topic Re: Field Calc, in Model Builder, using strings in Python Snippets Questions

Field Calc, in Model Builder, using strings

ChristopherSmith10 — Wed, 30 Oct 2019 19:38:41 GMT

Hello All, and thanks for any help.

I am trying to take this block of Python and insert it into the Field Calculator. What it is supposed to do is take a concatenated field (LF_Concat) and from within that field parse out the number of Greens (G), Yellows (Y), and Reds (R) and then apply a If/Than/elif to number of colors.

Right now I am at a loss on how to get it in. I have this too show for it. The parser while it shows VB is actually in Python (I maybe slow, but I ain't that slow)

I suspect that the delineatorList should be changed to the concatenated field (LF_concat) but I am lost beyond that.

Thanks,

Chris

delineatorList = [" ", ".", ","]

ColorClass = ['GGGY','GGYY','GYYR']

for letters in ColorClass:
    #print letters, " G = ", letters.count('G')
    #print letters, " Y = ", letters.count('Y')
    #print letters, " R = ", letters.count('R')

    Count_G = letters.count('G')
    Count_Y = letters.count('Y')
    Count_R = letters.count('R')

    if Count_R > 0:
        print "Red"
    elif Count_R == 0 and Count_Y >= Count_G:
        print "Yellow"
    elif Count_R == 0 and Count_Y < Count_G:
        print "Green"

Re: Field Calc, in Model Builder, using strings

FC_Basson — Sun, 12 Dec 2021 10:49:50 GMT

First change your parser to Python. The definition function declaration should be small caps - "def". And remember Python is strict on indentation. You can also simplify your "checking" code to something like this:

for c in ColorClass:
    if c.count('R') > 0:
       print 'Red'
    elif c.count('R') == 0 and (c.count('Y') >= c.count('G')):
       print 'Yellow'
    elif c.count('R') == 0 and (c.count('Y') < c.count('G')):
       print 'Green'‍‍‍‍‍‍‍

Re: Field Calc, in Model Builder, using strings

DanPatterson_Retired — Thu, 31 Oct 2019 13:49:51 GMT

The first line should be to make it a generic script

def Classify(ColorClass):‍

and you 'call' it as you have done. The field name will be passed to the function as ColorClass so there is no need for you ColorClass = … assignment statement at all

And replace FC Basson print statement with return.

(BTW, the print statement is python 2.7 if you are using ArcGIS Pro print('Red') would be needed)

Re: Field Calc, in Model Builder, using strings

ChristopherSmith10 — Fri, 01 Nov 2019 18:56:23 GMT

Expression:

Classify2(ColorClass)

Code Block:

def Classify2(ColorClass):
for c in ColorClass:
    if c.count('R') > 0:
       return 'Red'
    elif c.count('R') == 0 and (c.count('Y') >= c.count('G')):
       return 'Yellow'
    elif c.count('R') == 0 and (c.count('Y') < c.count('G')):
       return 'Green'

Executing (Calculate Field): CalculateField WorkLF_P2 LanduseFeas Classify2(ColorClass) PYTHON_9.3 "def Classify2(ColorClass):\n for c in ColorClass:\n if c.count('R') > 0:\n return 'Red'\n elif c.count('R') == 0 and (c.count('Y') >= c.count('G')):\n return 'Yellow'\n elif c.count('R') == 0 and (c.count('Y') < c.count('G')):\n return 'Green'\n"

Start Time: Fri Nov 1 13:52:28 2019

ERROR 000539: Error running expression: Classify2(ColorClass)

Traceback (most recent call last):

File "<expression>", line 1, in <module>

NameError: name 'ColorClass' is not defined

Failed to execute (Calculate Field).

Failed at Fri Nov 1 13:52:28 2019 (Elapsed Time: 0.01 seconds)

So I got it into the code block and now I am attempting debugging.

"ColorClass" is not defined, ...but it is the first line of the code

Thanks again.

Re: Field Calc, in Model Builder, using strings

DanPatterson_Retired — Fri, 01 Nov 2019 19:06:12 GMT

expression

Classify2(!LF_concat!)

Re: Field Calc, in Model Builder, using strings

ChristopherSmith10 — Fri, 01 Nov 2019 19:08:36 GMT

D'OH!

Thank you. Works great now.