Age <- c(90,56,51,64,67,59,51,55,48,50,43,57,44,55,60,39,62,66,49,61,58,55,45,47,54,56,52,54,50,62,48,52,50,65,59,68,55,78,62,56)
Tenure <- c(2,2,3,4,3,3,2,2,2,3,3,2,4,3,2,4,1,3,4,2,2,4,3,4,1,2,2,3,3,1,3,4,3,2,2,2,2,3,1,1)
df <- data.frame(Age, Tenure)
I'm trying to count the unique values of Tenure, thus I've used the table() function to look at the frequencies
table(df$Tenure)
1 2 3 4
5 15 13 7
However I'm curious to know what the aggregate() function is showing?
aggregate(Age~Tenure , df, function(x) length(unique(x)))
Tenure Age
1 1 3
2 2 13
3 3 11
4 4 7
What's the difference between these two outputs?
The reason for the difference is your inclusion of unique in the aggregate. You are counting the number of distinct Ages by Tenure, not the count of Ages by Tenure. To get the analogous output with aggregate try
aggregate(Age~Tenure , df, length)
Tenure Age
1 1 5
2 2 15
3 3 13
4 4 7
Related
I'm pretty new to R and hope i'll make myself clear enough.
I have a table of several columns which are factors. I want to make a score for each of these columns. Then I want to calculate the mean of each score, and display the list of columns ranked by their mean scores, is that possible ?
Table would be:
head(musico[,69:73])
AVIS1 AVIS2 AVIS3 AVIS4 AVIS5
1 2 1 2 3 2
2 2 5 2 3 2
3 3 2 5 5 1
4 1 2 5 5 5
5 1 5 1 3 1
6 4 1 4 5 4
I want to make a score for each:
musico$score1<-0
musico$score1[musico$AVIS1==1]<-1
musico$score1[musico$AVIS1==2]<-0.5
then do the mean of each column score: mean of score1, mean of score2, ...:
mean(musico$score1), mean(musico$score2), ...
My goal is to have a list of titles (avis1, avis2,...) ranked by their mean score.
Any advice appreciated !
Here's one way using base although it is somewhat unclear what you want. What does score1 have to do with AVIS1? I think you may be missing some of the data from musico.
Based on the example provided, here's a base R solution. vapply loops through the data.frame and produces the mean for each column. Then the stack and order are only there to make the output a dataframe that looks nice.
music <- read.table(text = "
AVIS1 AVIS2 AVIS3 AVIS4 AVIS5
1 2 1 2 3 2
2 2 5 2 3 2
3 3 2 5 5 1
4 1 2 5 5 5
5 1 5 1 3 1
6 4 1 4 5 4", header = TRUE)
means <- vapply(music, mean, 1)
stack(means[order(means, decreasing = TRUE)])
values ind
4 4.000000 AVIS4
3 3.166667 AVIS3
2 2.666667 AVIS2
5 2.500000 AVIS5
1 2.166667 AVIS1
This is how I would do it by first introducing a scores vector to be used as a lookup. I assume that scores are decreasing by 0.5 and that the number of scores needed are according to the maximum number of levels found in your columns (i.e. 6 seen in AVIS1).
Then using tidyr you can organise your data set such that you have to variables (i.e. AVIS and Value) containing the respective levels. Then add a score variable with the mutate function from dplyr in which the position of the score in the score vector matches the value in the Value variable. From here you can find the mean scores corresponding to the AVIS levels, arrange them accordingly and put them in a list.
music <- read.table(text = "
AVIS1 AVIS2 AVIS3 AVIS4 AVIS5
1 2 1 2 3 2
2 2 5 2 3 2
3 3 2 5 5 1
4 1 2 5 5 5
5 1 5 1 3 1
6 4 1 4 5 4", header = TRUE) # your data
scores <- seq(1, by = -0.5, length.out = 6) # vector of scores
library(tidyr)
library(dplyr)
music2 <- music %>%
gather(AVIS, Value) %>% # here you tidy the data
mutate(score = scores[Value]) %>% # match score to value
group_by(AVIS) %>% # group AVIS levels
summarise(score.mean = mean(score)) %>% # find mean scores for AVIS levels
arrange(desc(score.mean))
list <- list(AVIS = music2$AVIS) # here is the list
> list$AVIS
[1] "AVIS1" "AVIS5" "AVIS2" "AVIS3" "AVIS4"
I have a long data frame with players' decisions who worked in groups.
I need to convert the data in such a way that each row (individual observation) would contain all group members decisions (so we basically can see whether they are interdependent).
Let's say the generating code is:
group_id <- c(rep(1, 3), rep(2, 3))
player_id <- c(rep(seq(1, 3), 2))
player_decision <- seq(10,60,10)
player_contribution <- seq(6,1,-1)
df <-
data.frame(group_id, player_id, player_decision, player_contribution)
So the initial data looks like:
group_id player_id player_decision player_contribution
1 1 1 10 6
2 1 2 20 5
3 1 3 30 4
4 2 1 40 3
5 2 2 50 2
6 2 3 60 1
But I need to convert it to wide per each group, but only for some of these variables, (in this example specifically for player_contribution, but in such a way that the rest of the data remains. So the head of the converted data would be:
data.frame(group_id=c(1,1),
player_id=c(1,2),
player_decision=c(10,20),
player_1_contribution=c(6,6),
player_2_contribution=c(5,5),
player_3_contribution=c(4,6)
)
group_id player_id player_decision player_1_contribution player_2_contribution player_3_contribution
1 1 1 10 6 5 4
2 1 2 20 6 5 6
I suspect I need to group_by in dplyr and then somehow gather per group but only for player_contribution (or a vector of variables). But I really have no clue how to approach it. Any hints would be welcome!
Here is solution using tidyr and dplyr.
Make a dataframe with the columns for the players contributions. Then join this dataframe back onto the columns of interest from the original Dataframe.
library(tidyr)
library(dplyr)
wide<-pivot_wider(df, id_cols= - player_decision,
names_from = player_id,
values_from = player_contribution,
names_prefix = "player_contribution_")
answer<-left_join(df[, c("group_id", "player_id", "player_decision") ], wide)
answer
group_id player_id player_decision player_contribution_1 player_contribution_2 player_contribution_3
1 1 1 10 6 5 4
2 1 2 20 6 5 4
3 1 3 30 6 5 4
4 2 1 40 3 2 1
5 2 2 50 3 2 1
6 2 3 60 3 2 1
I am a nebie to R.I have a data table DT as
id time day type
1 1 9 10
2 2 3 10
1 3 6 12
3 8 9 10
6 9 9 10
8 2 6 18
9 3 5 10
9 1 4 12
From this I initially wanted the count group by day time type.SO i did
DT[,.N,by=list(day,time,type)]
which gives the count for each group.
Now I need to calculate the variance for each group. So I tried
DT[,var(.N),by=list(day,time,type)]
But this gave NA for all fields.Any help is appreciated.
In the example given, there is only a single unique value for many of the combinations, so there is no variance for those rows.
DT <- data.frame (id = c(1,2,1,3,6,8,9,9),
time = c(1,2,3,8,9,2,3,1),
day = c(9,3,6,9,9,6,5,4),
type = c(10,10, 12, 10,10,18,10,12))
aggregate(DT, list(DT$id), FUN = var)
I am new to the R programming, so wanted to learn that if it is possible to perform merging and grouping of the data with a single function or within a single step in R.
I'm not sure if I've understood your question correctly. It's possible to group and merge data via the aggregate function:
df <- data.frame(a=1:40, b=rbinom(40, 10, 0.5), n=rnorm(40), p=rpois(40, lambda=4), group=gl(4,10), even=rep(c(1,2),20))
require(plyr)
aggregate(b ~ group, df, sum) #aggregate/sum over group
aggregate(b ~ group + even, df, sum) #aggregate/sum over group & even
Results:
> aggregate(b ~ group, df, sum)
group b
1 1 51
2 2 49
3 3 49
4 4 47
> aggregate(b ~ group + even, df, sum)
group even b
1 1 1 27
2 2 1 23
3 3 1 25
4 4 1 23
5 1 2 24
6 2 2 26
7 3 2 24
8 4 2 24
I have data arranged like this in R:
indv time mass
1 10 7
2 5 3
1 5 1
2 4 4
2 14 14
1 15 15
where indv is individual in a population. I want to add columns for initial mass (mass_i) and final mass (mass_f). I learned yesterday that I can add a column for initial mass using ddply in plyr:
sorted <- ddply(test, .(indv, time), sort)
sorted2 <- ddply(sorted, .(indv), transform, mass_i = mass[1])
which gives a table like:
indv mass time mass_i
1 1 1 5 1
2 1 7 10 1
3 1 10 15 1
4 2 4 4 4
5 2 3 5 4
6 2 8 14 4
7 2 9 20 4
However, this same method will not work for finding the final mass (mass_f), as I have a different number of observations for each individual. Can anyone suggest a method for finding the final mass, when the number of observations may vary?
You can simply use length(mass) as the index of the last element:
sorted2 <- ddply(sorted, .(indv), transform,
mass_i = mass[1], mass_f = mass[length(mass)])
As suggested by mb3041023 and discussed in the comments below, you can achieve similar results without sorting your data frame:
ddply(test, .(indv), transform,
mass_i = mass[which.min(time)], mass_f = mass[which.max(time)])
Except for the order of rows, this is the same as sorted2.
You can use tail(mass, 1) in place of mass[1].
sorted2 <- ddply(sorted, .(indv), transform, mass_i = head(mass, 1), mass_f=tail(mass, 1))
Once you have this table, it's pretty simple:
t <- tapply(test$mass, test$ind, max)
This will give you an array with ind. as the names and mass_f as the values.