Advanced Formula Challenge #12: Results and Discussion 5

Last week I set readers the challenge which can be found here.

Such was the number and variety of responses to this challenge that presenting a detailed breakdown of one such solution – as has been the case for all of the first eleven in this series of challenges – would, I feel, be somewhat inappropriate.

For the majority of these challenges, it could be argued that there has been one solution which is indisputably “better” than the rest. Perhaps such an adjudication can also be made here, though to do so would certainly not be a straightforward exercise. What’s more, to pick just one of the many solutions would be to leave the rest – unfairly in my opinion – left on the sidelines.

As such, I would refer the readers to the many solutions in that post and to enjoy dissecting the varied and wonderful constructions therein. And to simply thank all those – Alex, aMareis, Maxim, John Jairo, sam, Jeff, Lori, Ron, Michael, Christian and XLarium – whose excellent contributions led to such a fruitful and inspiring discussion.

There’s evidently still much to be discovered in the world of worksheet formulas!

Another challenge to follow shortly. Watch this space!

Counting Rows Where Condition Is Met In At Least One Column 63

In this post I would like to present a solution to the situation in which we wish to count the number of rows for which a stipulated condition is met in at least one of several columns.

To illustrate what is meant by this, consider the extract below:

which details levels of scrap nickel exports for various countries and for various years (you can download the workbook here).

Which numbers add up to total? (2): Multiple Solutions 14

Note to readers: this post has been updated due to the inclusion – at the request of Torstein – of a further version of this solution, in which the number of values to be considered is dynamic and so may be set by the user. This version may be found at the very end of this post.

This post, inspired by a question from Patrick MacKay, from Belgium – thanks, Patrick! ðŸ™‚ – is a (rather belated) follow-up to that which I made here, in which, to recap, I presented a formula-based set-up which, given a target figure plus a series of values, determined which, if any, combination of those values had a sum equal to the target.

The only slight drawback to that solution was the caveat that, if more than one combination of values existed which satisfied that condition, then only one of those combinations was given.

Here I would like to improve upon that set-up by presenting a refined version which will return all such combinations. What’s more, at the very end of this deconstruction I will give a further version of the solution in which the number of values to be considered is a variable which may be set by the user.

In fact, that early post was also one of the very few in which I did not give an explanation as to how the solution works, which I would like to do here.

As an example of the output, imagine that our target value – Â£1054.35, for example – is here in A1, and that we have a list of 10 values in A2:A11, as below:

Unique, Ordered List of Most Frequent Numbers in a Two-Dimensional Range 9

I recently received a request from James, who was interested in a formula-based solution to the following problem: given a two-dimensional range containing a mixture of numbers and empty cells (which I am defining as being either “genuinely” empty or as containing the null string “” as a result of formulas in those cells), generate a unique list of those numbers in order of their frequency within that range, with the most frequent first. What’s more, if two or more numbers occur the same number of times within that range, then they should be listed in order of their size from smallest to largest.

For example, for the dataset in A1:F6 below, we would return the list as given beginning in I1.

Non-Array TRANSPOSE 4

We sometimes look for non-array (i.e. non-CSE) versions of constructions which would normally require array-entry. Our reasons for doing so may be varied:

1) We may feel that it improves spreadsheet performance (sometimes true, sometimes not)

2) We perhaps have a dislike for having to use the required keystroke combination necessary for committing array formulas

3) We may simply be interested from a theoretical point of view

List of unique entries from column of space-separated strings 6

Given the list below in A1:A10, we may wish to create a list of unique, single words from that list, as per column B here.

We can do this with the following set-up: More…

COUNTIFS: Multiple “OR” criteria for one or two criteria_Ranges 111

In this post I would like to clear up what appears to me to be a rather widespread misunderstanding of how COUNTIFS/SUMIFS operate, in particular when we pass arrays consisting of more than one element as the Criteria to one or even two of the Criteria_Ranges.

This latter technique is used when the criteria in question are to be considered as “OR” criteria, which is not to be confused with cases where we wish the criteria passed to be calculated rather as “AND” criteria.

For example, given the following data:

Advanced Formula Challenge #5: Results and Discussion 5

Last week I set readers the challenge which can be found here.

This is a reasonably complex problem, and certainly so if we want to present a solution which is relatively concise. However, despite its complexity (and arguably lack of practical use), the solution demonstrates some important techniques for working with strings, and so is not without merit.

The required set-up is as follows:

Last week I set readers the challenge which can be found here.

This one turns out to be a good deal more complex than it at first appears, and so perhaps not surprisingly no correct results were received..

GreasySpot at first thought that Advanced Filter would be a viable solution, but quickly realised that it wasn’t actually appropriate here. Besides, as I mentioned, the idea of this (and of all these challenges in fact) is to try to achieve the results using worksheet formulas alone.

So how can we achieve our desired results?

Which numbers add up to total? 28

Sometimes we are in a situation where we have a target figure plus a series of values and we want to know which, if any, combination of those values has a sum which is equal to the target.

This can be done as follows:

Edit: this post has now been revised here to account for multiple returns, should that be a requirement.

Using the above set-up, with our target value in A2 and our (in this case 9) values in C1:K1, we will place formulas in C2:K2 which will contain an “X” if the value in the row above forms part of our solution.

Advanced Formula Challenge #3: Results and Discussion 8

Last week I posed readers the challenge which can be found here.

One solution was received, again from Bill, and this time it was not only correct, but a very good solution indeed. So congratulations again to Bill!

In fact, rather than dissect my own solution this week (which in any case differs only in minor details from Bill’s), I would like to present a breakdown of the solution given by Bill, as follows:

Advanced Formula Challenge #2: Results and Discussion 3

Last week I set readers the challenge which can be found here.

Three solutions were offered, two of which from the same person, and both of which were correct! So many congratulations to Bill on successfully solving what was quite a complex challenge!

Indeed, as Ben Schwartz pointed out, this challenge appears to have been set previously on the internet, and seems to have been only partially solved on those occasions. In any case, thanks also to Ben for his suggestion, which he confesses was cobbled together from those previous solutions he found, and which worked in all but a few exceptional cases.