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Python - Negative value with field Calculator

Pierre-LucBoivin

Hi,

I'm trying to subtract two integer columns and put the result into a text field. I got good result with positive valuesâ??â??, but I have a problem with negative valuesâ??â??.

[ATTACH=CONFIG]16352[/ATTACH]

Variation column is the result I get with this formula :

def Resultat(x,y):
 diff =  x-y
 return '%d:%02d' % divmod(diff, 60)

Variation = Resultat( !ScePUPbr! , !SitPUPbr!)

Testvari column is the result if I reverse the values â??â??xy

Testvari2 column is the result I should get

Thanks you for help

Pierre-Luc

Field_Calculator.jpg

EricIronside_15-1622050916722.png

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Accepted answers

Pierre-LucBoivin

Here the final solution !

def Resultat(x,y):
 diff =  x-y
 if diff < 0:
  sign = '-'
 else:
  sign = ''
 return sign+'%d:%02d' % divmod(abs(diff), 60)

Thanks everyone for your help

All comments

SolomonPulapkura

What expression are you using to calculate Testvari2?

The expression you provided is for Variation.

Pierre-LucBoivin

I populate the field manually using the field calculator !!

BruceNielsen

Take the absolute value of the difference:

diff = abs(x-y)

Pierre-LucBoivin

Take the absolute value of the difference:
diff = abs(x-y)

I've tried this solution but it's isn't returning the minus before my number !

BruceNielsen

This should add the negative sign, if needed:

def Resultat(x,y):
    diff =  x-y
    if diff < 0:
        sign = '-'
    else:
        sign = '' #These are two single quotes (an empty string)
    return '%s%d:%02d' % (sign, divmod(diff, 60))

Pierre-LucBoivin

Thanks a lot Bruce for helping !

I think we are almost there but i'm still having an error !!

ERROR 000539: Error running expression: Resultat( -2000 , 0) <type 'exceptions.TypeError'>: %d format: a number is required, not tuple

I tried to fix the error but it wasn't successful .

BruceNielsen

We'll try it without using string formatting for the sign then:

def Resultat(x,y):
    diff =  x-y
    if diff < 0:
        sign = '-'
    else:
        sign = '' #These are two single quotes (an empty string)
    return sign + '%d:%02d' % divmod(diff, 60)

Pierre-LucBoivin

I thought this code will work too but it returning the same value as the Variation column but I have two minus before the number ex:(--1:56)

I'm try to put the abs() but it was evident it wouldn't work.

def Resultat(x,y):
    diff =  abs(x-y) #abs doesn't make any difference without a positive and negative value
    if diff < 0:
        sign = '-'
    else:
        sign = '' #These are two single quotes (an empty string)
    return sign + '%d:%02d' % divmod(diff, 60)

is it because of the %02d that it always calculate the difference from 2:00 ?

BruceNielsen

If you go back to your original formula, but set diff = y - x, does that give you the results you expect?

Pierre-LucBoivin

Well still doesn't work ! I'm gonna populate my field testvari2 with an selection and concatenation of the column Variation and testVari !!!

For sure, this is not an optimal solution but it'll work.

!Variation![0:1] + !testvari!

Thanks for your help

ChristopherThompson

I guess I am not understanding the value you are trying to obtain - subtracting a smaller number from a larger number will always return a positive value, and it stands to reason that nearly any action taken using that value will also return another positive value. Absolute value doesn't work because by definition absolute value returns a positive value, it doesn't simply reverse the sign. Now, if the minus sign is simply symbolic of the the difference between the two numbers, then thats somewhat different though misleading. What was the expression you put into the field calculator to obtain the negative value? If you're set on getting a negative value there then this will do it i think:

'-%d:%02d' % divmod(4, 60)

Pierre-LucBoivin

I absolutely need the negative value because i'm calculating difference between 2 streets network for Emergency travel time. So I need to know the variation(negative or positive) between the different network.

Pierre-LucBoivin

Here the final solution !

def Resultat(x,y):
 diff =  x-y
 if diff < 0:
  sign = '-'
 else:
  sign = ''
 return sign+'%d:%02d' % divmod(abs(diff), 60)

Thanks everyone for your help

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