Let's say I have a data table like this:
smalldat <- data.table(group1 = rep(1:2, each = 3),
group2 = rep(c('a','b'), times = 3,
value = 1:6)
That looks as follows:
group1 group2 value
1 a 1
1 b 2
1 a 3
2 b 4
2 a 5
2 b 6
I want to calculate the number of observed combinations of group1 and group2.
The dplyr way would be (possibly not the most optimal):
nrow(smalldat %>% select(group1, group2) %>% distinct())
What would be the data.table way?
Use uniqueN along with .SD and .SDcols:
smalldat[, uniqueN(.SD), .SDcols=group1:group2]
# [1] 4
Or even more efficient, as #DavidArenburg shows under comment:
uniqueN(smalldat, by=c("group1", "group2"))
# [1] 4
We can use unique with the by option.
nrow(unique(smalldat, by = c('group1', 'group2')))
Or
length(smalldat[,.GRP ,.(group1, group2)]$GRP)
Related
How do I add a column to a data frame consisting of the minimum values from other columns? So in this case, to create a third column that will have the values 1, 2 and 2?
df = data.frame(A = 1:3, B = 4:2)
You can use apply() function to do this. See below.
df$C <- apply(df, 1, min)
The second argument allows you to choose the dimension in which you want min to be applied, in this case 1, applies min to all columns in each row separately.
You can choose specific columns from the dataframe, as follows:
df$newCol <- apply(df[c('A','B')], 1, min)
You can call the parallel minimum function with do.call to apply it on all your columns:
df$C <- do.call(pmin, df)
df %>%
rowwise() %>%
mutate(C = min(A, B))
# A tibble: 3 × 3
# Rowwise:
A B C
<int> <int> <int>
1 1 4 1
2 2 3 2
3 3 2 2
Using input with equal values across rows:
df = data.frame(A = 1:10, B = 11:2)
df %>%
rowwise() %>%
mutate(C = min(A, B))
# A tibble: 10 × 3
# Rowwise:
A B C
<int> <int> <int>
1 1 11 1
2 2 10 2
3 3 9 3
4 4 8 4
5 5 7 5
6 6 6 6
7 7 5 5
8 8 4 4
9 9 3 3
10 10 2 2
You do simply:
df$C <- apply(FUN=min,MARGIN=1,X=df)
Or:
df[, "C"] <- apply(FUN=min,MARGIN=1,X=df)
or:
df["C"] <- apply(FUN=min,MARGIN=1,X=df)
Instead of apply, you could also use data.farme(t(df)), where t transposes df, because sapply would traverse a data frame column-wise applying the given function. So the rows must be made columns. Since t outputs always a matrix, you need to make it a data.frame() again.
df$C <- sapply(data.frame(t(df)), min)
Or one could use the fact that ifelse is vectorized:
df$C <- with(df, ifelse(A<B,A,B))
Or:
df$C <- ifelse(df$A < df$B, df$A, df$B)
matrixStats
# install.packages("matrixStats")
matrixStats::rowMins(as.matrix(df))
According to this SO answer the fastest.
apply-type functions use lists and are always quite slow.
You can use transform() to add the min column as the output of pmin(a, b) and access the elements of df without indexing:
df <- transform(df, min = pmin(a, b))
or
In data.table
library(data.table)
DT = data.table(a = 1:3, b = 4:2)
DT[, min := pmin(a, b)]
I have a data frame where column "A" has 6 distinct values. Column "B" has float values. By using dplyr, I can group by column "A" and find mean of column "B" of each group as follows:
mydf %>% group_by(A) %>% summarize(Mean = mean(B, na.rm=TRUE))
My utter aim is to find rows in each group whose "B" values are higher than the group average. How can I achieve this (using base R or dplyr)?
A simple alternative with base R ave would be
df[df$b > ave(df$b, df$a) , ]
# a b
#4 1 4
#5 1 5
#9 2 9
#10 2 10
The default argument for ave is mean so no need to mention it explicitly, if there are NA values present in b modify it to
df[df$b > ave(df$b, df$a, FUN = function(x) mean(x,na.rm = TRUE)) , ]
Another solution with subset and ave as suggested by #Onyambu
subset(df,b>ave(b,a))
# a b
#4 1 4
#5 1 5
#9 2 9
#10 2 10
data
df <- data.frame(a = rep(c(1, 2), each = 5), b = 1:10)
df
# a b
#1 1 1
#2 1 2
#3 1 3
#4 1 4
#5 1 5
#6 2 6
#7 2 7
#8 2 8
#9 2 9
#10 2 10
You can just group and then filter:
mydf %>%
group_by(A) %>%
filter(B > mean(B, na.rm = TRUE)) %>%
ungroup()
Using Base R, I would go for this. It is not as elegant as dplyr.
mean.df <- aggregate(mydf$b, by =list(a = mydf$a), FUN = mean)
names(mean.df)[2] <- "mean"
mydf <- merge(mydf, mean.df, by = "a")
# Rows whose values are higher than mean
new.df <- subset(mydf, b > mean, select = -mean)
I like working with Data tables. So a data.table solution would be,
mydt <- data.table(mydf)
mydt[, mean := mean(b), by = a]
new.dt <- mydt[b > mean, -c("mean"), with = TRUE]
Another way to do it using base R and tapply:
mydf = cbind.data.frame(A=sample(6,20,rep=T),B=runif(20))
mydf.ave = tapply(mydf$B,mydf$A,mean)
newdf = mydf[mydf$B > mydf.ave[as.character(mydf$A)],]
(thus the one liner would be:mydf[mydf$B > tapply(mydf$B,mydf$A,mean)[as.character(mydf$A)],])
I am trying to select the maximum value in a dataframe's third column based on the combinations of the values in the first two columns.
My problem is similar to this one but I can't find a way to implement what I need.
EDIT: Sample data changed to make the column names more obvious.
Here is some sample data:
library(tidyr)
set.seed(1234)
df <- data.frame(group1 = letters[1:4], group2 = letters[1:4])
df <- df %>% expand(group1, group2)
df <- subset(df, subset = group1!=group2)
df$score <- runif(n = 12,min = 0,max = 1)
df
# A tibble: 12 × 3
group1 group2 score
<fctr> <fctr> <dbl>
1 a b 0.113703411
2 a c 0.622299405
3 a d 0.609274733
4 b a 0.623379442
5 b c 0.860915384
6 b d 0.640310605
7 c a 0.009495756
8 c b 0.232550506
9 c d 0.666083758
10 d a 0.514251141
11 d b 0.693591292
12 d c 0.544974836
In this example rows 1 and 4 are 'duplicates'. I would like to select row 4 as the value in the score column is larger than in row 1. Ultimately I would like a dataframe to be returned with the group1 and group2 columns and the maximum value in the score column. So in this example, I expect there to be 6 rows returned.
How can I do this in R?
I'd prefer dealing with this problem in two steps:
library(dplyr)
# Create function for computing group IDs from data frame of groups (per column)
get_group_id <- function(groups) {
apply(groups, 1, function(row) {
paste0(sort(row), collapse = "_")
})
}
group_id <- get_group_id(select(df, -score))
# Perform the computation
df %>%
mutate(groupId = group_id) %>%
group_by(groupId) %>%
slice(which.max(score)) %>%
ungroup() %>%
select(-groupId)
I didn't find a solution for this common grouping problem in R:
This is my original dataset
ID State
1 A
2 A
3 B
4 B
5 B
6 A
7 A
8 A
9 C
10 C
This should be my grouped resulting dataset
State min(ID) max(ID)
A 1 2
B 3 5
A 6 8
C 9 10
So the idea is to sort the dataset first by the ID column (or a timestamp column). Then all connected states with no gaps should be grouped together and the min and max ID value should be returned. It's related to the rle method, but this doesn't allow the calculation of min, max values for the groups.
Any ideas?
You could try:
library(dplyr)
df %>%
mutate(rleid = cumsum(State != lag(State, default = ""))) %>%
group_by(rleid) %>%
summarise(State = first(State), min = min(ID), max = max(ID)) %>%
select(-rleid)
Or as per mentioned by #alistaire in the comments, you can actually mutate within group_by() with the same syntax, combining the first two steps. Stealing data.table::rleid() and using summarise_all() to simplify:
df %>%
group_by(State, rleid = data.table::rleid(State)) %>%
summarise_all(funs(min, max)) %>%
select(-rleid)
Which gives:
## A tibble: 4 × 3
# State min max
# <fctr> <int> <int>
#1 A 1 2
#2 B 3 5
#3 A 6 8
#4 C 9 10
Here is a method that uses the rle function in base R for the data set you provided.
# get the run length encoding
temp <- rle(df$State)
# construct the data.frame
newDF <- data.frame(State=temp$values,
min.ID=c(1, head(cumsum(temp$lengths) + 1, -1)),
max.ID=cumsum(temp$lengths))
which returns
newDF
State min.ID max.ID
1 A 1 2
2 B 3 5
3 A 6 8
4 C 9 10
Note that rle requires a character vector rather than a factor, so I use the as.is argument below.
As #cryo111 notes in the comments below, the data set might be unordered timestamps that do not correspond to the lengths calculated in rle. For this method to work, you would need to first convert the timestamps to a date-time format, with a function like as.POSIXct, use df <- df[order(df$ID),], and then employ a slight alteration of the method above:
# get the run length encoding
temp <- rle(df$State)
# construct the data.frame
newDF <- data.frame(State=temp$values,
min.ID=df$ID[c(1, head(cumsum(temp$lengths) + 1, -1))],
max.ID=df$ID[cumsum(temp$lengths)])
data
df <- read.table(header=TRUE, as.is=TRUE, text="ID State
1 A
2 A
3 B
4 B
5 B
6 A
7 A
8 A
9 C
10 C")
An idea with data.table:
require(data.table)
dt <- fread("ID State
1 A
2 A
3 B
4 B
5 B
6 A
7 A
8 A
9 C
10 C")
dt[,rle := rleid(State)]
dt2<-dt[,list(min=min(ID),max=max(ID)),by=c("rle","State")]
which gives:
rle State min max
1: 1 A 1 2
2: 2 B 3 5
3: 3 A 6 8
4: 4 C 9 10
The idea is to identify sequences with rleid and then get the min and max of IDby the tuple rle and State.
you can remove the rle column with
dt2[,rle:=NULL]
Chained:
dt2<-dt[,list(min=min(ID),max=max(ID)),by=c("rle","State")][,rle:=NULL]
You can shorten the above code even more by using rleid inside by directly:
dt2 <- dt[, .(min=min(ID),max=max(ID)), by=.(State, rleid(State))][, rleid:=NULL]
Here is another attempt using rle and aggregate from base R:
rl <- rle(df$State)
newdf <- data.frame(ID=df$ID, State=rep(1:length(rl$lengths),rl$lengths))
newdf <- aggregate(ID~State, newdf, FUN = function(x) c(minID=min(x), maxID=max(x)))
newdf$State <- rl$values
# State ID.minID ID.maxID
# 1 A 1 2
# 2 B 3 5
# 3 A 6 8
# 4 C 9 10
data
df <- structure(list(ID = 1:10, State = c("A", "A", "B", "B", "B",
"A", "A", "A", "C", "C")), .Names = c("ID", "State"), class = "data.frame",
row.names = c(NA,
-10L))
I have, for example, a vector with 1000 obs and 3 levels (A, B, C). I want to count how many times level A occurs for every 5 rows and produce another vector of the count values, ie with 200obs. Is anyone able to help? I've found how to count based on another variable but not number of rows. Thank you!
df <- data.frame(test=factor(sample(c("A","B", "C" ),1000,replace=TRUE)))
head(df, 10)
test
1 A
2 A
3 B
4 C
5 B
6 A
7 C
8 B
9 C
10 C
Here are a couple of options you might find useful:
a) count all entries per 5 rows and return a list:
head(lapply(split(df$test, rep(1:200, each = 5)), table), 2)
# $`1` # <- result for rows 1:5
#
# A B C
# 1 0 4
#
# $`2` # <- result for rows 6:10
#
# A B C
# 3 0 2
b) count all entries per 5 rows and return a matrix:
head(t(sapply(split(df$test, rep(1:200, each = 5)), table)), 2)
# A B C
# 1 1 0 4
# 2 3 0 2
c) count number of As per 5 rows and return a list:
head(lapply(split(df$test == "A", rep(1:200, each = 5)), sum), 2)
# $`1`
# [1] 1
#
# $`2`
# [1] 3
d) count number of As per 5 rows and return a vector:
head(sapply(split(df$test == "A", rep(1:200, each = 5)), sum), 2)
#1 2
#1 3
Each of the results will be 200 entries long / have 200 rows.
Here is a solution with dplyr and tidyr
library(dplyr)
library(tidyr)
df %>%
mutate(Set = (seq_along(test) - 1) %/% 5) %>%
group_by(Set, test) %>%
summarise(N = n()) %>%
spread(key = test, value = N, fill = 0)
We can use data.table
library(data.table)
setDT(df)[, .N , .(grp= gl(nrow(df), 5, nrow(df)), test)]
If you prefer dplyr, you could use
c1 <- df %>%
mutate(group = rep(paste0("G", seq(1, 200)), each = 5)) %>%
# count each level
count(group, test)
Note that this method doesn't include levels with no values for a certain group (i.e. no 0 values)