More Pythonic Way To Write This Block (and Elimnate: Valueerror: Dictionary Update Sequence Element #0 Has Length 1; 2 Is Required)

I have this (which works): def make_key(x): return '{0}/{1}/{2}'.format(x.f1, x.f2, x.f3) def make_value(x): return (x.f7, x.f1, x.f4, x.f9, x.f2, x.f8, x.f10

Solution 1:

You need to drop the list wrapper; just produce the tuples:

row_data = dict((make_key(x), make_value(x)) for x in records)

You were producing a list with one element each, a tuple.

If you are using Python 2.7 or newer, you can also use a dictionary comprehension:

row_data = {make_key(x): make_value(x) for x in records}

Your make_value could be expressed by using a operator.itemgetter() object:

fromoperator import itemgetter

make_value = itemgetter(
    'f7', 'f1', 'f4', 'f9', 'f2', 'f8', 'f10', 'f11', 'f12', 'f17', 
    'f18', 'f19', 'f14', 'f15', 'f16', 'f17', 'f20')

The make_key() function can make use of the fact that you can pull out attributes from an object; {0.f1} would interpolate the f1 attribute of the first positional argument to the str.format() method; make use of this to create a bound str.format() method that takes just one positional argument:

make_key = '{0.f1}/{0.f2}/{0.f3}'.format

Solution 2:

A few different things you can do to make it simpler :)

No need for a separate make_key function definition, you already have one:

make_key = '{0.f1}/{0.f2}/{0.f3}'.format

Beyond that, depending on your Python version you can also use dict comprehensions:

row_data = {make_key(x): make_value(x) for x in records}