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I have a dataframe in R that I'd like to repeat several times, and I want to add in a new variable to index those repetitions. The best I've come up with is using mutate + rbind over and over, and I feel like there has to be an efficient dataframe method I could be using here.
Here's an example: df <- data.frame(x = 1:3, y = letters[1:3]) gives us the dataframe
x
y
1
a
2
b
3
c
I'd like to repeat that say 3 times, with an index that looks like this:
x
y
index
1
a
1
2
b
1
3
c
1
1
a
2
2
b
2
3
c
2
1
a
3
2
b
3
3
c
3
Using the rep function, I can get the first two columns, but not the index column. The best I've come up with so far (using dplyr) is:
df2 <-
df %>%
mutate(index = 1) %>%
rbind(df %>% mutate(index = 2)) %>%
rbind(df %>% mutate(index = 3))
This obviously doesn't work if I need to repeat my dataframe more than a handful of times. It feels like the kind of thing that should be easy to do using dataframe methods, but I haven't been able to find anything.
Grateful for any tips!
You can use this code for as many data frames as you would like. You just have to set the n argument:
replicate function takes 2 main arguments. We first specify the number of time we would like to reproduce our data set by n. Then we specify our data set as expr argument. The result would be a list whose elements are instances of our data set
After that we pass it along to imap function from purrr package to define the unique id for each of our data set. .x represents each element of our list (here a data frame) and .y is the position of that element which amounts to the number of instances we created. So for example we assign value 1 to the first id column of the first data set as .y is equal to 1 for that and so on.
library(dplyr)
library(purrr)
replicate(3, df, simplify = FALSE) %>%
imap_dfr(~ .x %>%
mutate(id = .y))
x y id
1 1 a 1
2 2 b 1
3 3 c 1
4 1 a 2
5 2 b 2
6 3 c 2
7 1 a 3
8 2 b 3
9 3 c 3
In base R you can use the following code:
do.call(rbind,
mapply(function(x, z) {
x$id <- z
x
}, replicate(3, df, simplify = FALSE), 1:3, SIMPLIFY = FALSE))
x y id
1 1 a 1
2 2 b 1
3 3 c 1
4 1 a 2
5 2 b 2
6 3 c 2
7 1 a 3
8 2 b 3
9 3 c 3
You can use rerun to repeat the dataframe n times and add an index column using bind_rows -
library(dplyr)
library(purrr)
n <- 3
df <- data.frame(x = 1:3, y = letters[1:3])
bind_rows(rerun(n, df), .id = 'index')
# index x y
#1 1 1 a
#2 1 2 b
#3 1 3 c
#4 2 1 a
#5 2 2 b
#6 2 3 c
#7 3 1 a
#8 3 2 b
#9 3 3 c
In base R, we can repeat the row index 3 times.
transform(df[rep(1:nrow(df), n), ], index = rep(1:n, each = nrow(df)))
One more way
n <- 3
map_dfr(seq_len(n), ~ df %>% mutate(index = .x))
x y index
1 1 a 1
2 2 b 1
3 3 c 1
4 1 a 2
5 2 b 2
6 3 c 2
7 1 a 3
8 2 b 3
9 3 c 3
I have a dataset with several variables, but I want to keep the rows that are the same based on two columns. Here is an example of what I want to do:
a <- c(rep('A',3), rep('B', 3), rep('C',3))
b <- c(1,1,2,4,4,4,5,5,5)
df <- data.frame(a,b)
a b
1 A 1
2 A 1
3 A 2
4 B 4
5 B 4
6 B 4
7 C 5
8 C 5
9 C 5
I know that if I use the duplicated function I can get:
df[!duplicated(df),]
a b
1 A 1
3 A 2
4 B 4
7 C 5
But since the level 'A' on column a does not have a unique value in b, I want to drop both observations to get a new data.frame as this:
a b
4 B 4
7 C 5
I don't mind to have repeated values across b, as long as for every same level on a there is the same value in b.
Is there a way to do this? Thanks!
This one maybe?
ag <- aggregate(b~a, df, unique)
ag[lengths(ag$b)==1,]
# a b
#2 B 4
#3 C 5
Maybe something like this:
> ind <- apply(sapply(with(df, split(b,a)), diff), 2, function(x) all(x==0) )
> out <- df[!duplicated(df),]
> out[out$a %in% names(ind)[ind], ]
a b
4 B 4
7 C 5
Here is another option with data.table
library(data.table)
setDT(df)[, if(uniqueN(b)==1) .SD[1L], by = a]
# a b
#1: B 4
#2: C 5
This question already has answers here:
How to select the rows with maximum values in each group with dplyr? [duplicate]
(6 answers)
Closed 6 years ago.
I have a data frame, say
df <- data.frame(x = c(1,2,5,6,3,3,3,6,8,8,8,8),
y = c(1,1,1,1,1,2,3,1,1,2,3,4),
z = c("a","b","c","d","e","f","g","h","i","j","k","l"))
it looks like this
x y z
1 1 1 a
2 2 1 b
3 5 1 c
4 6 1 d
5 3 1 e
6 3 2 f
7 3 3 g
8 6 1 h
9 8 1 i
10 8 2 j
11 8 3 k
12 8 4 l
I would like pick unique elements from column x, based on column y such that y should be maximum (in this case say for row number 5 to 7 are 3'3, I would like to pick the x = 3 corresponding to y = 3 (maximum value) similarly for x = 8 I d like to pick y = 4 row )
the output should look like this
x y z
1 1 1 a
2 2 1 b
3 5 1 c
4 6 1 d
5 3 3 g
6 6 1 h
7 8 4 l
I have a solution for that, which I am posting in the solution, but if there is there any better method to achieve this, My solution only works in this specific case (picking the largest) what is the general case solution for this?
One solution using dplyr
library(dplyr)
df %>%
group_by(x) %>%
slice(max(y))
# x y z
# (dbl) (dbl) (chr)
#1 1 1 a
#2 2 1 b
#3 3 3 g
#4 5 1 c
#5 6 1 d
#6 8 4 l
The base R alternative is using aggregate
aggregate(y~x, df, max)
You can achieve the same result using a dplyr chain and dplyr's group_by function. Once you use a group_by function the rest of the functions in the chain are applied within group as opposed to the whole data.frame. So here I filter to where the only rows left are the max(y) per the grouping value of x. This can be extended to be used for the min of y or a particular value.
I think its generally good practice to ungroup the data at the end of a chain using group_by to avoid any unexpected behavior.
library(dplyr)
df <- data.frame(x = c(1,2,5,6,3,3,3,6,8,8,8,8),
y = c(1,1,1,1,1,2,3,1,1,2,3,4),
z = c("a","b","c","d","e","f","g","h","i","j","k","l"))
df %>%
group_by(x) %>%
filter(y==max(y)) %>%
ungroup()
To make it more general... say instead you wanted the mean of y for a given x as opposed to the max. You could then use the summarise function instead of the filter as shown below.
df %>%
group_by(x) %>%
summarise(y=mean(y)) %>%
ungroup()
Using data.table we can use df[order(z), .I[which.max(y)], by = x] to get the rownumbers of interest, eg:
library(data.table)
setDT(df)
df[df[order(z), .I[which.max(y)], by = x][, V1]]
x y z
1: 1 1 a
2: 2 1 b
3: 5 1 c
4: 6 1 d
5: 3 3 g
6: 8 4 l
Here is my solution using dplyr package
library(dplyr)
df <- data.frame(x = c(1,2,5,6,3,3,3,6,8,8,8,8),
y = c(1,1,1,1,1,2,3,1,1,2,3,4),
z = c("a","b","c","d","e","f","g","h","i","j","k","l"))
df <- arrange(df,desc(y))
df_out <- df[!duplicated(df$x),]
df_out
Printing df_out
x y z
1 8 4 l
2 3 3 g
6 1 1 a
7 2 1 b
8 5 1 c
9 6 1 d
Assuming the data frame is ordered by df[order(df$x, df$y),] as it is in the example, you can use base R functions, split, lapply, and do.call/rbind to extract your desired rows using the "split / apply / combine" methodology.
do.call(rbind, lapply(split(df, df$x), function(i) i[nrow(i),]))
x y z
1 1 1 a
2 2 1 b
3 3 3 g
5 5 1 c
6 6 1 h
8 8 4 l
split breaks up the data.frame into a list based on x. This list is fed to lapply which selects the last row of each data.frame, and returns these one row data.frames as a list. This list is then rbinded into a single data frame using do.call.
Suppose I have a dataframe like so
a b c
1 2 3
1 3 4
1 4 5
2 5 6
2 6 7
3 7 8
4 8 9
What I want is the following:
a b c d
1 2 3 a
1 3 4 b
1 4 5 c
2 5 6 a
2 6 7 b
3 7 8 a
4 8 9 a
Essentially, I want to do a cycling, for each group by the column a, I want to create a new column which cycles the letters from a to z in order. Group 1 has three elements, so the letter goes from 'a' to 'c'. Group 3 and 4 has only 1 element, so the letter only gets assigned 'a'.
A data.table option is
library(data.table)
setDT(dd)[, d:= letters[seq_len(.N)], by = a]
One way to do this is with a split-apply-combine paradigm, as in plyr (or dplyr or data.table or ...
Create data:
dd <- data.frame(a=rep(1:4,c(3,2,1,1)),
b=2:8,c=3:9)
Use ddply to split the data frame by variable a, transforming each piece by adding an appropriate variable, then recombine:
library("plyr")
ddply(dd,"a",
transform,
d=letters[1:length(b)])
Or in dplyr:
library("dplyr")
dd %>% group_by(a) %>%
mutate(d=letters[1:n()])
Or in base R (thanks #thelatemail):
dd$d <- ave(rownames(dd), dd$a,
FUN=function(x) letters[seq_along(x)] )
I have a data.frame mydf with about 2500 rows. These rows correspond to 69 classes of objects in colum 1 mydf$V1, and I want to count how many rows per object class I have.
I can get a factor of these classes with:
objectclasses = unique(factor(mydf$V1, exclude="1"));
What's the terse R way to count the rows per object class? If this were any other language I'd be traversing an array with a loop and keeping count but I'm new to R programming and am trying to take advantage of R's vectorised operations.
Or using the dplyr library:
library(dplyr)
set.seed(1)
dat <- data.frame(ID = sample(letters,100,rep=TRUE))
dat %>%
group_by(ID) %>%
summarise(no_rows = length(ID))
Note the use of %>%, which is similar to the use of pipes in bash. Effectively, the code above pipes dat into group_by, and the result of that operation is piped into summarise.
The result is:
Source: local data frame [26 x 2]
ID no_rows
1 a 2
2 b 3
3 c 3
4 d 3
5 e 2
6 f 4
7 g 6
8 h 1
9 i 6
10 j 5
11 k 6
12 l 4
13 m 7
14 n 2
15 o 2
16 p 2
17 q 5
18 r 4
19 s 5
20 t 3
21 u 8
22 v 4
23 w 5
24 x 4
25 y 3
26 z 1
See the dplyr introduction for some more context, and the documentation for details regarding the individual functions.
Here 2 ways to do it:
set.seed(1)
tt <- sample(letters,100,rep=TRUE)
## using table
table(tt)
tt
a b c d e f g h i j k l m n o p q r s t u v w x y z
2 3 3 3 2 4 6 1 6 5 6 4 7 2 2 2 5 4 5 3 8 4 5 4 3 1
## using tapply
tapply(tt,tt,length)
a b c d e f g h i j k l m n o p q r s t u v w x y z
2 3 3 3 2 4 6 1 6 5 6 4 7 2 2 2 5 4 5 3 8 4 5 4 3 1
Using plyr package:
library(plyr)
count(mydf$V1)
It will return you a frequency of each value.
Using data.table
library(data.table)
setDT(dat)[, .N, keyby=ID] #(Using #Paul Hiemstra's `dat`)
Or using dplyr 0.3
res <- count(dat, ID)
head(res)
#Source: local data frame [6 x 2]
# ID n
#1 a 2
#2 b 3
#3 c 3
#4 d 3
#5 e 2
#6 f 4
Or
dat %>%
group_by(ID) %>%
tally()
Or
dat %>%
group_by(ID) %>%
summarise(n=n())
We can use summary on factor column:
summary(myDF$factorColumn)
One more approach would be to apply n() function which is counting the number of observations
library(dplyr)
library(magrittr)
data %>%
group_by(columnName) %>%
summarise(Count = n())
In case I just want to know how many unique factor levels exist in the data, I use:
length(unique(df$factorcolumn))
Use the package plyr with lapply to get frequencies for every value (level) and every variable (factor) in your data frame.
library(plyr)
lapply(df, count)
This is an old post, but you can do this with base R and no data frames/data tables:
sapply(levels(yTrain), function(sLevel) sum(yTrain == sLevel))