I am struggling with the following. Basically I have a list:
dolist = [(1280, ['A1'], ['A2']), (1278, ['A1'], ['A2']), (1276, ['A1'], ['A2']), (1274, ['B1'], ['B2']), (1272, ['A1'], ['A2']), (1270, [], ['A2'])]
Now I want to have lists sorted sorted by element 2 and 3.
uniqdo = [ (['A1'],['A2']), (['B1'],['B2']),([],['A2']) ]
dorange = [ "1280-1276,1272","1274","1270" ]
I have tried to do with straightforward comparisons but the code becomes very long with several tests and looks a bit messy. There must be library functions which can do this reasonable quick.
It looks like itertools.groupby could help you:
>>> dolist = [ (1280,['A1'],['A2']),(1278,['A1'],['A2']),(1276,['A1'],['A2']),(1274,['B1'],['B2']),(1272,['A1'],['A2']) ]
>>> from itertools import groupby
>>> [[v, [i for i,*_ in g]] for v, g in groupby(dolist, key= lambda l: (l[1][0], l[2][0]))]
[[('A1', 'A2'), [1280, 1278, 1276]], [('B1', 'B2'), [1274]], [('A1', 'A2'), [1272]]]
It shouldn't be hard to convert the above data structure to the one you want.
Here's a start. You cannot leave any list as input because a Python list cannot be used as a dict key. So get_value
returns None
instead of an empty list:
from itertools import groupby
dolist = [(1280, ['A1'], ['A2']), (1278, ['A1'], ['A2']), (1276, ['A1'], ['A2']), (1274, ['B1'], ['B2']), (1272, ['A1'], ['A2']), (1270, [], ['A2'])]
ranges = {}
def get_value(l):
if l:
return l[0]
else:
return None
def get_values(t):
return (get_value(t[1]), get_value(t[2]))
for v, g in groupby(dolist, get_values):
ids = [str(t[0]) for t in g]
if len(ids) > 1:
range_str = ids[0] + '-' + ids[-1]
else:
range_str = ids[0]
ranges.setdefault(v, []).append(range_str)
print(ranges)
# {('A1', 'A2'): ['1280-1276', '1272'], ('B1', 'B2'): ['1274'], (None, 'A2'): ['1270']}