根据给定列中的公共值聚合 R 中相同 data.frame 的多行

Question

I have a data.frame that looks like this:

# set example data
df <- read.table(textConnection("item\tsize\tweight\tvalue
A\t2\t3\t4
A\t2\t3\t6
B\t1\t2\t3
C\t3\t2\t1
B\t1\t2\t4
B\t1\t2\t2"), header = TRUE)

# print example data
df

  item size weight value
1    A    2      3     4
2    A    2      3     6
3    B    1      2     3
4    C    3      2     1
5    B    1      2     4
6    B    1      2     2

As you can see the size and weight columns do not add any complexity since they are the same for each item . However, there can be multiple value s for the same item .

I want to collapse the data.frame to have one row per item using the mean value :

  item size weight value
1    A    2      3     5
3    B    1      2     3
4    C    3      2     1

I guess I have to use the aggregate function but I could not figure out how exactly I can get the above result.

Answer 1

aggregate(value ~ item + size + weight, FUN = mean, data=df)

  item size weight value
1    B    1      2     3
2    C    3      2     1
3    A    2      3     5

Answer 2

Here is the solution using the ddply from plyr package:

library(plyr)
ddply(df,.(item),colwise(mean))
  item size weight value
1    A    2      3     5
2    B    1      2     3
3    C    3      2     1

Answer 3

df$value <- ave(df$value,df$item,FUN=mean)
df[!duplicated(df$item),]

  item size weight value
1    A    2      3     5
3    B    1      2     3
4    C    3      2     1

Answer 4

The data.table solution...

require(data.table)
DT <- data.table(df)

DT[ , lapply(.SD , mean ) , by = item ]
   item size weight value
1:    A    2      3     5
2:    B    1      2     3
3:    C    3      2     1

Answer 5

Nowadays, this is what I would do:

library(dplyr)

df %>%
  group_by(item, size, weight) %>%
  summarize(value = mean(value)) %>%
  ungroup

This yields the following result:

# A tibble: 3 x 4
   item  size weight value
  <chr> <int>  <int> <dbl>
1     A     2      3     5
2     B     1      2     3
3     C     3      2     1

I will leave the accepted answer as such as I specifically asked for aggregate , but I find the dplyr solution the most readable.