Conditions in vectors

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Jackson White

• November 24, 2025 23:44

Hi there,

I'm trying to create a vector with an if statement in it. However, it has a bunch of conditions as shown below:

if the pH is less than -4.5, then set the alkalinity to 0 mg/L, else use the alkalinity from the look up table. 

The expression that I'm trying to define is:

vector(if(Vector1[H] < (1.008*10**-4.5) mg/L then Vector1[Alkalinity] = 0 mg/L else Vector1[Alkalinity] = Alkalinity_lookup))

This leads to an error; the vector type is a value not a condition. If I set the type to a condition, this returns a True and False statement. Really all I'm trying to do is replace the alkalinity value in the vector. The intention is to use the output of this expression to define the chemistry of a cell pathway (contaminant transport module).

If there are any suggestions out there, please reach out :)

Kind regards,

Jackson 

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  Jason

  • November 25, 2025 23:13

  Jackson, 

  If I understand correctly, you want to create a new vector that is essentially a copy of Vector1 with a row changed based on some criteria. 

  To accomplish this, you should set up a vector that assigns the row index with each name. Like this: Vrows = [1,2,3,4,...] so that you can then get the row value by doing row = Vrows[Alkalinity]. Once you are able to do that, please try the following:

  V2 = vector(if(row == Vrows[Alkalinity], if(V1[H] < (1.008*10**-4.5) then 0 else Alkalinity_Lookup), V1[row]))

  We make sure to stop at the index of the item we want to modify with the row == vindex() condition. This will let us single that part out so we can make assignments. The next if statement is just the condition used to know if we need to update the Alkalinity row.

  Let me know if that works for you.

  -Jason

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  Jackson White

  • November 26, 2025 03:41

  Hi Jason, 

  Thanks for your message! It's working now. It took me a moment to understand what the trick was with defining a data element (Vrows), to then use in the expression. I forgot to use the same row labels! 

  Thanks for your help again, 

  Jackson

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  Jason

  • December 06, 2025 00:32

  You are welcome, Jackson. Good luck with your model!

  -Jason

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