Make map() return a dictionary

Question

I have the following function:

def heading_positions(self):
    return map(
            lambda h:
                {'{t}.{c}'.format(t=h.table_name,c=h.column_name) : h.position },
                self.heading_set.all()
            )

It gives me output like this:

[{'customer.customer_number': 0L}, {'customer.name': 2L}, ... ]

I would prefer just a single dictionary like this:

{'customer.customer_number': 0L, 'customer.name': 2L, ...

Is there a way to make map (or something similar) return just a single dictionary instead of an array of dictionaries?

Answer 1

Yes. The basic problem is that you don't create a dictionary out of single-entry dicts, but out of a a sequence of length-two sequences (key, value).

So, rather than create an independent single-entry dict with the function, create a tuple and then you can use the dict() constructor:

dict(map(lambda h: ('{t}.{c}'.format(t=h.table_name, c=h.column_name), h.position), 
         self.heading_set.all()))

Or directly use a generator or list comprehension inside the dict constructor:

dict(('{t}.{c}'.format(t=h.table_name, c=h.column_name), h.position) 
     for h in self.heading_set.all())

Or, on the latest versions (2.7, 3.1) a dictionary comprehension directly:

{'{t}.{c}'.format(t=h.table_name : c=h.column_name), h.position) 
     for h in self.heading_set.all()}