topic Re: dict(zip()) in for loop with Calc Field codeblock in Python Questions

dict(zip()) in for loop with Calc Field codeblock

coolflippers — Sun, 08 Oct 2023 21:54:39 GMT

Good evening, some trouble with the below example code. This is my first time putting a loop in a function for Calculate Field. I've tried a number of ways (putting brackets after row in the condition of the for loop, putting a 0 inside the brackets, changing the equals sign, trying the brackets with Key, etc).

KeyList = ['a','b','c'] ValueList = ['chair','lamp','rug'] Dict = dict(zip(KeyList,ValueList))

codeblock = """ def output(Key,Value): for row in KeyList: if row == Key: return Dict[KeyList] else: return Value"""

TableExample:

ID	NAME	(NewField)
a		(chair)
z	desk	(desk)
c		(rug)

arcpy.management.CalculateField( in_table = "TableExample", field = "NewField", expression = "output(!ID!,!NAME!)", expression_type = "PYTHON3", code_block = codeblock )

Re: dict(zip()) in for loop with Calc Field codeblock

coolflippers — Sun, 08 Oct 2023 22:12:55 GMT

I am noticing something about this. I can get it to run, but only for the first row. In my scenario, I have a few hundred rows between ID cell 'a' and ID cell 'c'. I run my code, and NewField populates correctly based on the ID in row 1 and the NAME and row 2. So, my NewField, as applied to this example, is correct for its first two cells. However, on its third cell, which, in this example, should be 'rug', in my scenario in Arc it's blank. It's as if the function was satisfied on the first row, and then went to the else statement for all the remaining rows. hm. maybe i should do two for loops. let me try that

Re: dict(zip()) in for loop with Calc Field codeblock

coolflippers — Sun, 08 Oct 2023 22:17:50 GMT

Symptom verified. I took out the else statement, and all the values in ValueList were calculated into the field. Now, I just need to figure out how to populate the remaining fields based off NAME without messing up my progress...

Re: dict(zip()) in for loop with Calc Field codeblock

coolflippers — Sun, 08 Oct 2023 22:18:39 GMT

This is what my table looks like before calc field:

Re: dict(zip()) in for loop with Calc Field codeblock

coolflippers — Sun, 08 Oct 2023 22:20:40 GMT

If I include the else statement, only field in the list populates correctly:

Re: dict(zip()) in for loop with Calc Field codeblock

coolflippers — Sun, 08 Oct 2023 22:23:21 GMT

But, when I delete the 'else' statement, the new field populates correctly off the list (of course omitting all the remaining fields)

Re: dict(zip()) in for loop with Calc Field codeblock

DanPatterson — Sun, 08 Oct 2023 22:28:45 GMT

Is this what you are trying to do?

KeyList = ['a','b','c'] ValueList = ['chair','lamp','rug'] Dict = dict(zip(KeyList,ValueList)) def output(id_val, Dict): if id_val in Dict.keys(): return (f"({Dict[id_val]})") else: return None for i in ['a', 'z', 'c', 'd', 'a']: print(output(i, Dict)) (chair) None (rug) None (chair)

Re: dict(zip()) in for loop with Calc Field codeblock

coolflippers — Sun, 08 Oct 2023 22:33:23 GMT

codeblock = """ def output(Key,Value): for row in KeyList: if row == Key: return Dict[KeyList] for row in KeyList: if row != Key: return Value"""

This is my fix. Not that neat though. I feel like somebody probably has a better idea...

Re: dict(zip()) in for loop with Calc Field codeblock

coolflippers — Sun, 08 Oct 2023 22:35:15 GMT

I'm not sure. I feel like what you posted is probably a better way to do it. I've never used the .keys function.

Re: dict(zip()) in for loop with Calc Field codeblock

DanPatterson — Sun, 08 Oct 2023 22:42:29 GMT

my line 6...

if the id_val is a key in the dictionary, then pull its value and format it for output.
- I wasn't sure why you wanted ( ) brackets around the output, but I put them in, just remove them if not needed
if it isn't, then return None (which will produce a <Null> in the table... this could be replaced with anything you want

Re: dict(zip()) in for loop with Calc Field codeblock

AlfredBaldenweck — Tue, 10 Oct 2023 12:58:06 GMT

So, first thing: the dict(zip()) with two lists is cool, but dictionaries can also just be manually made if that's easier.

keyDict = {a:red, b:blue, ...} #Code

The other thing that you can try is a dictionary Get() method. This will return the value from the dictionary if the key is there OR else return a default value specified in the second parameter of the method. If you leave the second parameter empty, it will default to None, but you can also manually make it that way.

keyDict = {"a":"red", "b":"blue", "c":"yellow"} res = keyDict.get("a", "white") print(res) # red res = keyDict.get("d", "white) print(res) # white # The None does not have any quotes on purpose. # It is truly no value, not the word "None". res = keyDict.get("d", None) print(res) # None res = keyDict.get("e") print(res) # None

So, give this a shot?

# I think I messed up with the actual function part here # but the code inside is good. def output(): KeyList = ['a','b','c'] ValueList = ['chair','lamp','rug'] # (Don't name your variables after functions, including "dict") keyDict = dict(zip(KeyList,ValueList)) for key in KeyList: return keyDict.get(key, None)