问题
I wonder whether there is a more straighforward way to calculate a certain type of variables than the approach i normally take....
The example below probably explains it best. I have a dataframe with 2 columns (fruit and whether the fruit is rotten or not). I would like to, for each row, add e.g. the percentage of fruit of the same category that is rotten. For example, there are 4 entries for apples, 2 of them are rotten, so each row for apple should read 0.5. The target values (purely as illustration) are included in the "desired outcome" column.
I have previously approached this problem by * using the "ddply" command on the fruit variable (with sum/lenght as function), creating a new 3*2 dataframe * use the "merge" command to link these values back into the old dataframe.
This feels like a roundabout way, and I was wondering whether there are better/faster way of doing this! ideallly a generic approach, that is easily adjusted if one instead of the percentage needs to determine whether e.g. all fruits are rotten, any fruits are rotten, etc. etc. etc....
Many thanks in advance,
W
Fruit Rotten Desired_Outcome_PercRotten
1 Apple 1 0.5
2 Apple 1 0.5
3 Apple 0 0.5
4 Apple 0 0.5
5 Pear 1 0.75
6 Pear 1 0.75
7 Pear 1 0.75
8 Pear 0 0.75
9 Cherry 0 0
10 Cherry 0 0
11 Cherry 0 0
#create example datagram; desired outcome columns are purely inserted as illustrative of target outcomes
Fruit=c(rep("Apple",4),rep("Pear",4),rep("Cherry",3))
Rotten=c(1,1,0,0,1,1,1,0,0,0,0)
Desired_Outcome_PercRotten=c(0.5,0.5,0.5,0.5,0.75,0.75,0.75,0.75,0,0,0)
df=as.data.frame(cbind(Fruit,Rotten,Desired_Outcome_PercRotten))
df
回答1:
You can do this with just ddply and mutate:
# changed summarise to transform on joran's suggestion
# changed transform to mutate on mnel's suggestion :)
ddply(df, .(Fruit), mutate, Perc = sum(Rotten)/length(Rotten))
# Fruit Rotten Perc
# 1 Apple 1 0.50
# 2 Apple 1 0.50
# 3 Apple 0 0.50
# 4 Apple 0 0.50
# 5 Cherry 0 0.00
# 6 Cherry 0 0.00
# 7 Cherry 0 0.00
# 8 Pear 1 0.75
# 9 Pear 1 0.75
# 10 Pear 1 0.75
# 11 Pear 0 0.75
回答2:
data.table is super fast as it updates by reference. What about using it?
library(data.table)
dt=data.table(Fruit,Rotten,Desired_Outcome_PercRotten)
dt[,test:=sum(Rotten)/.N,by="Fruit"]
#dt
# Fruit Rotten Desired_Outcome_PercRotten test
# 1: Apple 1 0.50 0.50
# 2: Apple 1 0.50 0.50
# 3: Apple 0 0.50 0.50
# 4: Apple 0 0.50 0.50
# 5: Pear 1 0.75 0.75
# 6: Pear 1 0.75 0.75
# 7: Pear 1 0.75 0.75
# 8: Pear 0 0.75 0.75
# 9: Cherry 0 0.00 0.00
#10: Cherry 0 0.00 0.00
#11: Cherry 0 0.00 0.00
回答3:
One solution in base R is to use ave.
within(df, {
## Because of how you've created your data.frame
## Rotten is actually a factor. So, we need to
## convert it to numeric before we can use mean
Rotten <- as.numeric(as.character(Rotten))
NewCol <- ave(Rotten, Fruit)
})
Fruit Rotten Desired_Outcome_PercRotten NewCol
1 Apple 1 0.5 0.50
2 Apple 1 0.5 0.50
3 Apple 0 0.5 0.50
4 Apple 0 0.5 0.50
5 Pear 1 0.75 0.75
6 Pear 1 0.75 0.75
7 Pear 1 0.75 0.75
8 Pear 0 0.75 0.75
9 Cherry 0 0 0.00
10 Cherry 0 0 0.00
or shorter:
transform(df, desired = ave(Rotten == 1, Fruit))
The default function applied with ave is mean, hence I have not included it here. However, you could specify a different function by appending FUN = some-function-here if you wanted to do something different.
回答4:
As ave is already out, let me add one solution using my base R function of choice: aggregate.
You can get the desired data simply with:
aggregate(as.numeric(as.character(Rotten)) ~ Fruit, df, mean)
However, you will need to still merge it afterwards (or in one piece):
merge(df, aggregate(as.numeric(as.character(Rotten)) ~ Fruit, df, mean))
来源：https://stackoverflow.com/questions/15467219/calculate-group-characteristics-without-ddply-and-merge