# Calculate float value in ModelBuilder and use as input for "Minus" tool

I'm using ModelBuilder and have three variables (var1, var2 and var3) which I want to use to calculate a new value. This value should be used as an input for the "Minus" tool. The calculation to be performed looks like this:

``````NewValue = var1 + (var2 / 2 / var3)
``````

When calculating this manually for var1 = 2, var2 = 2 and var3 = 3 I get 2.333... (2 + (2 / 2 / 3))

When using the "Calculate Value" tool I get: 2

I was forced to use the data type "Formulated Raster" as output for "Calculate Value" because it seems to be impossible to use the output as an input for the "Minus" tool otherwise. Maybe this is the reason for the rounding.

So I tried using the "Raster Calculator" instead but this just leads me to an error message I don't understand:

How can I get my Model to calculate and use the correct value of 2.333...?

EDIT:

Changing the Rastercalculator Expression like suggested in the comments doesn't fix the problem:

• Try float(var1 + (var2 / 2 / var3)) Jul 31, 2016 at 14:56
• When you populate the Calculate Value tool what are you setting the parameter Data Type to? In your case you must set it to Double as well as following @Ali's advice. Aug 5, 2016 at 11:13
• @ Hornbydd This was my first guess but as I wrote above "I was forced to use the data type "Formulated Raster" as output for "Calculate Value" because it seems to be impossible to use the output as an input for the "Minus" tool otherwise. " Aug 5, 2016 at 11:19

For python division to result in a float data type you must either have the numerator or demoninator as a float. Consider 1 divided by 2:

``````>>> 1/2
0
``````

The answer is 0.5 but is output as the integer component 0. In your case the answer 2 results from 2 + (0) - even though we know it should be 2 + (0.3333).

Now, back to 1 divided by 2. Making either number a float, explicitly or implicitly, will give the result as a float:

``````>>> 1/float(2) # explicit
0.5

>>>float(1)/2 # explicit
0.5

>>>1.0/2 # implicit
0.5
``````

Therefore, changing your expression to something along the lines of:

``````2 + (2/2/float(3))
``````