method function
K_means <- function(x, centers, euclid, nItter) {
clusterHistory <- vector(nItter, mode="list")
centerHistory <- vector(nItter, mode="list")
for(i in 1:nItter) {
distsToCenters <- euclid(x, centers)
clusters <- apply(distsToCenters, 1, which.min)
centers <- apply(x, 2, tapply, clusters, mean)
# Saving history
clusterHistory[[i]] <- clusters
centerHistory[[i]] <- centers
}
structure(list(clusters = clusterHistory, centers = centerHistory))
}
res <- K_means(x, centers, euclid, 5)
#To use the same plot operations I had to use unlist, since the resulting object in my function is a list of lists,
#and default object is just a list. And also i store the history of each iteration in that object.
res <- unlist(res, recursive = FALSE)
plot(x, col = res$clusters5)
points(res$centers5, col = 1:5, pch = 8, cex = 2)
It works fine on this simple matrix. But I've been asked to use it on iris:
head(iris)
a <-data.frame(iris$Sepal.Length, iris$Sepal.Width, iris$Petal.Length, iris$Petal.Width)
centers <- a[sample(nrow(a),3),]
iris_clusters <- K_means(a, centers, euclid, 3)
iris_clusters <- unlist(iris_clusters, recursive = FALSE)
head(iris_clusters)
And the problem is that it doesn't work. The error is:
Error in distanceMatrix[, i] <- sqrt(rowSums(t(t(points1) - points2[i, :
number of items to replace is not a multiple of replacement length
I understand that dimensions of objects don't match, but I don't understand why. That's why i'm asking for help. I apologize for all the stupidity there may be in this code in advance, but I'm not really familiar with the language yet, so don't judge me too harsh. Thank you!
Your implementation should work with simple typecasts
iris_clusters <- K_means(as.matrix(a), as.matrix(centers), euclid, 3) # 3 iterations
iris_clusters <- unlist(iris_clusters, recursive = FALSE)
# plotting the clusters obtained on the first two dimensions at the end of 3rd iteration
plot(a[,1:2], col = iris_clusters$clusters3, pch=19)
points(iris_clusters$centers3, col = 1:5, pch = 8, cex = 2)
head(iris_clusters)
# cluster assignments and centroids computed at different iterations
$clusters1
[1] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 3 2 3 2 3 2 3 3 3 3 2 3 3 3 3 3 3 2 3 2 2 3 3
[77] 2 2 3 3 3 3 3 2 3 3 2 3 3 3 3 2 3 3 3 3 3 3 3 3 1 2 1 2 1 1 3 1 1 1 2 2 2 2 2 2 2 1 1 2 1 2 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 2 2 2 1 2 2 1 1 2 2 2 2 2
$clusters2
[1] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 3 2 3 3 2 2 2 3 2 2 2 2 3 2 2 2 2 2 2
[77] 2 2 2 3 3 3 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 3 2 1 2 1 2 1 1 2 1 1 1 2 2 1 2 2 2 2 1 1 2 1 2 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 2 2 1 1 2 2 1 1 2 2 2 2 2
$clusters3
[1] 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
[77] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 3 2 1 2 1 2 1 1 2 1 1 1 2 2 1 2 2 2 2 1 1 2 1 2 1 2 1 1 2 2 1 1 1 1 1 2 2 1 1 2 2 1 1 1 2 1 1 1 2 2 2 2
$centers1
iris.Sepal.Length iris.Sepal.Width iris.Petal.Length iris.Petal.Width
1 7.150000 3.120000 6.090000 2.1350000
2 6.315909 2.915909 5.059091 1.8000000
3 5.297674 3.115116 2.550000 0.6744186
$centers2
iris.Sepal.Length iris.Sepal.Width iris.Petal.Length iris.Petal.Width
1 7.122727 3.113636 6.031818 2.1318182
2 6.123529 2.852941 4.741176 1.6132353
3 5.056667 3.268333 1.810000 0.3883333
$centers3
iris.Sepal.Length iris.Sepal.Width iris.Petal.Length iris.Petal.Width
1 7.014815 3.096296 5.918519 2.155556
2 6.025714 2.805714 4.588571 1.518571
3 5.005660 3.369811 1.560377 0.290566
I have done an experiment in which participants have solved a task in pairs, with another participant. Each participant has then received a score for how well they did the task. Pairs have gone through different amounts of trials.
I have a data frame similar to the one below:
participant <- c(1,1,2,2,3,3,3,4,4,4,5,6)
pair <- c(1,1,1,1,2,2,2,2,2,2,3,3)
trial <- c(1,2,1,2,1,2,3,1,2,3,1,1)
score <- c(2,3,6,3,4,7,3,1,8,5,4,3)
data <- data.frame(participant, pair, trial, score)
participant pair trial score
1 1 1 2
1 1 2 3
2 1 1 6
2 1 2 3
3 2 1 4
3 2 2 7
3 2 3 3
4 2 1 1
4 2 2 8
4 2 3 5
5 3 1 4
6 3 1 3
I would like to add a new vector to the data frame, where each participant gets the numeric difference between their own score and the other participant's score within each trial.
Does someone have an idea about how one might do that?
It should end up looking something like this:
participant pair trial score difference
1 1 1 2 4
1 1 2 3 0
2 1 1 6 4
2 1 2 3 0
3 2 1 4 3
3 2 2 7 1
3 2 3 3 2
4 2 1 1 3
4 2 2 8 1
4 2 3 5 2
5 3 1 4 1
6 3 1 3 1
Here's a solution that involves first reordering data such that each sequential pair of rows corresponds to a single pair within a single trial. This allows us to make a single call to diff() to extract the differences:
data <- data[order(data$trial,data$pair,data$participant),];
data$diff <- rep(diff(data$score)[c(T,F)],each=2L)*c(-1L,1L);
data;
## participant pair trial score diff
## 1 1 1 1 2 -4
## 3 2 1 1 6 4
## 5 3 2 1 4 3
## 8 4 2 1 1 -3
## 11 5 3 1 4 1
## 12 6 3 1 3 -1
## 2 1 1 2 3 0
## 4 2 1 2 3 0
## 6 3 2 2 7 -1
## 9 4 2 2 8 1
## 7 3 2 3 3 -2
## 10 4 2 3 5 2
I assumed you wanted the sign to capture the direction of the difference. So, for instance, if a participant has a score 4 points below the other participant in the same trial-pair, then I assumed you would want -4. If you want all-positive values, you can remove the multiplication by c(-1L,1L) and add a call to abs():
data$diff <- rep(abs(diff(data$score)[c(T,F)]),each=2L);
data;
## participant pair trial score diff
## 1 1 1 1 2 4
## 3 2 1 1 6 4
## 5 3 2 1 4 3
## 8 4 2 1 1 3
## 11 5 3 1 4 1
## 12 6 3 1 3 1
## 2 1 1 2 3 0
## 4 2 1 2 3 0
## 6 3 2 2 7 1
## 9 4 2 2 8 1
## 7 3 2 3 3 2
## 10 4 2 3 5 2
Here's a solution built around ave() that doesn't require reordering the whole data.frame first:
data$diff <- ave(data$score,data$trial,data$pair,FUN=function(x) abs(diff(x)));
data;
## participant pair trial score diff
## 1 1 1 1 2 4
## 2 1 1 2 3 0
## 3 2 1 1 6 4
## 4 2 1 2 3 0
## 5 3 2 1 4 3
## 6 3 2 2 7 1
## 7 3 2 3 3 2
## 8 4 2 1 1 3
## 9 4 2 2 8 1
## 10 4 2 3 5 2
## 11 5 3 1 4 1
## 12 6 3 1 3 1
Here's how you can get the score of the other participant in the same trial-pair:
data$other <- ave(data$score,data$trial,data$pair,FUN=rev);
data;
## participant pair trial score other
## 1 1 1 1 2 6
## 2 1 1 2 3 3
## 3 2 1 1 6 2
## 4 2 1 2 3 3
## 5 3 2 1 4 1
## 6 3 2 2 7 8
## 7 3 2 3 3 5
## 8 4 2 1 1 4
## 9 4 2 2 8 7
## 10 4 2 3 5 3
## 11 5 3 1 4 3
## 12 6 3 1 3 4
Or, assuming the data.frame has been reordered as per the initial solution:
data$other <- c(rbind(data$score[c(F,T)],data$score[c(T,F)]));
data;
## participant pair trial score other
## 1 1 1 1 2 6
## 3 2 1 1 6 2
## 5 3 2 1 4 1
## 8 4 2 1 1 4
## 11 5 3 1 4 3
## 12 6 3 1 3 4
## 2 1 1 2 3 3
## 4 2 1 2 3 3
## 6 3 2 2 7 8
## 9 4 2 2 8 7
## 7 3 2 3 3 5
## 10 4 2 3 5 3
Alternative, using matrix() instead of rbind():
data$other <- c(matrix(data$score,2L)[2:1,]);
data;
## participant pair trial score other
## 1 1 1 1 2 6
## 3 2 1 1 6 2
## 5 3 2 1 4 1
## 8 4 2 1 1 4
## 11 5 3 1 4 3
## 12 6 3 1 3 4
## 2 1 1 2 3 3
## 4 2 1 2 3 3
## 6 3 2 2 7 8
## 9 4 2 2 8 7
## 7 3 2 3 3 5
## 10 4 2 3 5 3
Here is an option using data.table:
library(data.table)
setDT(data)[,difference := abs(diff(score)), by = .(pair, trial)]
data
# participant pair trial score difference
# 1: 1 1 1 2 4
# 2: 1 1 2 3 0
# 3: 2 1 1 6 4
# 4: 2 1 2 3 0
# 5: 3 2 1 4 3
# 6: 3 2 2 7 1
# 7: 3 2 3 3 2
# 8: 4 2 1 1 3
# 9: 4 2 2 8 1
#10: 4 2 3 5 2
#11: 5 3 1 4 1
#12: 6 3 1 3 1
A slightly faster option would be:
setDT(data)[, difference := abs((score - shift(score))[2]) , by = .(pair, trial)]
If we need the value of the other pair:
data[, other:= rev(score) , by = .(pair, trial)]
data
# participant pair trial score difference other
# 1: 1 1 1 2 4 6
# 2: 1 1 2 3 0 3
# 3: 2 1 1 6 4 2
# 4: 2 1 2 3 0 3
# 5: 3 2 1 4 3 1
# 6: 3 2 2 7 1 8
# 7: 3 2 3 3 2 5
# 8: 4 2 1 1 3 4
# 9: 4 2 2 8 1 7
#10: 4 2 3 5 2 3
#11: 5 3 1 4 1 3
#12: 6 3 1 3 1 4
Or using dplyr:
library(dplyr)
data %>%
group_by(pair, trial) %>%
mutate(difference = abs(diff(score)))
# participant pair trial score difference
# <dbl> <dbl> <dbl> <dbl> <dbl>
#1 1 1 1 2 4
#2 1 1 2 3 0
#3 2 1 1 6 4
#4 2 1 2 3 0
#5 3 2 1 4 3
#6 3 2 2 7 1
#7 3 2 3 3 2
#8 4 2 1 1 3
#9 4 2 2 8 1
#10 4 2 3 5 2
#11 5 3 1 4 1
#12 6 3 1 3 1
I'm sure this has been asked before but for the life of me I can't figure out what to search for!
I have the following data:
x y
1 3
1 3
1 3
1 2
1 2
2 2
2 4
3 4
3 4
And I would like to output a running count that resets everytime either x or y changes value.
x y o
1 3 1
1 3 2
1 3 3
1 2 1
1 2 2
2 2 1
2 4 1
3 4 1
3 4 2
Try something like
df<-read.table(header=T,text="x y
1 3
1 3
1 3
1 2
1 2
2 2
2 4
3 4
3 4")
cbind(df,o=sequence(rle(paste(df$x,df$y))$lengths))
> cbind(df,o=sequence(rle(paste(df$x,df$y))$lengths))
x y o
1 1 3 1
2 1 3 2
3 1 3 3
4 1 2 1
5 1 2 2
6 2 2 1
7 2 4 1
8 3 4 1
9 3 4 2
After seeing #ttmaccer's I see my first attempt with ave was wrong and this is perhaps what is needed:
> dat$o <- ave(dat$y, list(dat$y, dat$x), FUN=seq )
# there was a warning but the answer is corect.
> dat
x y o
1 1 3 1
2 1 3 2
3 1 3 3
4 1 2 1
5 1 2 2
6 2 2 1
7 2 4 1
8 3 4 1
9 3 4 2