Conditional average or replace value if zero in dataframe - r

Question

I have a problem finding the right code for the following problem.
Here is a simplified and short version of my dataframe df :
Line Id Amount
1 1 10
2 2 12
3 2 13
4 2 0
5 3 11
6 4 12
7 4 14
8 5 0
9 6 11
10 6 0
I would like to create another colum Amount_Avrg with the folowing conditions:
-if several lines have the same Id and an Amount that is different from zero the case for lines 2 and 3 and for lines 6 and 7, calculate the average of the different amounts
-if one line has an amount that is equal with 0 then:
A/ erase it if it is alone (if there is no other line with the same Id and a value different from 0) (the case of line 8)
B/ if there is one line with the same Id and a value different from 0 (the case for lines 9 and 10), replace 0 with the value of the other
C/ if there are two lines or more with a value different from zero (the case for lines 2 and 3), replace 0 with the average of the other amounts
The final dataframe I am expecting would then look like this one:
Line Id Amount Amount_Avrg
1 1 10 10
2 2 12 12.5
3 2 13 12.5
4 2 0 12.5
5 3 11 11
6 4 12 13
7 4 14 13
9 6 11 11
10 6 0 11
I have read in many answers that if loops were not efficient on R so if you could help me with another solution, that would be fantastic :-)

Answer 1

Using dplyr, we can group_by ID and take mean of non-zero Amount and remove rows with NA in them.
library(dplyr)
df %>%
group_by(Id) %>%
mutate(mn = mean(Amount[Amount > 0])) %>%
filter(!is.na(mn))
# Line Id Amount mn
# <int> <int> <int> <dbl>
#1 1 1 10 10
#2 2 2 12 12.5
#3 3 2 13 12.5
#4 4 2 0 12.5
#5 5 3 11 11
#6 6 4 12 13
#7 7 4 14 13
#8 9 6 11 11
#9 10 6 0 11
Or with data.table
library(data.table)
setDT(df)[, mn := mean(Amount[Amount > 0]), by = Id][!is.na(mn)]
data
df <- structure(list(Line = 1:10, Id = c(1L, 2L, 2L, 2L, 3L, 4L, 4L,
5L, 6L, 6L), Amount = c(10L, 12L, 13L, 0L, 11L, 12L, 14L, 0L,
11L, 0L)), class = "data.frame", row.names = c(NA, -10L))

Answer 2

You can use ave to calculate the mean per Id and then subset with !is.na to remove the rows where you have only 0 per Id.
x$Amount_Avrg <- ave(x$Amount, x$Id, FUN=function(x) mean(x[x>0]))
x <- x[!is.na(x$Amount_Avrg),]
x
# Line Id Amount Amount_Avrg
#1 1 1 10 10.0
#2 2 2 12 12.5
#3 3 2 13 12.5
#4 4 2 0 12.5
#5 5 3 11 11.0
#6 6 4 12 13.0
#7 7 4 14 13.0
#9 9 6 11 11.0
#10 10 6 0 11.0
Or with within and na.omit:
na.omit(within(x, mount_Avrg <- ave(Amount, Id, FUN=function(x) mean(x[x>0]))))
Or using aggregate and merge:
merge(x, aggregate(cbind(Amount_Avrg = Amount) ~ Id, data=x[x$Amount>0,], mean))
Data:
x <- read.table(header=TRUE, text="Line Id Amount
1 1 10
2 2 12
3 2 13
4 2 0
5 3 11
6 4 12
7 4 14
8 5 0
9 6 11
10 6 0")

Answer 3

If you create a summary table of all the nonzero-means, you can right-join that to the original table to get the result displayed in the question.
library(data.table)
setDT(df)
nonzero_means <- df[Amount > 0, .(Amount_Avg = mean(Amount)), Id]
df[nonzero_means, on = .(Id)]
# Line Id Amount Amount_Avg
# 1: 1 1 10 10.0
# 2: 2 2 12 12.5
# 3: 3 2 13 12.5
# 4: 4 2 0 12.5
# 5: 5 3 11 11.0
# 6: 6 4 12 13.0
# 7: 7 4 14 13.0
# 8: 9 6 11 11.0
# 9: 10 6 0 11.0