Python 3 - How to convert dictionary to a counter ?

I have  a list (mylist)  which I iterate - each iteration tests the list value against two dictionaries. It can happen that the  end_cnt dictionary may not contain the mylist item being iterated and then returns an error. If I were to convert end_cnt to a counter  it would return a 0, which is exactly what I want.  The code is attached and this is the error message when I run it.

I thought it would be easy to convert a dictionary to a counter, but cant find anything on it .  Can someone show me how to do this, or suggest an alternative approach ?

many thanks

ERROR
<H3>Traceback (most recent call last):</H3>
<PRE>  File &quot;C:\xampp2\htdocs\OFFICE_15\adding_dicts.cgi&quot;, line 43, in &lt;module&gt;
    if drop_cnt[each] == 2 and end_cnt[each] == 0:
<B>KeyError: 20
</B></PRE>

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PAGE
#!C:\Python34\python.exe
import cgi,cgitb
import mysql.connector as conn
import collections
import datetime
import json
import traceback
def htmlTop():
    print("""Content-type:text/html\n\n
        <!DOCTYPE html>
        <html lang="en">
            <head>
                #<meta http-equiv="Content-Type" content="text/html;charset=iso-8859-1">
                <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
                <title> My Server-side template</title>
        </head>
        <body>""")

def htmlTail():
    print(""" its done</body>
        </html>""")

def connectDB():
    db=conn.connect(host='localhost' ,user='root' ,passwd='844cheminduplan' ,db='office')
    cursor = db.cursor()
    return db, cursor 


#main program
if __name__== "__main__":
    try:
        htmlTop()
        
        mylist = [10,9,20]
        drop_cnt = ({10: 1, 9: 1, 20: 2})
        end_cnt = ({9: 2, 11: 2, 10: 1, 12: 1}) 
       
        
        
        for each in mylist:
            if drop_cnt[each] == 1 and end_cnt[each] == 2:
                print(each, "no rebate")
            if drop_cnt[each] == 1 and end_cnt[each] == 1:
                print(each, "rebate")               
            if drop_cnt[each] == 2 and end_cnt[each] == 0: #THIS IS THE PROBLEM
                print(each, "rebatex2")            
        
        
        
        
       
            
            
            
        
             
        htmlTail()
    except:
        cgi.print_exception()

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jameskaneAsked:
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gelonidaCommented:
if I understand correctly you'd like to return the value from the dict if the key exists and otherwise a zero, right?

To get a default value from a dict in case the key doesn't exist use get()

value = yourdict.get(key, default_value)

Your changed code would look like.

        mylist = [10,9,20]
        drop_cnt = ({10: 1, 9: 1, 20: 2})
        end_cnt = ({9: 2, 11: 2, 10: 1, 12: 1})  
        drop_defaut_value = 0
        end_default_value = 0
        
        for each in mylist:
            drop_value = drop_cnt.get(each, drop_default_value)
            end_value = end_cnt.get(each, end_default_value)

            if drop_value == 1 and end_value == 2:
                print(each, "no rebate")
            elif drop_value == 1 and end_value == 1:
                print(each, "rebate")               
            elif drop_value == 2 and end_value == 0: #THIS IS THE PROBLEM
                print(each, "rebatex2")
           else:
               print("not sure what to do here")
           

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Just a general suggestion for future questions:
It might be a good idea to make you example code as small as possible, such, that the experts don't have
to read through code lines, that are irrelevant to the problem.
In your case (if I understood the question correctly) html_top / html_tail and connect_db are irrelevant for the given problem.
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jameskaneAuthor Commented:
Thanks very much, that works and now I better understand the relationship between dictionaries and collections.

Noted your point about the info supplied. Reason it happened is that I  use a standard template when trying out some new code - lot of the time this involves database connections.  I'll remember to strip that out in the future if its not relevant to the problem !

thanks again
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