仅使用元组的前两个部分将3元组与3元组列表进行比较

Question

I have a list of 3-tuples in a Python program that I'm building while looking through a file (so one at a time), with the following setup:

(feature,combination,durationOfTheCombination),

such that if a unique combination of feature and combination is found, it will be added to the list. The list itself holds a similar setup, but the durationOfTheCombination is the sum of all duration that share the unique combination of (feature,combination). Therefore, when deciding if it should be added to the list, I need to only compare the first two parts of the tuple, and if a match is found, the duration is added to the corresponding list item.

Here's an example for clarity. If the input is (ABC,123,10);(ABC,123,10);(DEF,123,5);(ABC,123,30);(EFG,456,30) The output will be (ABC,123,50);(DEF,123,5);(EFG,456,30) .

Is there any way to do this comparison?

Answer 1

You can do this with Counter ,

In [42]: from collections import Counter
In [43]: lst = [('ABC',123,10),('ABC',123,10),('DEF',123,5)]
In [44]: [(i[0],i[1],i[2]*j) for i,j in Counter(lst).items()]
Out[44]: [('DEF', 123, 5), ('ABC', 123, 20)]

As per the OP suggestion if it's have different values, use groupby

In [26]: lst = [('ABC',123,10),('ABC',123,10),('ABC',123,25),('DEF',123,5)]

In [27]: [tuple(list(n)+[sum([i[2] for i in g])]) for n,g in groupby(sorted(lst,key = lambda x:x[:2]), key = lambda x:x[:2])]
Out[27]: [('ABC', 123, 45), ('DEF', 123, 5)]

Answer 2

If you don't want to use Counter, you can use a dict instead.

setOf3Tuples = dict()

def add3TupleToSet(a):
  key = a[0:2]
  if key in setOf3Tuples:
    setOf3Tuples[a[0:2]] += a[2]
  else:
    setOf3Tuples[a[0:2]] = a[2]

def getRaw3Tuple():
  for k in setOf3Tuples:
    yield k + (setOf3Tuples[k],)

if __name__ == "__main__":
  add3TupleToSet(("ABC",123,10))
  add3TupleToSet(("ABC",123,10))
  add3TupleToSet(("DEF",123,5))
  print([i for i in getRaw3Tuple()])

Answer 3

It seems a dict is more suited than a list here, with the first 2 fields as key. And to avoid checking each time if the key is already here you can use a defaultdict.

from collections import defaultdict

d = defaultdict(int)
for t in your_list:
    d[t[:2]] += t[-1]

Answer 4

Assuming your input is collected in a list as below, you can use pandas groupby to accomplish this quickly:

import pandas as pd
input = [('ABC',123,10),('ABC',123,10),('DEF',123,5),('ABC',123,30),('EFG',456,30)]
output = [tuple(x) for x in pd.DataFrame(input).groupby([0,1])[2].sum().reset_index().values]