How to average columns based on ID in R? [duplicate]

Answer 1

Using sqldf:

library(sqldf)
sqldf("SELECT ID, avg(Value) Mean 
      FROM df 
      GROUP BY ID")

Output:

    ID      Mean
1 1000 0.2600000
2 1001 0.6133333
3 1002 0.4166667
4 1003 0.1200000

With dplyr, instead of summarise_each as Cleb pointed out, we can just use summarise:

df %>% group_by(ID) %>% summarise(mean = mean(Value))
#or 
summarise(group_by(df, ID), mean = mean(Value))

Output:

    ID      mean
  (int)     (dbl)
1  1000 0.2600000
2  1001 0.6133333
3  1002 0.4166667
4  1003 0.1200000

Answer 2

You can try by():

> with(df, by(Value, ID, mean))
# ID: 1000
# [1] 0.26
# ------------------------------------------------------------ 
# ID: 1001
# [1] 0.6133333
# ------------------------------------------------------------ 
# ID: 1002
# [1] 0.4166667
# ------------------------------------------------------------ 
# ID: 1003
# [1] 0.12

or aggregate():

> aggregate( Value ~ ID, df, mean)
#     ID     Value
# 1 1000 0.2600000
# 2 1001 0.6133333
# 3 1002 0.4166667
# 4 1003 0.1200000

or using data.table (if you need fast calculation on large data sets):

> library(data.table)
> setDT(df)[, mean(Value), by = ID]
#      ID        V1
# 1: 1000 0.2600000
# 2: 1001 0.6133333
# 3: 1002 0.4166667
# 4: 1003 0.1200000

data

df <- structure(list(ID = c(1000L, 1000L, 1001L, 1001L, 1001L, 1002L, 
1002L, 1002L, 1003L), Value = c(0.51, 0.01, 0.81, 0.41, 0.62, 
0.98, 0.12, 0.15, 0.12)), .Names = c("ID", "Value"), 
class = "data.frame", row.names = c(NA, -9L))

Answer 3

You could use the package dplyr and the function summarise_each:

df=data.frame(ID=c(1000,1000,1001,1001,1001,1002,1002,1002,1003), Value=c(0.51,0.01,0.81,0.41,0.62,0.98,0.12,0.15,0.12))

library(dplyr)
newdf <- df %>% group_by(ID) %>% summarise_each(funs(mean))

which gives you:

     ID     Value
  (dbl)     (dbl)
1  1000 0.2600000
2  1001 0.6133333
3  1002 0.4166667
4  1003 0.1200000

If you deal with large datasets this should be the most efficient way of doing this task.