Selecting the first nth rows by group with number of rows varied - r

I like to select the first (2,3,0,4) rows of each group in a data frame.
> f<-data.frame(group=c(1,1,1,2,2,3,4),y=c(1:7))
>
> group y
> 1 1
> 1 2
> 1 3
> 2 4
> 2 5
> 3 6
> 4 7
and obtain a data frame as follows
group y
1 1
1 2
2 4
2 5
4 7
I tried to use by and head but head does not take a vector.
Thank you for your help.

With the more traditional lapply:
k <- c(2,3,0,4)
fs <- split(f, f$group)
do.call(rbind,lapply(seq_along(k), function(i) head(fs[[i]], k[i])))
result is:
group y
1 1 1
2 1 2
4 2 4
5 2 5
7 4 7

Using plyr:
library(plyr)
rows <- c(2,3,0,4)
ddply(f,.(group),function(x)head(x,rows[x[1,1]]))
group y
1 1 1
2 1 2
3 2 4
4 2 5
5 4 7
edit:
misunderstood the question so updated answer

Version of function with indexes.
fun1 <- function(){
idx <- c(0,which(diff(f$group)!=0))+1
idx2 <- unlist(lapply(1:length(nf),function(x) seq.int(from=idx[x],length.out=nf[x])),use.names=F)
f1 <- f[idx2,]
return(f1)
}
fun2 <- function(){
ddply(f,.(group),function(x) head(x,nf[x[1,1]]))
}
Test data (size suggested by author of question)
f<-data.frame(group=sample(1:1000,50000,T),y=c(1:50000))
f <- f[order(f$group),]
nf <- rpois(length(unique(f$group)),3)
system.time(fun1())
system.time(fun2())
On my system ~60 times faster is fun1.

Related

How to get the index column in a data frame in R [duplicate]

Starting with a data.frame...
df = data.frame(k=c(1,5,4,7,6), v=c(3,1,4,1,5))
> df
k v
1 1 3
2 5 1
3 4 4
4 7 1
5 6 5
I might run some number of arbitrary manipulations...
> foo1 = df[df$k>3,]
> foo2 = head(foo1[order(foo1$v),], 2)
> foo2
k v
2 5 1
4 7 1
At this point foo2 has somehow retained the original row numbers fromdf (in this case 2 and 4).
How do I extract these?
> insert_magic_function_here(foo2)
[1] 2 4
I think you're looking for rownames.

How can I subset a dataframe according to group membership?

I am wanting to write a function so that a (potentially large) dataframe can be subsetted according to group membership, where a 'group' is a unique combination of a set of column values.
For example, I would like to subset the following data frame according to unique combination of the first two columns (Loc1 and Loc2).
Loc1 <- c("A","A","A","A","B","B","B")
Loc2 <- c("a","a","b","b","a","a","b")
Dat1 <- c(1,1,1,1,1,1,1)
Dat2 <- c(1,2,1,2,1,2,2)
Dat3 <- c(2,2,4,4,6,5,3)
DF=data.frame(Loc1,Loc2,Dat1,Dat2,Dat3)
Loc1 Loc2 Dat1 Dat2 Dat3
1 A a 1 1 2
2 A a 1 2 2
3 A b 1 1 4
4 A b 1 2 4
5 B a 1 1 6
6 B a 1 2 5
7 B b 1 2 3
I want to return (i) the number of groups (i.e. 4), (ii) the number in each group (i.e. c(2,2,2,1), and (iii) to relabel the rows so that I can further analyse the data frame according to group membership (e.g. for ANOVA and MANOVA) (i.e.
Group<-as.factor(c(1,1,2,2,3,3,4))
Data <- cbind(Group,DF[,-1:-2])
Group Dat1 Dat2 Dat3
1 1 1 1 2
2 1 1 2 2
3 2 1 1 4
4 2 1 2 4
5 3 1 1 6
6 3 1 2 5
7 4 1 2 3
).
So far all I have managed is to get the number of groups, and I'm suspicious that there's a better way to do even this:
nrow(unique(DF[,1:2]))
I was hoping to avoid for-loops as I am concerned about the function being slow.
I have tried converting to a data matrix so that I could concatenate the row values but I couldn't get that to work either.
Many thanks
You could try:
Create Group column by using unique level combination of Loc1 and Loc2.
indx <- paste(DF[,1], DF[,2])
DF$Group <- as.numeric(factor(indx, unique(indx))) #query No (iii)
DF1 <- DF[-(1:2)][,c(4,1:3)]
# Group Dat1 Dat2 Dat3
#1 1 1 1 2
#2 1 1 2 2
#3 2 1 1 4
#4 2 1 2 4
#5 3 1 1 6
#6 3 1 2 5
#7 4 1 2 3
table(DF$Group) #(No. ii)
#1 2 3 4
#2 2 2 1
length(unique(DF$Group)) #(i)
#[1] 4
Then, if you need to subset the datasets by group, you could split the dataset using the Group to create a list of 4 list elements
split(DF1, DF1$Group)
Update
If you have multiple columns, you could still try:
ColstoGroup <- 1:2
indx <- apply(DF[,ColstoGroup], 1, paste, collapse="")
as.numeric(factor(indx, unique(indx)))
#[1] 1 1 2 2 3 3 4
You could create a function;
fun1 <- function(dat, GroupCols){
FactGroup <- dat[, GroupCols]
if(length(GroupCols)==1){
dat$Group <- as.numeric(factor(FactGroup, levels=unique(FactGroup)))
}
else {
indx <- apply(FactGroup, 1, paste, collapse="")
dat$Group <- as.numeric(factor(indx, unique(indx)))
}
dat
}
fun1(DF, "Loc1")
fun1(DF, c("Loc1", "Loc2"))
This gets all three of your queries.
Begin with a table of the first two columns and then work with that data.
> (tab <- table(DF$Loc1, DF$Loc2))
#
# a b
# A 2 2
# B 2 1
#
> (ct <- c(tab)) ## (ii)
# [1] 2 2 2 1
> length(unlist(dimnames(tab))) ## (i)
# [1] 4
> cbind(Group = rep(seq_along(ct), ct), DF[-c(1,2)]) ## (iii)
# Group Dat1 Dat2 Dat3
# 1 1 1 1 2
# 2 1 1 2 2
# 3 2 1 1 4
# 4 2 1 2 4
# 5 3 1 1 6
# 6 3 1 2 5
# 7 4 1 2 3
Borrowing a bit from this answer and using some dplyr idioms:
library(dplyr)
Loc1 <- c("A","A","A","A","B","B","B")
Loc2 <- c("a","a","b","b","a","a","b")
Dat1 <- c(1,1,1,1,1,1,1)
Dat2 <- c(1,2,1,2,1,2,2)
Dat3 <- c(2,2,4,4,6,5,3)
DF <- data.frame(Loc1, Loc2, Dat1, Dat2, Dat3)
emitID <- local({
idCounter <- -1L
function(){
idCounter <<- idCounter + 1L
}
})
DF %>% group_by(Loc1, Loc2) %>% mutate(Group=emitID())
## Loc1 Loc2 Dat1 Dat2 Dat3 Group
## 1 A a 1 1 2 0
## 2 A a 1 2 2 0
## 3 A b 1 1 4 1
## 4 A b 1 2 4 1
## 5 B a 1 1 6 2
## 6 B a 1 2 5 2
## 7 B b 1 2 3 3

Find index of removed unique pairs

I have a data frame of 2 columns and a vector of the same length. I am trying to remove all duplicated pairs in the data frame and at the same index, remove it from the vector.
I have a data frame:
> from <- c(1,1,2,4,3)
> to <- c(1,1,2,3,5)
> ft <- data.frame(from,to)
> ft
from to
1 1 1
2 1 1
3 2 2
4 4 3
5 3 5
And vector:
> dist <- c(1,2,3,4,5)
> dist
[1] 1 2 3 4 5
I used the function unique() to remove all duplicated pairs:
> unique(ft)
from to
1 1 1
3 2 2
4 4 3
5 3 5
How can I get the index of where every pair from "ft" has been removed so that I can remove it from "dist" which would be the 2 in this case.
As #eddi notes, you can get a logical vector that indicates which rows are duplicates with duplicated(). I combined that with which(), which returns the number associated with the logical that is TRUE (i.e., the duplicated row). You can then create a new data.frame (vector, etc.) by using - to not include the indicated rows in the subscript of your object.
Edit: In the comments, #DWin points out a better way than using -. If we negate the duplicated() function with !, we will get a vector that we can use to determine which rows to retain:
> from <- c(1,1,2,4,3)
> to <- c(1,1,2,3,5)
> ft <- data.frame(from,to)
> ft
from to
1 1 1
2 1 1
3 2 2
4 4 3
5 3 5
> dist <- c(1,2,3,4,5)
> dist
[1] 1 2 3 4 5
> remove <- !duplicated(ft)
> remove
[1] TRUE FALSE TRUE TRUE TRUE
> ft.new <- ft[which(remove), ]
> ft.new
from to
1 1 1
3 2 2
4 4 3
5 3 5
> dist.new <- dist[which(remove)]
> dist.new
[1] 1 3 4 5

Count and label observations per participant using loop

I have repeated-measures data.
I need to create a loop that will incrementally count each observation, within a participant, and label it.
I am new to writing loops. My logic was to say, for each item in the list of unique ids, count each row in that, and apply some function to that row.
Could someone point our what I am doing wrong?
data$Ob <- 0
for (i in unique(data$id)) {
count <- 1
for (u in data[data$id == i,]) {
data[data$id ==u,]$Ob <- count
count <- count + 1
print(count)
}
}
Thanks!
Justin
You can also use ave:
set.seed(1)
data <- data.frame(id = sample(4, 10, TRUE))
data$Ob = ave(data$id, data$id, FUN=seq_along)
data
id Ob
1 2 1
2 2 2
3 3 1
4 4 1
5 1 1
6 4 2
7 4 3
8 3 2
9 3 3
10 1 2
# Generate some dummy data
data <- data.frame(Ob=0, id=sample(4,20,TRUE))
# Go through every id value
for(i in unique(data$id)){
# Label observations
data$Ob[data$id == i] = 1:sum(data$id == i)
}
Be aware though that for loops are notoriously slow in R. In this simple case they work fine, but should you have millions and millions of rows in your data frame you'd better do something purely vectorized.
But you don't need a loop...
data <- data.frame (id = sample (4, 10, TRUE))
## id
## 1 3
## 2 4
## 3 1
## 4 3
## 5 3
## 6 4
## 7 2
## 8 1
## 9 1
## 10 4
data$Ob [order (data$id)] <- sequence (table (data$id))
## id Ob
## 1 3 1
## 2 4 1
## 3 1 1
## 4 3 2
## 5 3 3
## 6 4 2
## 7 2 1
## 8 1 2
## 9 1 3
## 10 4 3
(works also with character or factor IDs)
(isn't R just cool!?)

extracting row labels (?) from a data.frame

Starting with a data.frame...
df = data.frame(k=c(1,5,4,7,6), v=c(3,1,4,1,5))
> df
k v
1 1 3
2 5 1
3 4 4
4 7 1
5 6 5
I might run some number of arbitrary manipulations...
> foo1 = df[df$k>3,]
> foo2 = head(foo1[order(foo1$v),], 2)
> foo2
k v
2 5 1
4 7 1
At this point foo2 has somehow retained the original row numbers fromdf (in this case 2 and 4).
How do I extract these?
> insert_magic_function_here(foo2)
[1] 2 4
I think you're looking for rownames.

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