Is there a way to filter columns in Google Spreadsheets:

  • Column (A) contains new data
  • Column (B) contains filter data, i.e. cells with text that should not be included in column C
  • Column (C) is the outcome, i.e. column A minus all the cell data in column B

If I cannot filter by column, is there another way to reach a similar result?

  • Could you give a short example of your table? I'm not sure what column A - cell data in column B would be if they are text. May 6, 2011 at 23:17
  • Ok, it could look something like this: goo.gl/idyXy /// Column A contains new data, Column B contains data that should be filtered and Column C contains the result.
    – Pongy
    May 7, 2011 at 7:09

1 Answer 1



Place this formula in cell C2:


What you want is the set difference. This formula will give you all the unique values of column A that do not appear in column B.


MATCH(search_criterion, lookup_array, [match_type]) checks if an element is found in the given array. In this case it checks the element from column A in the entire array of B. If it is not found it returns NA.

IF(ISNA(...), A2:A, "") returns the contents of cell from column A if it was not matched, or an empty string if it was.

SORT(UNIQUE(...)) is optional, but will filter the results in alphabetical order and remove any duplicates which is probably what you want.

Finally ARRAYFORMULA returns the result as an array so that it populates the entire C column.

More information:

  • Thank you, that seems to work very well! Much appreciated!
    – Pongy
    May 12, 2011 at 14:36
  • Is there a way to also filter strings found in both A and B? Like, I have a domain name in column B (example.com) so all e-mail addresses ending with example.com are excluded from column C.
    – Pongy
    May 13, 2011 at 15:43
  • You'll need to have the raw data in a separate sheet or column from the filtered data. Afther that for the filtered dat you can use something like: =FILTER(rawData!A:A, isError(search("example.com", rawData!A:A))) If you need more info open a new question. May 15, 2011 at 7:06
  • Thanks for your advice so far! I created a new question as per your suggestion: goo.gl/H33PW
    – Pongy
    May 16, 2011 at 15:43

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