Merge Lists with same first index but other second index - list

Question

I am working on a search algorithm in python but there is something I don't get to work..
I have a list which looks like this [["A","1.txt"],["A","2.txt"],["A","3.txt"],["B","1.txt"],["B","3.txt"]]
Now I want to merge the sub-lists that have the same first index. So the result would be:
[["A",["1.txt","2.txt",3.txt"]],["B",["1.txt"],["3.txt"]]]
Anyone who knows how to do this...
Kinda got a sort (on mergesort basis) but this does not merge the tuples
def merge_pairs(data):
if len(data) <= 1 :
return data[:]
else:
mid = len(data) // 2
fst = merge_pairs(data[:mid])
snd = merge_pairs(data[mid:])
res = []
fi = 0
si = 0
while fi < len(fst) and si < len(snd):
if fst[fi][0] < snd[si][0] or fst[fi][0] == snd[si][0] and fst[fi][1] < snd[si][1]:
res.append(fst[fi])
fi = fi + 1
else:
res.append(snd[si])
si = si + 1
if fi < len(fst) :
res.extend(fst[fi:])
elif si < len(snd) :
res.extend(snd[si:])
return res
So i'd like not to use the dict() function of python
Thanks in advance

Answer 1

The easiest way (which may or may not be slower than the hard way) is to use a defaultdict:
>>> from collections import defaultdict
>>> result = defaultdict(list)
>>> mylist = [["A","1.txt"],["A","2.txt"],["A","3.txt"],["B","1.txt"],["B","3.txt"]]
>>> for key, value in mylist:
... result[key].append(value)
...
>>> print(sorted(result.items()))
[('A', ['1.txt', '2.txt', '3.txt']), ('B', ['1.txt', '3.txt'])]
The hard way (if your data is truly already sorted):
>>> src = [["A","1.txt"],["A","2.txt"],["A","3.txt"],["B","1.txt"],["B","3.txt"]]
>>> prev = None
>>> dst = []
>>> for key, value in src:
... if key != prev:
... prev = key
... dst.append((key, []))
... dst[-1][-1].append(value)
...
>>> print(dst)
[('A', ['1.txt', '2.txt', '3.txt']), ('B', ['1.txt', '3.txt'])]
But note that Python sort is really, really fast, and Python loops like this... Not so much.
Edit According to your comment below, you also want counts. Again there is a dictionary way:
>>> from collections import defaultdict
>>> result = defaultdict(lambda: defaultdict(int))
>>> mylist = [["A","1.txt"],["A","2.txt"],["A", "2.txt"],["A","3.txt"],["B","1.txt"],["B","3.txt"]]
>>> for key, value in mylist:
... result[key][value] += 1
...
>>> print(sorted((x, sorted(y.items())) for (x, y) in result.items()))
[('A', [('1.txt', 1), ('2.txt', 2), ('3.txt', 1)]), ('B', [('1.txt', 1), ('3.txt', 1)])]
and a loop way:
>>> src = [["A","1.txt"],["A","2.txt"],["A", "2.txt"],["A","3.txt"],["B","1.txt"],["B","3.txt"]]
>>> prevkey, prevvalue = None, None
>>> dst = []
>>> for key, value in src:
... if key != prevkey:
... prevkey = key
... prevvalue = None
... dst.append((key, []))
... if value != prevvalue:
... prevvalue = value
... dst[-1][-1].append([value, 0])
... dst[-1][-1][-1][-1] += 1
...
>>> dst
[('A', [['1.txt', 1], ['2.txt', 2], ['3.txt', 1]]), ('B', [['1.txt', 1], ['3.txt', 1]])]
You'd really want to run timeit to be sure, but in this instance, the loop way almost looks guaranteed to be slower (and of course, the dictionary way doesn't require you to do a pre-sort.)