Averaging a Non-Contiguous Range

Figuring out how to average data that is in a contiguous range of cells is easy. When the data is spread over a group of non-contiguous cells then getting the average can be a bit more challenging. Throw into the mix the need to exclude zero values from the average and the solution can be downright elusive.

Sometimes you need to average a series of non-contiguous cells, excluding any cells that may contain zero values. For example, average cells A1, C1, E1, G1, and J1, but only counting those cells that don't contain zero.

Before getting into what works, let's take a look at what doesn't work. First of all, it doesn't work to simply add the cells and divide by 5; that doesn't take zero values into account. Second, it doesn't work to use COUNTIF in the denominator of your formula, as shown here:

=(A1+C1+E1+G1+J1) / COUNTIF(A1:J1,"<>0")

This doesn't work because it examines and counts cells within the entire range of A1:J1, not just the five cells you want considered in the average. You might also think that you could select your five non-contiguous cells, give them a name, and then use the name in your formula. While Excel allows you to create the name, the following gives an error:

=SUM(MyCells) / COUNTIF(MyCells,"<>0")

It appears that COUNTIF will only work with a single contiguous range, so the non-contiguous nature of the MyCells range throws the function into a tailspin. A similar problem occurs if you try to use a non-contiguous range with the AVERAGEIF function:

=AVERAGEIF(MyCells, "<>0")

Since you can't use any of the functions you might want to use, you are left to rely on a bit longer formula to calculate the average. You can calculate the average of these five cells by applying a bit of "trickery" to your denominator, in this manner:

=(A1+C1+E1+G1+J1) / ((A1<>0)+(C1<>0)+(E1<>0)+(G1<>0)+(J1<>0))

The evaluation done on each cell in the denominator returns either a 1 (for True) or a 0 (for False) depending on whether the cell contains a non-zero value or not. This series of values is added together, providing the necessary count of non-zero cells for the denominator.

Notice that the discussion here has been all about the denominator in the formula, not the numerator. The reason is simple—you can add all five values into the numerator; zero values there don't really matter. The only place they matter is in the denominator, which is what makes calculating this average so tricky.

Posted courtesy of Tips.Net