I have dict
of nested list
s:
d = {'a': [[('a1', 1, 1), ('a2', 1, 2)]], 'b': [[('b1', 2, 1), ('b2', 2, 2)]]}
print (d)
{'b': [[('b1', 2, 1), ('b2', 2, 2)]], 'a': [[('a1', 1, 1), ('a2', 1, 2)]]}
I need create list
of tuple
s like:
[('b', 'b1', 2, 1), ('b', 'b2', 2, 2), ('a', 'a1', 1, 1), ('a', 'a2', 1, 2)]
I tried:
a = [[(k, *y) for y in v[0]] for k,v in d.items()]
a = [item for sublist in a for item in sublist]
I think my solution is a bit over-complicated. Is there some better, more pythonic, maybe one line solution?
You were almost there:
[(k, *t) for k, v in d.items() for t in v[0]]
The v[0]
is needed because your values are just single-element lists with another list contained. The above can be expanded to the following nested for
loops, if you wanted to figure out what it does:
for key, value in d.items(): # value is [[(...), (...), ...]]
for tup in value[0]: # each (...) from value[0]
(key, *tup) # produce a new tuple
Demo:
>>> d = {'a': [[('a1', 1, 1), ('a2', 1, 2)]], 'b': [[('b1', 2, 1), ('b2', 2, 2)]]}
>>> [(k, *t) for k, v in d.items() for t in v[0]]
[('a', 'a1', 1, 1), ('a', 'a2', 1, 2), ('b', 'b1', 2, 1), ('b', 'b2', 2, 2)]
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