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I want to calculate an equation in R. I don't want to use the function sum because it's returning 1 value. I want the full vector of values.
x = 1:10
y = c(21:29,NA)
x+y
[1] 22 24 26 28 30 32 34 36 38 NA
x = 1:10
y = c(21:30)
x+y
[1] 22 24 26 28 30 32 34 36 38 40
I don't want:
sum(x,y, na.rm = TRUE)
[1] 280
Which does not return a vector.
This is a toy example but I have a more complex equation using multiple vector of length 84647 elements.
Here is another example of what I mean:
x = 1:10
y = c(21:29,NA)
z = 11:20
a = c(NA,NA,NA,30:36)
5 +2*(x+y-50)/(x+y+z+a)
[1] NA NA NA 4.388889 4.473684 4.550000 4.619048 4.681818 4.739130 NA
797
First, if we want to exclude missing values from mathematical operations use the na. rm = TRUE argument. If you do not exclude these values most functions will return an NA . We may also desire to subset our data to obtain complete observations, those observations (rows) in our data that contain no missing data.
We can remove those NA values from the vector by using is.na(). is.na() is used to get the na values based on the vector index. !
To remove all rows having NA, we can use na. omit function. For Example, if we have a data frame called df that contains some NA values then we can remove all rows that contains at least one NA by using the command na. omit(df).
Replace NA with 0 in R Data Frame To replace NA with 0 in an R data frame, use is.na() function and then select all those values with NA and assign them to 0. myDataframe is the data frame in which you would like replace all NAs with 0.
1) %+% Define a custom + operator:
`%+%` <- function(x, y) mapply(sum, x, y, MoreArgs = list(na.rm = TRUE))
5 + 2 * (x %+% y - 50) / (x %+% y %+% z %+% a)
giving:
[1] 3.303030 3.555556 3.769231 4.388889 4.473684 4.550000 4.619048 4.681818
[9] 4.739130 3.787879
Here are some simple examples:
1 %+% 2
## [1] 3
NA %+% 2
## [1] 2
2 %+% NA
## [1] 2
NA %+% NA
## [1] 0
2) na2zero Another possibility is to define a function which maps NA to 0 like this:
na2zero <- function(x) ifelse(is.na(x), 0, x)
X <- na2zero(x)
Y <- na2zero(y)
Z <- na2zero(z)
A <- na2zero(a)
5 + 2 * (X + Y - 50) / (X + Y + Z + A)
giving:
[1] 3.303030 3.555556 3.769231 4.388889 4.473684 4.550000 4.619048 4.681818
[9] 4.739130 3.787879
3) combine above A variation combining (1) with the idea in (2) is:
X <- x %+% 0
Y <- y %+% 0
Z <- z %+% 0
A <- a %+% 0
5 + 2 * (X + Y - 50) / (X + Y + Z + A)
4) numeric0 class We can define a custom class "numeric0" with its own + operator:
as.numeric0 <- function(x) structure(x, class = "numeric0")
`+.numeric0` <- `%+%`
X <- as.numeric0(x)
Y <- as.numeric0(y)
Z <- as.numeric0(z)
A <- as.numeric0(a)
5 + 2 * (X + Y - 50) / (X + Y + Z + A)
Note: The inputs used were those in the question, namely:
x = 1:10
y = c(21:29,NA)
z = 11:20
a = c(NA,NA,NA,30:36)
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