How can I write a list comprehension to enumerate key: value pairs into 3-membered tuples from a Python dictionary?
d = {'one': 'val1', 'two': 'val2', 'three': 'val3', 'four': 'val4', 'five': 'val5'}
When I try this:
li = [(index, k, val) for index, k, val in enumerate(d.items())]
I get a ValueError: need more than 2 values to unpack
The desired output would be:
[(0, 'one', 'val1'),
(1, 'two', 'val2'),
(2, 'three', 'val3'),
(3, 'four', 'val4'),
(4, 'five', 'val5')]
Nest your tuples. But the order may not be as desired.
li = [(index, k, val) for index, (k, val) in enumerate(d.items())]
The output of enumerate() is a 2-element tuple. Since the value of that tuple is another 2-element tuple, you need to use parentheses to group them.
li = [(index, k, val) for index, (k, val) in enumerate(d.items())]
However, since a dict is by default unordered, you need to create an OrderedDict.
odict = collections.OrderedDict()
odict['one'] = 'val1'
odict['two'] = 'val2'
odict['three'] = 'val3'
odict['four'] = 'val4'
li = [(index, k, val) for (index, (k, val)) in enumerate(odict.items())]
This would give you the following value for li
[(0, 'one', 'val1'), (1, 'two', 'val2'), (2, 'three', 'val3'), (3, 'four', 'val4')]
