Question

I think it is pretty straightforward. All I am trying to do is update the original dictionary’s 'code' with that of another dictionary which has the value. I get a feeling 2 for loops and an IF loop can be further shortened to get the answer. In my actual problem, I have few 1000’s of dicts that I have to update. Thanks guys!

Python:

referencedict = {'A': 'abc', 'B': 'xyz'}

mylistofdict = [{'name': 'John', 'code': 'A', 'age': 28}, {'name': 'Mary', 'code': 'B', 'age': 32}, {'name': 'Joe', 'code': 'A', 'age': 43}]

for eachdict in mylistofdict:
    for key, value in eachdict.items():
        if key == 'code':
            eachdict[key] = referencedict[value]

print mylistofdict

Output:

[{'age': 28, 'code': 'abc', 'name': 'John'}, {'age': 32, 'code': 'xyz', 'name': 'Mary'}, {'age': 43, 'code': 'abc', 'name': 'Joe'}]

Answer 1

There is no need to loop over all values of eachdict, just look up code directly:

for eachdict in mylistofdict:
    if 'code' not in eachdict:
        continue
    eachdict['code'] = referencedict[eachdict['code']]

You can probably omit the test for code being present, your example list always contains a code entry, but I thought it better to be safe. Looking up the code in the referencedict structure assumes that all possible codes are available.

I used if 'code' not in eachdict: continue here; the opposite is just as valid (if 'code' in eachdict), but this way you can more easily remove the line if you do not need it, and you save yourself an indent level.