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Ranking Rows based on values in specific column

AJ_vf
AJ_vf
edited 12/09/19 in Archived 2017 Posts

I would like to formulate a ranking system in a column that will base the rank on an existing column. Are there any functions that can do this?

Comments

  • Not exactly sure what you're looking for when creating your ranking system, but you may want to check out our Using Formulas article in the Help Center to see a list of our available functions.

     

    Check out the IF statement, as that's likely the one you'll want to use. Others that come to mind are the LOOKUP function (you might create a table for your rankings and use LOOKUP to return values) and possibly the COUNT, COUNTIF, SUM, and SUMIF functions if you're looking to compare the count or sum of a range of values in your other column.

  • AJ_vf
    AJ_vf
    edited 01/17/17

    @shaine, I have three columns. Name, Rank, and Score. Based on Score, I want to give each Name a Rank. The Scores are updating constantly so manually sorting and assigning Rank is out of the question.

  • J. Craig Williams
    J. Craig Williams ✭✭✭✭✭✭

    AJ_vf,

     

    Then you are probably looking for nested IF statements

     

    =IF([Score]23 < 10, "Low", IF([Score]23 < 80, "Medium", "High"))

     

    This will look at the Score column, row 23 for a number value.

    If it is less than 10, it will return the text "Low".

    If it is less than 80, it will return the text "Medium" - Note that a 9 will be found by the first check and the system stops looking.

    Otherwise, it will return "High"

     

    Depending on what your ranking system is (numbers, text, symbols), will determine what the final formula will be, but this should give you the gist of it.

     

    Craig

  • Burto
    Burto
    edited 01/20/18

    You may also be able to use the =SMALL() and =LARGE() functions to create a ranked list view.

    Forgot to mention that if you have duplicate values you may want to refine the formula to add a very small value based on a different criteria onto it until it becomes unique

This discussion has been closed.