Suppose I have the following data
df = data.frame(name=c("A", "B", "C", "D"), score = c(10, 10, 9, 8))
I want to add a new column with the ranking. This is what I'm doing:
df %>% mutate(ranking = rank(score, ties.method = 'first'))
# name score ranking
# 1 A 10 3
# 2 B 10 4
# 3 C 9 2
# 4 D 8 1
However, my desired result is:
# name score ranking
# 1 A 10 1
# 2 B 10 1
# 3 C 9 2
# 4 D 8 3
Clearly rank does not do what I have in mind. What function should I be using?
It sounds like you're looking for dense_rank from "dplyr" -- but applied in a reverse order than what rank normally does.
Try this:
df %>% mutate(rank = dense_rank(desc(score)))
# name score rank
# 1 A 10 1
# 2 B 10 1
# 3 C 9 2
# 4 D 8 3
Other solution when you need to apply the rank to all variables (not just one).
df = data.frame(name = c("A","B","C","D"),
score=c(10,10,9,8), score2 = c(5,1,9,2))
select(df, -name) %>% mutate_all(funs(dense_rank(desc(.))))
#user101089 --- you can try out with this alternative way:
df = data.frame(name = c("A","B","C","D"),
score=c(10,10,9,8), score2 = c(5,1,9,2))
df %>% mutate(rank_score = dense_rank(desc(score)),
rank_score2 = dense_rank(desc(score2)))
Related
I have a R DataFrame that has a structure similar to the following:
df <- data.frame(var1 = c(1, 1), var2 = c(0, 2), var3 = c(3, 0), f1 = c('a', 'b'), f2=c('c', 'd') )
So visually the DataFrame would look like
> df
var1 var2 var3 f1 f2
1 1 0 3 a c
2 1 2 0 b d
What I want to do is the following:
(1) Treat the first C=3 columns as counts for three different classes. (C is the number of classes, given as an input variable.) Add a new column called "class".
(2) For each row, duplicate the last two entries of the row according to the count of each class (separately); and append the class number to the new "class" column.
For example, the output for the above dataset would be
> df_updated
f1 f2 class
1 a c 1
2 a c 3
3 a c 3
4 a c 3
5 b d 1
6 b d 2
7 b d 2
where row (a c) is duplicated 4 times, 1 time with respect to class 1, and 3 times with respect to class 3; row (b d) is duplicated 3 times, 1 time with respect to class 1 and 2 times with respect to class 2.
I tried looking at previous posts on duplicating rows based on counts (e.g. this link), and I could not figure out how to adapt the solutions there to multiple count columns (and also appending another class column).
Also, my actual dataset has many more rows and classes (say 1000 rows and 20 classes), so ideally I want a solution that is as efficient as possible.
I wonder if anyone can help me on this. Thanks in advance.
Here is a tidyverse option. We can use uncount from tidyr to duplicate the rows according to the count in value (i.e., from the var columns) after pivoting to long format.
library(tidyverse)
df %>%
pivot_longer(starts_with("var"), names_to = "class") %>%
filter(value != 0) %>%
uncount(value) %>%
mutate(class = str_extract(class, "\\d+"))
Output
f1 f2 class
<chr> <chr> <chr>
1 a c 1
2 a c 3
3 a c 3
4 a c 3
5 b d 1
6 b d 2
7 b d 2
Another slight variation is to use expandrows from splitstackshape in conjunction with tidyverse.
library(splitstackshape)
df %>%
pivot_longer(starts_with("var"), names_to = "class") %>%
filter(value != 0) %>%
expandRows("value") %>%
mutate(class = str_extract(class, "\\d+"))
base R
Row order (and row names) notwithstanding:
tmp <- subset(reshape2::melt(df, id.vars = c("f1","f2"), value.name = "class"), class > 0, select = -variable)
tmp[rep(seq_along(tmp$class), times = tmp$class),]
# f1 f2 class
# 1 a c 1
# 2 b d 1
# 4 b d 2
# 4.1 b d 2
# 5 a c 3
# 5.1 a c 3
# 5.2 a c 3
dplyr
library(dplyr)
# library(tidyr) # pivot_longer
df %>%
pivot_longer(-c(f1, f2), values_to = "class") %>%
dplyr::filter(class > 0) %>%
select(-name) %>%
slice(rep(row_number(), times = class))
# # A tibble: 7 x 3
# f1 f2 class
# <chr> <chr> <dbl>
# 1 a c 1
# 2 a c 3
# 3 a c 3
# 4 a c 3
# 5 b d 1
# 6 b d 2
# 7 b d 2
I'm using group by funciton in a dataset using R software. But the target of the id would duplicate. Here is the sample dataset:
ID Var1
A 1
A 3
B 2
C 3
C 1
D 2
In tradtional groupby function by each id, I can do
DT<- data.table(dataset )
DT[,sum(Var1),by = ID]
and get the result:
ID V1
A 4
B 2
C 4
D 2
However, I've to group ID by A+B and B+C and D
(PS. say that F=A+B ,G=B+C)
and the target result dataset below:
ID V1
F 6
G 6
D 2
IF I use recoding technique on ID, the duplicate B would be covered twice.
IS there any one have the solution?
MANY THANKS!
library(dplyr)
library(tidyr)
df <- df %>% mutate(F=ifelse(ID %in% c("A", "B"), 1, 0),
G = ifelse(ID %in% c("B", "C"), 1, 0),
D = ifelse(ID == "D", 1, 0))
df %>%
gather(var, val, F:D) %>%
filter(val==1) %>%
group_by(var) %>%
summarise(V1=sum(V1))
# # A tibble: 3 x 2
# var V1
# <chr> <dbl>
# 1 D 2
# 2 F 6
# 3 G 6
I have some data like this:
X Y
-----
A 1
A 2
B 3
B 4
C 5
C 6
I would like to add a new column with values equal to the mean of all Ys in rows where X is not euqal to X of the current observation.
In this particlar case we would get
X Y Mean
-------------------
A 1 (3+4+5+6)/4
A 2 (3+4+5+6)/4
B 3 (1+2+5+6)/4
B 4 (1+2+5+6)/4
C 5 (1+2+3+4)/4
C 6 (1+2+3+4)/4
Thanks in advance!
You can likely do this more succinctly, but this will get you the result.
You essentially create a column which contains the total observations and sum of records for the whole data.frame. Then you group by the X column and repeat the process, by taking the difference you can calculate your mean.
data
df <- data.frame(X = c("A", "A", "B", "B", "C", "C"),
Y = c(1:6))
solution
library(tidyverse)
df %>%
mutate(total_sum = sum(Y),
total_obs = n()) %>%
group_by(X) %>%
mutate(group_sum = sum(Y),
group_obs = n()) %>%
ungroup() %>%
mutate(other_group_sum = total_sum - group_sum,
other_group_obs = total_obs - group_obs,
other_mean = other_group_sum/other_group_obs) %>%
select(X, Y, other_mean)
result
# A tibble: 6 x 3
X Y other_mean
<fct> <int> <dbl>
1 A 1 4.50
2 A 2 4.50
3 B 3 3.50
4 B 4 3.50
5 C 5 2.50
6 C 6 2.50
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))
Suppose I have the following data
df = data.frame(name=c("A", "B", "C", "D"), score = c(10, 10, 9, 8))
I want to add a new column with the ranking. This is what I'm doing:
df %>% mutate(ranking = rank(score, ties.method = 'first'))
# name score ranking
# 1 A 10 3
# 2 B 10 4
# 3 C 9 2
# 4 D 8 1
However, my desired result is:
# name score ranking
# 1 A 10 1
# 2 B 10 1
# 3 C 9 2
# 4 D 8 3
Clearly rank does not do what I have in mind. What function should I be using?
It sounds like you're looking for dense_rank from "dplyr" -- but applied in a reverse order than what rank normally does.
Try this:
df %>% mutate(rank = dense_rank(desc(score)))
# name score rank
# 1 A 10 1
# 2 B 10 1
# 3 C 9 2
# 4 D 8 3
Other solution when you need to apply the rank to all variables (not just one).
df = data.frame(name = c("A","B","C","D"),
score=c(10,10,9,8), score2 = c(5,1,9,2))
select(df, -name) %>% mutate_all(funs(dense_rank(desc(.))))
#user101089 --- you can try out with this alternative way:
df = data.frame(name = c("A","B","C","D"),
score=c(10,10,9,8), score2 = c(5,1,9,2))
df %>% mutate(rank_score = dense_rank(desc(score)),
rank_score2 = dense_rank(desc(score2)))