Eliminating data from table

Hello,
I want to delete all rows in my table which have no values over 8 or under -8. How can I do this?

Best regards
wbart

Hi,
you can use the Row Filter node for that. It has the option to define upper and lower bounds for a column. There is also the new Row Filter (Labs) which allows for more complex filter expressions.
Kind regards,
Alexander

Hi

This may work for one column. I have several columns to check. I tried it like this:

image

with the rule based row splitter like this:

To make it more clear hear a table to show:
image

In this example only Row 1, 2 and 3 should persist.

Hi,
you can do it with the Row Filter (Labs), I think. It allows you to combine multiple conditions on different columns using AND. Otherwise, it might be easiest to first calculate minimum and maximum for each row using the Column Aggregator node and then filter based on those two columns.
Kind regards,
Alexander

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The columns are of type string. Not sure if the row filters treats that correctly when doing a range filter.

The solution would be to convert these columns to integer or double (our example looks like integers but then it might just be an example). After conversion you can use the column loop and a simple row filter with ranged based row filtering.

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Sorry the table was an example in the original columns are double.
My problem is to filter the hole row. In Knime i can select single columns but what I want to do is "Check if in the complete row is one or more values greater than 8 or lower then -8. If it is so I will keep the row. If not I will delete the row.

Got it. Then go with Alexanders suggestion and use the column aggregator and filter on that output (min, max)

Sorry sorry sorry The range is over -8 and under 8 ! It is a midrange

Hi,
you can still do the same. Calculate min and max over all columns, then check if the min is over -8 and the max is less than 8. That can be done by two sequential row filters.
Kind regards,
Alexander

It took a time until the knot burst. Now it worked.

Thank you

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