Brain afunctional today: How do I tell acast to return different aggregations?
# the rows and columns have integer names
Rgames> foo
1 2
1 1 1
2 2 2
3 3 3
4 4 4
1 1 4
2 2 8
3 3 2
4 4 1
Rgames> mfoo<-melt(foo)
Rgames> mfoo
Var1 Var2 value
1 1 1 1
2 2 1 2
3 3 1 3
4 4 1 4
5 1 1 1
6 2 1 2
7 3 1 3
8 4 1 4
9 1 2 1
10 2 2 2
11 3 2 3
12 4 2 4
13 1 2 4
14 2 2 8
15 3 2 2
16 4 2 1
Rgames> acast(mfoo,Var1~Var2,function(x)x[1]-x[2])
1 2
1 0 -3
2 0 -6
3 0 1
4 0 3
# what I would like is the casting formula to return
1 2
1 1 -3
2 2 -6
3 3 1
4 4 3
With the caveat that this is a simple example. In the general case, there will be rows with unique names -- but never more than two rows with a given name, so my x[1]-x[2] won't ever fail.
Or should I just use this:
aggregate(foo[,2],by=list((foo[,1])),function(x)x[1]-x[2])
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 ?
How can I compare values within a variable dependent on another variable with dplyr?
The df is based on choice data (long format) from a survey. It has one variable that indicates a participants id, another that indicates the choice instance and one that indicates which alternative was chosen.
In my data I have the feeling that a lot of people tend to get bored of the task and therefore stick to one alternative for every instance. I would therefore like to identify people who always selected the same option from a certain instance onwards till the end.
Here is an example df:
set.seed(0)
df <- tibble(
id = rep(1:5,each=12),
inst = rep(1:12,5),
alt = sample(1:3, size =60, replace=T),
)
That looks like the following:
id inst alt
1 1 1 3
2 1 2 1
3 1 3 2
4 1 4 2
5 1 5 3
6 1 6 1
7 1 7 3
8 1 8 3
9 1 9 2
10 1 10 2
11 1 11 1 <-
12 1 12 1 <-
13 2 1 1
14 2 2 3
...
I would like to create two new variables count and count_alt. The new variable count should indicate how often the same value appeared in alt based on id and inst, only counting values from the end of id. So for participant (id==1) the count variable should be 2, since alternative 1 was chosen in the last two instances (11 & 12). The count_alt would take the value 1 (always the same as inst == 12)
The new df schould look like the following
id inst alt count count_alt
1 1 1 3 2 1
2 1 2 1 2 1
3 1 3 2 2 1
4 1 4 2 2 1
5 1 5 3 2 1
6 1 6 1 2 1
7 1 7 3 2 1
8 1 8 3 2 1
9 1 9 2 2 1
10 1 10 2 2 1
11 1 11 1 2 1
12 1 12 1 2 1
...
I would prefer to solve this with dplyr and not with a loop since I want to incooperate it into further data wrangling steps.
See if that solves it:
library(dplyr)
df %>%
group_by(id) %>%
mutate(
count = cumsum(alt != lag(alt, default = "rndm")),
count = sum(count == max(count)),
count_alt = alt[n()]
)
Output:
id inst alt count count_alt
1 1 1 3 2 1
2 1 2 1 2 1
3 1 3 2 2 1
4 1 4 2 2 1
5 1 5 3 2 1
6 1 6 1 2 1
7 1 7 3 2 1
8 1 8 3 2 1
9 1 9 2 2 1
10 1 10 2 2 1
11 1 11 1 2 1
12 1 12 1 2 1
13 2 1 1 1 2
14 2 2 3 1 2
15 2 3 2 1 2
16 2 4 3 1 2
17 2 5 2 1 2
18 2 6 3 1 2
19 2 7 3 1 2
20 2 8 2 1 2
21 2 9 3 1 2
22 2 10 3 1 2
23 2 11 1 1 2
24 2 12 2 1 2
25 3 1 1 1 3
26 3 2 1 1 3
27 3 3 2 1 3
28 3 4 1 1 3
29 3 5 2 1 3
30 3 6 3 1 3
31 3 7 2 1 3
32 3 8 2 1 3
33 3 9 2 1 3
34 3 10 2 1 3
35 3 11 1 1 3
36 3 12 3 1 3
37 4 1 3 1 1
38 4 2 3 1 1
39 4 3 1 1 1
40 4 4 3 1 1
41 4 5 2 1 1
42 4 6 3 1 1
43 4 7 2 1 1
44 4 8 3 1 1
45 4 9 2 1 1
46 4 10 2 1 1
47 4 11 3 1 1
48 4 12 1 1 1
49 5 1 2 2 2
50 5 2 3 2 2
51 5 3 3 2 2
52 5 4 2 2 2
53 5 5 3 2 2
54 5 6 2 2 2
55 5 7 1 2 2
56 5 8 1 2 2
57 5 9 1 2 2
58 5 10 1 2 2
59 5 11 2 2 2
60 5 12 2 2 2
If I have a data frame with a column of monotonically increasing values such as:
x
1
2
3
4
1
2
3
1
2
3
4
5
6
1
2
How do I add a column to group each increasing sequence that results in:
x y
1 1
2 1
3 1
4 1
1 2
2 2
3 2
1 3
2 3
3 3
4 3
5 3
6 3
1 4
2 4
I can only think of using a loop which will be slow.
You may choose cumsum function to do it.
> x <- c(1,2,3,4,1,2,3,1,2,4,5,1,2)
> cumsum(x==1)
[1] 1 1 1 1 2 2 2 3 3 3 3 4 4
I would use diff and compute the cumulative sum:
df$y <- c(1, cumsum(diff(df$x) < 0 ) + 1)
> df
x y
1 1 1
2 2 1
3 3 1
4 4 1
5 1 2
6 2 2
7 3 2
8 1 3
9 2 3
10 3 3
11 4 3
12 5 3
13 6 3
14 1 4
15 2 4
I have done an experiment in which participants have solved a task in pairs, with another participant. Each participant has then received a score for how well they did the task. Pairs have gone through different amounts of trials.
I have a data frame similar to the one below:
participant <- c(1,1,2,2,3,3,3,4,4,4,5,6)
pair <- c(1,1,1,1,2,2,2,2,2,2,3,3)
trial <- c(1,2,1,2,1,2,3,1,2,3,1,1)
score <- c(2,3,6,3,4,7,3,1,8,5,4,3)
data <- data.frame(participant, pair, trial, score)
participant pair trial score
1 1 1 2
1 1 2 3
2 1 1 6
2 1 2 3
3 2 1 4
3 2 2 7
3 2 3 3
4 2 1 1
4 2 2 8
4 2 3 5
5 3 1 4
6 3 1 3
I would like to add a new vector to the data frame, where each participant gets the numeric difference between their own score and the other participant's score within each trial.
Does someone have an idea about how one might do that?
It should end up looking something like this:
participant pair trial score difference
1 1 1 2 4
1 1 2 3 0
2 1 1 6 4
2 1 2 3 0
3 2 1 4 3
3 2 2 7 1
3 2 3 3 2
4 2 1 1 3
4 2 2 8 1
4 2 3 5 2
5 3 1 4 1
6 3 1 3 1
Here's a solution that involves first reordering data such that each sequential pair of rows corresponds to a single pair within a single trial. This allows us to make a single call to diff() to extract the differences:
data <- data[order(data$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