I want to calculate the mean in a row if at least three out of six observations in the row are != NA. If four or more NA´s are present, the mean should show NA.
Example which gives me the mean, ignoring the NA´s:
require(dplyr)
a <- 1:10
b <- a+10
c <- a+20
d <- a+30
e <- a+40
f <- a+50
df <- data.frame(a,b,c,d,e,f)
df[2,c(1,3,4,6)] <- NA
df[5,c(1,4,6)] <- NA
df[8,c(1,2,5,6)] <- NA
df <- df %>% mutate(mean = rowMeans(df[,1:6], na.rm=TRUE))
I thought about the use of
case_when
but i´m not sure how to use it correctly:
df <- df %>% mutate(mean = case_when( ~ rowMeans(df[,1:6], na.rm=TRUE), TRUE ~ NA))
You can try a base R solution saving the number of non NA values in a new variable and then use ifelse() for the mean:
#Data
a <- 1:10
b <- a+10
c <- a+20
d <- a+30
e <- a+40
f <- a+50
df <- data.frame(a,b,c,d,e,f)
df[2,c(1,3,4,6)] <- NA
df[5,c(1,4,6)] <- NA
df[8,c(1,2,5,6)] <- NA
#Code
#Count number of non NA
df$count <- rowSums( !is.na( df [,1:6]))
#Compute mean
df$Mean <- ifelse(df$count>=3,rowMeans(df [,1:6],na.rm=T),NA)
Output:
a b c d e f count Mean
1 1 11 21 31 41 51 6 26.00000
2 NA 12 NA NA 42 NA 2 NA
3 3 13 23 33 43 53 6 28.00000
4 4 14 24 34 44 54 6 29.00000
5 NA 15 25 NA 45 NA 3 28.33333
6 6 16 26 36 46 56 6 31.00000
7 7 17 27 37 47 57 6 32.00000
8 NA NA 28 38 NA NA 2 NA
9 9 19 29 39 49 59 6 34.00000
10 10 20 30 40 50 60 6 35.00000