How do I concisely convert a python dictionary to a list of tuples, given a dictionary in which the values may be different types?

Question

Important: If the value is a list, then I want to get multiple key, value tuples for that key and each item in the list. The values in the dictionary will be of multiple types, including at least strings, ints, and lists.

I have a solution, below, but I'm wondering if there's a more concise way to do this.

# Sample data
d = {'key1': 'string1', 
    'key2': 15, 
    'key3': ['item1', 'item2', 'item3', 'item4']}

# This code does what I want
parameters = []
for k, v in d.iteritems():
    if isinstance(v, str):
        parameters.append((k ,v))
    elif isinstance(v, int):
        parameters.append((k, v))
    elif isinstance(v, list):
        for x,y in zip(itertools.repeat(k), v):
            parameters.append((x, y))
    else:
        continue

parameters
# [('key1', 'string1'),
#  ('key2', 15),
#  ('key3', 'item1'),
#  ('key3', 'item2'),
#  ('key3', 'item3'),
#  ('key3', 'item4')]

Answer 1

I like list comprehensions for readability. Also, I don't see why you need a separate statement for strings and ints. Either you have a list or not.

parameters = []
for k, v in d.iteritems():
  if isinstance(v, list):
    parameters.extend([(k, i) for i in v])
  else:
    parameters.append((k, v))

Results:

>>> parameters
[('key3', 'item1'), ('key3', 'item2'), ('key3', 'item3'), ('key3', 'item4'), ('key2', 15), ('key1', 'string1')]

Answer 2

Here is a shorter version with list comprehension:

>>> d = {'key1': 'string1', 'key2': 15, 'key3': ['item1', 'item2', 'item3', 'item4']}
>>> parameters = [(k, v1) for k, v in d.iteritems() for v1 in (v if isinstance(v, list) else [v])]
>>> parameters
[('key3', 'item1'), ('key3', 'item2'), ('key3', 'item3'), ('key3', 'item4'), ('key2', 15), ('key1', 'string1')]