Calculate Field with Python 3 / <Null> values

JustinWolff

I'm attempting to concatenate several fields, where one of the source fields is a double, the rest are string, and all of them have some <Null> values.  If I use the following syntax in Calculate Field in Pro 24.1/Python 3:

str(!AddrPrimaryHighNo!) + " " + str(!StreetPreDrctnAbbrev!) + " " + str(!StreetName!) + " " + str(!StreetSuffixAbbrev!) + " " + str(!StreetPostDrctnAbbrev!)

I get values like:

1498.0 None AVENUE K NONE NONE

where I would expect:

1498  AVENUE K

AddrPrimaryHighNo is Double

StreetPreDrctnAbbrev is Text

StreetName is Text

StreetSuffixAbbrev is Text

StreetostDrctnAbbrev is Text

If I perform the field calculation below in ArcMap 10.5.1/VB Script, it works out just fine:

[AddrPrimaryHighNo]&" "& [StreetPreDrctnAbbrev]&" "& [StreetName]&" "& [StreetSuffixAbbrev]&" "& [StreetPostDrctnAbbrev]

Meanwhile I'm just going to go back to ArcMap.  Thanks

-Justin

LanceCole

https://community.esri.com/migrated-users/3420 ,

You are correct, I should not assume Justin would understand how to place a def in a code block.  I was also using my phone to reply while traveling.  Below are the call, def and a Calculate Field screen shot to use a function called "joinAddress" with the needed address fields which returns the assembled address string less any None values. 

You call the function using:

joinAddress(!AddrPrimaryHighNo!, !StreetPreDrctnAbbrev!, !StreetName!, !StreetSuffixAbbrev!, !StreetPostDrctnAbbrev!)

The Code Block:

<SPAN class="keyword token">def</SPAN> <SPAN class="token function">joinAddress</SPAN><SPAN class="punctuation token">(</SPAN>a<SPAN class="punctuation token">,</SPAN>b<SPAN class="punctuation token">,</SPAN>c<SPAN class="punctuation token">,</SPAN>d<SPAN class="punctuation token">,</SPAN>e<SPAN class="punctuation token">)</SPAN><SPAN class="punctuation token">:</SPAN>
    flds <SPAN class="operator token">=</SPAN> <SPAN class="punctuation token">[</SPAN>a<SPAN class="punctuation token">,</SPAN>b<SPAN class="punctuation token">,</SPAN>c<SPAN class="punctuation token">,</SPAN>d<SPAN class="punctuation token">,</SPAN>e<SPAN class="punctuation token">]</SPAN>
    <SPAN class="keyword token">return</SPAN> <SPAN class="string token">" "</SPAN><SPAN class="punctuation token">.</SPAN>join<SPAN class="punctuation token">(</SPAN><SPAN class="punctuation token">[</SPAN>str<SPAN class="punctuation token">(</SPAN>i<SPAN class="punctuation token">)</SPAN> <SPAN class="keyword token">for</SPAN> i <SPAN class="keyword token">in</SPAN> flds <SPAN class="keyword token">if</SPAN> i <SPAN class="keyword token">is</SPAN> <SPAN class="operator token">not</SPAN> None<SPAN class="punctuation token">]</SPAN><SPAN class="punctuation token">)</SPAN>‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍<SPAN class="line-numbers-rows"><SPAN>‍</SPAN><SPAN>‍</SPAN><SPAN>‍</SPAN></SPAN>

Side note: If all your fields were string you could also use " ".join(filter(None, flds)).  You can still use this by changing the following; flds = [str(a),b,c,d,e].  I prefer the method above as you can also change the str(i) to str(i).strip() to remove any leading or training spaces that may be in your fields.

Calculate Field:

If you do not want to use the code block with a call, you can use the following as a calculate statement:

" ".join([str(i) for i in [!AddrPrimaryHighNo!, !StreetPreDrctnAbbrev!, !StreetName!, !StreetSuffixAbbrev!, !streetPostDrctnAbbrev!] if i is not None])