I have this kind of data frame :
df<- data.frame(cluster=c('1','1','2','3','3','3'), class=c('A','B','C','B','B','C'))
I would like to get for each cluster (1,2,3), the class which appears the most often. In case of a tie, it would also be great to get an info, as for example the combination of the classes (or if not possible just have NA).
So for my example, I would like to have something like this as result:
cluster class.max
1 'A B' (or NA)
2 'C'
3 'B'
Maybe I should use aggregate() but don't know how.
rank has ways of dealing with ties:
aggregate(class~cluster,df,function(x) paste(names(table(x)[rank(-1*table(x),ties.method="min")==1]),collapse=" "))
cluster class
1 1 A B
2 2 C
3 3 B
Related
How do I draw the sum value of each class represented like the table:
a a,b a,b,c c
5 2 1 2
E.g for the above example expected result is:
a b c
8 3 3
I'm asking this since I couldn't even find a close solution anywhere in the stackoverflow, the closest solution represented was the dcast function, but that only checks for equality, not presence.
One way using base R,
sapply(unique(unlist(strsplit(names(df), '\\.'))), function(i)
sum(df[grepl(i, names(df))]))
#a b c
#8 3 3
Note: I used \\. for strsplit instead of , since names were read like that.
this is probably a simple one, but I somehow got stuck...
I need to many loops to get the result of every sample in my support like the usual stacked loops:
for (a in 1:N1){
for (b in 1:N2){
for (c in 1:N3){
...
}
}
}
but the number of the for loops needed in this messy system depends on another random variable, let's say,
for(f in 1:N.for)
so how can I write a for loop to do deal with this? Or are there more elegant ways to do this?
note that the difference is that the nested for loops above (the variables a,b,c,...) do matter in my calculations, but the variable f of the for loop that controls for the number of for loops needed does not go into any of my calculations for my real purpose - all it does is count/ensure the number of for loops needed is correct.
Did I make it clear?
So what I am actually trying to do is generate all the possible combinations of a number of peoples preferences towards others.
Let's say I have 6 people (the simplest case for my purpose): Abi, Bob, Cath, Dan, Eva, Fay.
Abi and Bob have preference lists of C D E F ( 4!=24 possible permutations for each of them);
Cath and Dan have preference lists of A B and E F, respectively (2! * 2! = 4 possible permutations for each of them);
Eva and Fay have preference lists of A B C D (4!=24 possible permutations for each of them);
So all together there should be 24*24*4*4*24*24 possible permutations of preferences when taking all six them together.
I am just wondering what is a clear, easy and systematic way to generate them all at once?
I'd want them in the format such as
c.prefs <- as.matrix(data.frame(Abi = c("Eva", "Fay", "Dan", "Cath"),Bob = c("Dan", "Eva", "Fay", "Cath"))
but any clear format is fine...
Thank you so much!!
I'll assume you have a list of each loop variable and its maximum value, ordered from the outermost to innermost variable.
loops <- list(a=2, b=3, c=2)
You could create a data frame with all the loop variable values in the correct order with:
(indices <- rev(do.call(expand.grid, lapply(rev(loops), seq_len))))
# a b c
# 1 1 1 1
# 2 1 1 2
# 3 1 2 1
# 4 1 2 2
# 5 1 3 1
# 6 1 3 2
# 7 2 1 1
# 8 2 1 2
# 9 2 2 1
# 10 2 2 2
# 11 2 3 1
# 12 2 3 2
If the code run at the innermost point of the nested loop doesn't depend on the previous iterations, you could use something like apply to process each iteration independently. Otherwise you could loop through the rows of the data frame with a single loop:
for (i in seq_len(nrow(indices))) {
# You can get "a" with indices$a[i], "b" with indices$b[i], etc.
}
For the way of doing the calculation, an option is to use the Reduce function or some other higher-order function.
Since your data is not inherently ordered (an individual is part of a set, its preferences are part of the set) I would keep indivudals in a factor and have eg preferences in lists named with the individuals. If you have large data you can store it in an environment.
The first code is just how to make it reproducible. the problem domain was akin for graph oriented naming. You just need to change in the first line and in runif to change the behavior.
#people
verts <- factor(c(LETTERS[1:10]))
#relations, disallow preferring yourself
edges<-lapply(seq_along(verts), function(ind) {
levels(verts)[-ind]
})
names(edges) <- levels(verts)
#directions
#say you have these stored in a list or something
pool <- levels(verts)
directions<-lapply(pool, function(vert) {
relations <- pool[unique(round(runif(5, 1, 10)))]
relations[!(vert %in% relations)]
})
names(directions) = pool
num_prefs <- (lapply(directions, length))
names(num_prefs) <- names(directions)
#First take factorial of each persons preferences,
#then reduce that with multiplication
combinations <-
Reduce(`*`,
sapply(num_prefs, factorial)
)
I hope this answers your question!
I have a large dataset "totaldata" containing multiple rows relating to each animal. Some of them are LactationNo 1 readings, and others are LactationNo 2 readings. I want to extract all animals that have readings from both LactationNo 1 and LactationNo 2 and store them in another dataframe "lactboth"
There are 16 other columns of variables of varying types in each row that I need to preserve in the new dataframe.
I have tried merge, aggregate and %in%, but perhaps I'm using them incorrectly eg.
(lactboth <- totaldata[totaldata$LactationNo %in% c(1,2), ])
Animal Id is column 1, and lactationno is column 2. I can't figure out how to select only those AnimalId with LactationNo=1&2
Have also tried
lactboth <- totaldata[ which(totaldata$LactationNo==1 & totaldata$LactationNo ==2), ]
I feel like this should be simple, but couldn't find an example to follow quite the same. Help appreciated!!
If I understand your question correctly, then your dataset looks something like this:
AnimalId LactationNo
1 A 1
2 B 2
3 E 2
4 A 2
5 E 2
and you'd like to select animals that happen to have both lactation numbers 1 & 2 (like A in this particular example). If that's the case, then you can simply use merge:
lactboth <- merge(totaldata[totaldata$LactationNo == 1,],
totaldata[totaldata$LactationNo == 2,],
by.x="AnimalId",
by.y="AnimalId")[,"AnimalId"]
I have a table that looks like this
uid gid score
1 a 5
1 a 8
1 a 9
1 b 2
1 b 7
2 a 5
2 a 9
.
.
.
But with many more entries for each user and group.
I want to get a table that has a row for each uid/gid pairing that is the mean of their bottom 5 scores.
This was trivial in Excel using pivot tables, but I need to do some analysis that R is much better for.
So I want my result to look like
uid gid top5avg
1 a 4.3
1 b 5.7
2 a 3.5
2 b 6.8
.
.
.
with one row for each uid gid pair and then the average of the top five scores for that uid/gid pair.
This is even more trivial in R, assuming your data frame is called dat and you really meant bottom 5 scores (even though your example suggests the top 5):
library(plyr)
ddply(dat,.(uid,gid),summarise,bottom5avg = mean(tail(sort(score),5)))
Note that this code assumes that there will be at least 5 observations in each group.
If your data was called dat this would work:
aggregate(score~uid+gid, data=dat, function(x) mean(sort(x)[1:5]))
EDIT:
If you meant the opposite (bottom 5) than what I had, as indicated by Joran (I was confused too), then use rev as in:
aggregate(score~uid+gid, data=dat, function(x) mean(rev(sort(x))[1:5]))
Or use the tail suggestion Joran made.
And the data.table solution
library(data.table)
setkey(dat,uid,gid,score)
sol <- dat[,list(avg5 = mean(tail(score,5)),by='uid,gid'])
I have done lot of googling but I didn't find satisfactory solution to my problem.
Say we have data file as:
Tag v1 v2 v3
A 1 2 3
B 1 2 2
C 5 6 1
A 9 2 7
C 1 0 1
The first line is header. The first column is Group id (the data have 3 groups A, B, C) while other column are values.
I want to read this file in R so that I can apply different functions on the data.
For example I tried to read the file and tried to get column mean
dt<-read.table(file_name,head=T) #gives warnings
apply(dt,2,mean) #gives NA NA NA
I want to read this file and want to get column mean. Then I want to separate the data in 3 groups (according to Tag A,B,C) and want to calculate mean(column wise) for each group. Any help
apply(dt,2,mean) doesn't work because apply coerces the first argument to an array via as.matrix (as is stated in the first paragraph of the Details section of ?apply). Since the first column is character, all elements in the coerced matrix object will be character.
Try this instead:
sapply(dt,mean) # works because data.frames are lists
To calculate column means by groups:
# using base functions
grpMeans1 <- t(sapply(split(dt[,c("v1","v2","v3")], dt[,"Tag"]), colMeans))
# using plyr
library(plyr)
grpMeans2 <- ddply(dt, "Tag", function(x) colMeans(x[,c("v1","v2","v3")]))