topic Re: Python Help needed in Python Questions

Python Help needed

StevenFama — Thu, 02 Nov 2017 19:09:24 GMT

I have a script that populates/updates a field based upon its content. If none of the conditions are true, I need it not to update / return <Null>.

def Reclass(StoryHeight_Copy):
if (StoryHeight_Copy == "1A"):
return "1.25"
elif (StoryHeight_Copy == "1H"):
return "1.50"
elif (StoryHeight_Copy == "1T"):
return "1.75"
elif (StoryHeight_Copy == "2A"):
return "2.25"
elif (StoryHeight_Copy == "2H"):
return "2.50"
elif (StoryHeight_Copy == "2T"):
return "2.75"
elif (StoryHeight_Copy == "3A"):
return "3.25"
elif (StoryHeight_Copy == "3H"):
return "3.50"
elif (StoryHeight_Copy == "3T"):
return "3.75"

Re: Python Help needed

DanPatterson_Retired — Sun, 12 Dec 2021 05:02:01 GMT

First, reformat your code then if add

else:
    return None‍‍‍‍‍‍

The above will return <null>

otherwise, if you don't want <null>

else:
    return "something else"‍‍‍‍

You have to return something, it is either None (aka ) or something

Re: Python Help needed

StevenFama — Thu, 02 Nov 2017 19:24:18 GMT

I want it to return what is in another field or just leave what is in the existing cell

Re: Python Help needed

curtvprice — Thu, 02 Nov 2017 19:29:14 GMT

Is this Pro or ArcMap? Are you using the Calculate Field tool?

Re: Python Help needed

StevenFama — Thu, 02 Nov 2017 19:31:09 GMT

AS you can see, I created a field *_Copy. when I run the script above, it works till the 'IF' is false. Then it updates to null. I want it to leave what is already there 1, or 2, or what ever else is their.

Re: Python Help needed

StevenFama — Thu, 02 Nov 2017 19:35:39 GMT

ArcMap using the field calc.

Re: Python Help needed

curtvprice — Sun, 12 Dec 2021 05:02:04 GMT

Sometimes https://community.esri.com/community/developers/gis-developers/python?sr=search&searchId=05a9d169-b0c4-4027-8973-8a0ea5cebc21&searchIndex=0 is simpler than one thinks.

Like Dan said, try:

else:
    return StoryHeight_Copy‍‍

Re: Python Help needed

StevenFama — Thu, 02 Nov 2017 19:43:10 GMT

Thank you. That worked!!!!

Re: Python Help needed

carlanderson2 — Fri, 03 Nov 2017 13:43:36 GMT

your code could be shorter and cleaner as

def Reclass(StoryHeight_Copy):
d = {"1A":"1.25","1H":"1.5","1T":"1.75"....}
if StoryHeight_Copy in d:
return d[StoryHeight_Copy]
return None

Re: Python Help needed

XanderBakker — Sun, 12 Dec 2021 05:02:06 GMT

Just an additional comment to put in my two cents.

When the list of possibilities gets a bit long you can use dictionaries: See the example below:

def Reclass(story_height):
    dct = {"1A": "1.25", "1H": "1.50", "1T": "1.75",
           "2A": "2.25", "2H": "2.50", "2T": "2.75",
           "3A": "3.25", "3H": "3.50", "3T": "3.75"}
    if story_height in dct:
        return dct[story_height]
    else:
        return story_height
‍‍‍‍‍‍‍‍

This allows you to search for a key in the dictionary and if it exists it will return the corresponding value.

Since I notice that there is a logic in the values returned, in this case you could also do this:

def Reclass2(story_height):
    dct = {"A": ".25", "H": ".50", "T": ".75"}
    result = story_height
    if len(story_height) == 2:
        if story_height[1] in dct:
            result = story_height[0] + dct[story_height[1]]
    return result‍‍‍‍‍‍‍

The A, H and T value represent the value (ok, text) behind the decimal and just put the first character in front of it.

When you run it (as shown below in a snippet of Python, not the Field Calculator) you can see the result:

def main():
    lst_story_height = ["2H", "3T", "1A", "4H", "5A", "6T"]

    print "\nOriginal methode: Reclass:"
    for story_height in lst_story_height:
        print story_height, Reclass(story_height)

    print "\nAlternative methode: Reclass2:"
    for story_height in lst_story_height:
        print story_height, Reclass2(story_height)


def Reclass(story_height):
    dct = {"1A": "1.25", "1H": "1.50", "1T": "1.75",
           "2A": "2.25", "2H": "2.50", "2T": "2.75",
           "3A": "3.25", "3H": "3.50", "3T": "3.75"}
    if story_height in dct:
        return dct[story_height]
    else:
        return story_height


def Reclass2(story_height):
    dct = {"A": ".25", "H": ".50", "T": ".75"}
    result = story_height
    if len(story_height) == 2:
        if story_height[1] in dct:
            result = story_height[0] + dct[story_height[1]]
    return result

if __name__ == '__main__':
    main()‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍

This will yield:

Original methode: Reclass:
2H 2.50
3T 3.75
1A 1.25
4H 4H
5A 5A
6T 6T

Alternative methode: Reclass2:
2H 2.50
3T 3.75
1A 1.25
4H 4.50
5A 5.25
6T 6.75‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍

Re: Python Help needed

DanPatterson_Retired — Sun, 12 Dec 2021 05:02:09 GMT

PEP 8 ... would help as well and going forward in Python 3.5+, PEP484... only a few covered here to improve readability.

Code Formatting... the basics ++ as well as others is useful to provide clean code with consistent indentation for the novice pythonista

def Reclass(StoryHeight_Copy):
    d = {"1A": "1.25", "1H": "1.5", "1T": "1.75"}
    if StoryHeight_Copy in d:
        return d[StoryHeight_Copy]
    return None‍‍‍‍‍

Re: Python Help needed

XanderBakker — Sat, 04 Nov 2017 18:16:43 GMT

I the examples shared by carl anderson and Dan Patterson there should be an else statement that returns the original value in case it is not in the dictionary as curtvprice showed earlier.