Reputation: 889
I have large data-frame
consists of two columns. I want to calculate the average of the second column values for each subset of the first column. The subset of the first column is based on a specified granularity. For example, for the following data-frame
, df
, I want to calculate the average of df$B
values for each subset of df$A
with an increment(granularity) of 1 for each subset. The results should be in two new columns.
A B expected results newA newB
0.22096 1 0 1.142857
0.33489 1 1 2
0.33655 1 2 4
0.43953 1
0.64933 2
0.86668 1
0.96932 1
1.09342 2
1.58314 2
1.88481 2
2.07654 4
2.34652 3
2.79777 5
This is a simple example, I'm not sure how to loop over the whole data-frame
and perform the calculation i.e. the average of the df$B
.
tried below to subset, but couldn't figure how to append the results and create final results:
Tried something like :
increment<-1
mx<-max(df$A)
i<-0
newDF<-data.frame()
while(i < mx){
tmp<-subset(df, (A >i & A< (i+increment)))
i<-i+granualrity
}
Not sure about the logic. But I'm sure there is a short way to do the required calculation. Any thoughts?
Upvotes: 0
Views: 352
Reputation: 25736
I would use findInterval
for the subset selection (In your example a simple ceiling
for each A
value should be sufficient, too. But if your increment
is different from 1 you need findInterval
.) and tapply
to calculate the mean
:
df <- read.table(textConnection("
A B
0.22096 1
0.33489 1
0.33655 1
0.43953 1
0.64933 2
0.86668 1
0.96932 1
1.09342 2
1.58314 2
1.88481 2
2.07654 4
2.34652 3
2.79777 5"), header=TRUE)
## sort data.frame by column A (needed for findInterval)
df <- df[order(df$A), ]
## define granuality
subsets <- seq(1, max(ceiling(df$A)), by=1) # change the "by" argument for different increments
df$subset <- findInterval(df$A, subsets)
tapply(df$B, df$subset, mean)
# 0 1 2
#1.142857 2.000000 4.000000
Upvotes: 1