R 中重复行之间的平均值

Question

I have a data frame df with rows that are duplicates for the names column but not for the values column:

name    value   etc1    etc2
A       9       1       X
A       10      1       X
A       11      1       X
B       2       1       Y
C       40      1       Y
C       50      1       Y

I need to aggregate the duplicate names into one row, while calculating the mean over the values column. The expected output is as follows:

name    value   etc1    etc2
A       10      1       X
B       2       1       Y
C       45      1       Y

I have tried to use df[duplicated(df$name),] but of course this does not give me the mean over the duplicates. I would like to use aggregate() , but the problem is that the FUN part of this function will apply to all the other columns as well, and among other problems, it will not be able to compute char content. Since all the other columns have the same content over the "duplicates", I need them to be aggregated as is just like the name column. Any hints...?

Answer 1

Here a data.table solution. The solution is general in the sense it will work even for a data.frame with 60 columns. Since I group the data by all variables different of value( See how I create keys below)

library(data.table)
dat <- read.table(text='name    value   etc1    etc2
A       9       1       X
A       10      1       X
A       11      1       X
B       2       1       Y
C       40      1       Y
C       50      1       Y',header=TRUE)
keys <- colnames(dat)[!grepl('value',colnames(dat))]
X <- as.data.table(dat)
X[,list(mm= mean(value)),keys]
  name etc1 etc2 mm
1:    A    1    X 10
2:    B    1    Y  2
3:    C    1    Y 45

EDIT extend to more than one value variable

In case you have more than one numeric variables on which you want to compute the mean , For example, if your data look like this

  name value etc1 etc2     value1
1    A     9    1    X  2.1763485
2    A    10    1    X -0.7954326
3    A    11    1    X -0.5839844
4    B     2    1    Y -0.5188709
5    C    40    1    Y -0.8300233
6    C    50    1    Y -0.7787496

The above solution can be extended like this :

X[,lapply(.SD,mean),keys]
   name etc1 etc2 value     value1
1:    A    1    X    10  0.2656438
2:    B    1    Y     2 -0.5188709
3:    C    1    Y    45 -0.8043865

This will compute the mean for all variables that don't exist in keys list.

Answer 2

您可以使用如下aggregate()函数：

aggregate(df$value,by=list(name=df$name,etc1=df$etc1,etc2=df$etc2),data=df,FUN=mean)

Answer 3

The code (written by Metrics) is almost working except in one place (.name). I slightly modified it:

sample<- structure(list(name = structure(c(1L, 1L, 1L, 2L, 3L, 3L), .Label = c("A", 
    "B", "C"), class = "factor"), value = c(9L, 10L, 11L, 2L, 40L, 
    50L), etc1 = c(1L, 1L, 1L, 1L, 1L, 1L), etc2 = structure(c(1L, 
    1L, 1L, 2L, 2L, 2L), .Label = c("X", "Y"), class = "factor")), .Names = c("name", 
    "value", "etc1", "etc2"), class = "data.frame", row.names = c(NA, 
    -6L))

sample.m <- ddply(sample, 'name', summarize, value =mean(value), etc1=head(etc1,1), etc2=head(etc2,1))

sample.m
      name value etc1 etc2
    1    A    10    1    X
    2    B     2    1    Y
    3    C    45    1    Y

Answer 4

Assuming your dataframe is df.

install.packages("plyr")
library(plyr)



df<- structure(list(name = structure(c(1L, 1L, 1L, 2L, 3L, 3L), .Label = c("A", 
    "B", "C"), class = "factor"), value = c(9L, 10L, 11L, 2L, 40L, 
    50L), etc1 = c(1L, 1L, 1L, 1L, 1L, 1L), etc2 = structure(c(1L, 
    1L, 1L, 2L, 2L, 2L), .Label = c("X", "Y"), class = "factor")), .Names = c("name", 
    "value", "etc1", "etc2"), class = "data.frame", row.names = c(NA, 
    -6L))

df.m<-ddply(df,.(name),summarize, value=mean(value),etc1=head(etc1,1),etc2=head(etc2,1))

df.m
 name value etc1 etc2
1    A      10    1    X
2    B       2    1    Y
3    C      45    1    Y

Answer 5

This simple one worked for me:

avg_data <- aggregate(. ~ name, df, mean)

Using the "aggregate" function: apply the formula method ( x ~ y ) for all variables (.) based on the naming variable ("name"), within the data.frame "df", to perform the "mean" function.