This question already has answers here:
Idiomatic R code for partitioning a vector by an index and performing an operation on that partition
(3 answers)
Closed 9 years ago.
Suppose I have a data frame like this:
set.seed(123)
df <- as.data.frame(cbind(y<-sample(c("A","B","C"),10,T), X<-sample(c(1,2,3),10,T)))
df <- df[order(df$V1),]
Is there a simply function to sum (or any FUN) V2 by V1 and add to df as a new column, such that:
df$sum <- c(6,6,8,8,8,8,6,6,6,6)
df
I may write a function for that, but I have to do that frequently and be better to know the simplest way to realize that.
I agree with #mnel at least on his first point. I didn't see ave demonstrated in the answers he cited and I think it's the "simplest" base-R method. Using that data.frame(cbind( ...)) construction should be outlawed and teachers who demonstrate it should be stripped of their credentials.
set.seed(123)
df<-data.frame(y=sample( c("A","B","C"), 10, T),
X=sample(c (1,2,3), 10, T))
df<-df[order(df$y),] # that step is not necessary for success.
df
df$sum <- ave(df$X, df$y, FUN=sum)
df
y X sum
1 A 3 6
6 A 3 6
3 B 3 8
7 B 1 8
9 B 1 8
10 B 3 8
2 C 2 6
4 C 2 6
5 C 1 6
8 C 1 6
Related
This question already has answers here:
Remove duplicated rows
(10 answers)
Closed 2 years ago.
I have tried various functions including compare and all.equal but I am having difficulty finding a test to see if variables are the same.
For context, I have a data.frame which in some cases has a duplicate result. I have tried copying the data.frame so I can compare it with itself. I would like to remove the duplicates.
One approach I considered was to look at row A from dataframe 1 and subtract it from row B from dataframe 2. If they equal to zero, I planned to remove one of them.
Is there an approach I can use to do this without copying my data?
Any help would be great, I'm new to R coding.
Suppose I had a data.frame named data:
data
Col1 Col2
A 1 3
B 2 7
C 2 7
D 2 8
E 4 9
F 5 12
I can use the duplicated function to identify duplicated rows and not select them:
data[!duplicated(data),]
Col1 Col2
A 1 3
B 2 7
D 2 8
E 4 9
F 5 12
I can also perform the same action on a single column:
data[!duplicated(data$Col1),]
Col1 Col2
A 1 3
B 2 7
E 4 9
F 5 12
Sample Data
data <- data.frame(Col1 = c(1,2,2,2,4,5), Col2 = c(3,7,7,8,9,12))
rownames(data) <- LETTERS[1:6]
This question already has answers here:
Idiomatic R code for partitioning a vector by an index and performing an operation on that partition
(3 answers)
Closed 9 years ago.
Suppose I have a data frame like this:
set.seed(123)
df <- as.data.frame(cbind(y<-sample(c("A","B","C"),10,T), X<-sample(c(1,2,3),10,T)))
df <- df[order(df$V1),]
Is there a simply function to sum (or any FUN) V2 by V1 and add to df as a new column, such that:
df$sum <- c(6,6,8,8,8,8,6,6,6,6)
df
I may write a function for that, but I have to do that frequently and be better to know the simplest way to realize that.
I agree with #mnel at least on his first point. I didn't see ave demonstrated in the answers he cited and I think it's the "simplest" base-R method. Using that data.frame(cbind( ...)) construction should be outlawed and teachers who demonstrate it should be stripped of their credentials.
set.seed(123)
df<-data.frame(y=sample( c("A","B","C"), 10, T),
X=sample(c (1,2,3), 10, T))
df<-df[order(df$y),] # that step is not necessary for success.
df
df$sum <- ave(df$X, df$y, FUN=sum)
df
y X sum
1 A 3 6
6 A 3 6
3 B 3 8
7 B 1 8
9 B 1 8
10 B 3 8
2 C 2 6
4 C 2 6
5 C 1 6
8 C 1 6
This question already has answers here:
Grouping functions (tapply, by, aggregate) and the *apply family
(10 answers)
Closed 7 years ago.
I have a data frame that looks like this:
data<-data.frame(y=c(1,1,2,2,3,4,5,5),x=c(5,5,10,10,5,10,5,5))
y x
1 1 5
2 1 5
3 2 10
4 2 30
5 3 5
6 4 10
7 5 4
8 5 8
How can a merge those rows with same value in y column and modify the x column value to the mean of them.
I would like something like this:
y x
1 1 5
2 2 20
3 3 5
4 4 10
7 5 6
I'm trying:
unique(data)
But it removes the values instead of doing the mean of same rows.
It is easy with dplyr. Like here:
library("dplyr")
data %>%
group_by(y) %>%
summarise(x=mean(x))
We can use aggregate
aggregate(x~y, data, mean)
User plyr.
# Create dummy data.
nel = 30
df <- data.frame(x = round(5*runif(nel)), y= round(10*runif(nel)))
# Summarise means
require(plyr)
df$x <- as.factor(df$x)
res <- ddply(df, .(x), summarise, mu=mean(y))
This question already has an answer here:
Compute the minimum of a pair of vectors
(1 answer)
Closed 7 years ago.
Say I have the following data frame:
dx=data.frame(id=letters[1:4], count=1:4)
# id count
# 1 a 1
# 2 b 2
# 3 c 3
# 4 d 4
And I would like to (grammatically) add a column that will get the count whenever count<3, otherwise 3, so I'll get the following:
# id count group
# 1 a 1 1
# 2 b 2 2
# 3 c 3 3
# 4 d 4 3
I thought to use
dx$group=if(dx$count<3){dx$count}else{3}
but it doesn't work on arrays. How can I do it?
In this particular case you can just use pmin (as I stated in the comments above):
df$group <- pmin(df$count, 3)
In general your if/else construction does not work on vectors, but you can use the function ifelse. It takes three arguments: First the condition, then the result if the condition is met and finally the result if the condition is not met. For your example you would write the following:
df$group <- ifelse(df$count < 3, df$count, 3)
Note that in your example the pmin solution is better. Just mentioning the ifelse solution for completeness.
I am trying to simulate n times the measuring order and see how measuring order effects my study subject. To do this I am trying to generate integer random numbers to a new column in a dataframe. I have a big dataframe and i would like to add a column into the dataframe that consists a random number according to the number of observations in a block.
Example of data(each row is an observation):
df <- data.frame(A=c(1,1,1,2,2,3,3,3,3),
B=c("x","b","c","g","h","g","g","u","l"),
C=c(1,2,4,1,5,7,1,2,5))
A B C
1 1 x 1
2 1 b 2
3 1 c 4
4 2 g 1
5 2 h 5
6 3 g 7
7 3 g 1
8 3 u 2
9 3 l 5
What I'd like to do is add a D column and generate random integer numbers according to the length of each block. Blocks are defined in column A.
Result should look something like this:
df <- data.frame(A=c(1,1,1,2,2,3,3,3,3),
B=c("x","b","c","g","h","g","g","u","l"),
C=c(1,2,4,1,5,7,1,2,5),
D=c(2,1,3,2,1,4,3,1,2))
> df
A B C D
1 1 x 1 2
2 1 b 2 1
3 1 c 4 3
4 2 g 1 2
5 2 h 5 1
6 3 g 7 4
7 3 g 1 3
8 3 u 2 1
9 3 l 5 2
I have tried to use R:s sample() function to generate random numbers but my problem is splitting the data according to block length and adding the new column. Any help is greatly appreciated.
It can be done easily with ave
df$D <- ave( df$A, df$A, FUN = function(x) sample(length(x)) )
(you could replace length() with max(), or whatever, but length will work even if A is not numbers matching the length of their blocks)
This is really easy with ddply from plyr.
ddply(df, .(A), transform, D = sample(length(A)))
The longer manual version is:
Use split to split the data frame by the first column.
split_df <- split(df, df$A)
Then call sample on each member of the list.
split_df <- lapply(split_df, function(df)
{
df$D <- sample(nrow(df))
df
})
Then recombine with
df <- do.call(rbind, split_df)
One simple way:
df$D = 0
counts = table(df$A)
for (i in 1:length(counts)){
df$D[df$A == names(counts)[i]] = sample(counts[i])
}