I try create dict
by nested list
:
groups = [['Group1', 'A', 'B'], ['Group2', 'C', 'D']]
L = [{y:x[0] for y in x if y != x[0]} for x in groups]
d = { k: v for d in L for k, v in d.items()}
print (d)
{'B': 'Group1', 'C': 'Group2', 'D': 'Group2', 'A': 'Group1'}
But it seems a bit complicated.
Is there a better solution?
What about:
d = {k:row[0] for row in groups for k in row[1:]}
This gives:
>>> {k:row[0] for row in groups for k in row[1:]}
{'D': 'Group2', 'B': 'Group1', 'C': 'Group2', 'A': 'Group1'}
So you iterate over every row
in the groups
. The first element of the row is taken as value (row[0]
) and you iterate over row[1:]
to obtain all the keys k
.
Weird as it might seem, this expression also works when you give it an empty row (like groups = [[],['A','B']]
). That is because row[1:]
will be empty and thus the row[0]
part is never evaluated:
>>> groups = [[],['A','B']]
>>> {k:row[0] for row in groups for k in row[1:]}
{'B': 'A'}
This is essentially a prettier version of Willem's:
>>> groups = [['Group1', 'A', 'B'], ['Group2', 'C', 'D']]
>>> {k:g for g,*tail in groups for k in tail}
{'B': 'Group1', 'A': 'Group1', 'C': 'Group2', 'D': 'Group2'}
But it won't work with an empty list:groups = [[],['A','B']]
>>> {k:head for head, *tail in grps for k in tail}
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 1, in <dictcomp>
ValueError: not enough values to unpack (expected at least 1, got 0)
I think one line solution is a bit confusion. I would write code like below
groups = [['Group1', 'A', 'B'], ['Group2', 'C', 'D']]
result = {}
for group in groups:
for item in group[1:]:
result[item] = group[0]
print result
I also like Willem's solution, but just for completeness...
another variation using itertools and a generator function (Python 3.x only)
def pairs(groups):
for value,*keys in groups:
for key_value in zip(keys, itertools.repeat(value)):
yield key_value
dict(pairs(groups))
{'A': 'Group1', 'B': 'Group1', 'C': 'Group2', 'D': 'Group2'}