I have two lists of tuples. I want a new list with every member of l2 and every member of l1 that does not begin with the same element from l2.
I used a for loop and my output is ok.
My question is: How can I use the filter function or a list comprehension?
def ov(l1, l2):
l3=l1.copy()
for i in l2:
for j in l1:
if i[0]==j[0]:
l3.pop(l3.index(j))
print (l3+l2)
ov([('c','d'),('c','e'),('a','b'),('a', 'd')], [('a','c'),('b','d')])
The output is:
[('c', 'd'), ('c', 'e'), ('a', 'c'), ('b', 'd')]
If I understand correctly, this should be the straight forward solution:
>>> l1 = [('c','d'),('c','e'),('a','b'),('a', 'd')]
>>> l2 = [('a','c'),('b','d')]
>>>
>>> starters = set(x for x, _ in l2)
>>> [(x, y) for x, y in l1 if x not in starters] + l2
[('c', 'd'), ('c', 'e'), ('a', 'c'), ('b', 'd')]
This can be generalized to work with longer tuples with extended iterable unpacking.
>>> starters = set(head for head, *_ in l2)
>>> [(head, *tail) for head, *tail in l1 if head not in starters] + l2
[('c', 'd'), ('c', 'e'), ('a', 'c'), ('b', 'd')]
Here is an approach using filter:
from operator import itemgetter
f = itemgetter(0)
zval = set(map(itemgetter(0), l2))
list(filter(lambda tup: f(tup) not in zval, l1)) + l2
[('c', 'd'), ('c', 'e'), ('a', 'c'), ('b', 'd')]
Or:
def parser(tup):
return f(tup) not in zval
list(filter(parser, l1)) + l2
[('c', 'd'), ('c', 'e'), ('a', 'c'), ('b', 'd')]
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