Which of the following would EXACTLY output the result below?

[1] 1 3 NA NA

Question

Which of the following would EXACTLY output the result below?

[1]  1  3 NA NA

BlackTom AI · Accepted Answer

This question asks which code would EXACTLY produce the given output: [1]  1  3 NA NA.
First, let's understand what the shown output represents. The vector [1] 1 3 NA NA indicates that there are four rows, and the sums for the rows are 1, 3, NA, and NA respectively. The presence of NA values in the third and fourth rows means that the function computing the sums did not ignore missing values by default.
Now, looking at the provided answer content: df <- data.frame(x = c(0,1,2,NA), y = c(1,2,NA,7), z = c(0,0,9,14))

rowSums(df)

This code creates a data frame with three columns x, y, and z, where the rows are:
- Row 1: (0, 1, 0) → sum = 1
- Row 2: (1, 2, 0) → sum = 3
- Row 3: (2, NA, 9) → sum involves NA, so the result is NA when na.rm is FALSE
- Row 4: (NA, 7, 14) → sum involves NA, so the result is NA when na.rm is FALSE

By default, rowSums uses na.rm = FALSE, which means any row containing NA yields NA in the result. Therefore, rowSums(df) would produce exactly c(1, 3, NA, NA), matching the target output.

Why the same result would not occur with na.rm = TRUE? If the code had rowSums(df, na.rm = TRUE), the NA-containing rows would be summed after removing NAs: Row 3 would be 2 + 9 = 11, and Row 4 would be 7 + 14 = 21, giving c(1, 3, 11, 21). That clearly does not match the provided output, so na.rm = TRUE would be incorrect for this target.
Since there are no alternative answer options presented here to compare, we focus on the given option and explain its alignment with the target output. The exact match hinges on using rowSums with the default na.rm = FALSE on this specific data frame, which produces the NA values in rows 3 and 4 as shown.

PSTAT 10 W25 Quiz 2

Which of the following would EXACTLY output the result below? [1] 1 3 NA NA

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