Is there a more "logic" form of programming this function?

Question

I finally make this works, but I dont like that my solution is like "someone else checking how many list exist" and writing the code.

It find common item in all list.

I tried to use recursion, but I coldn't make it works

def comunes (*lista):
  word = ""
  for i in lista:
    word = "set({})".format(i) + " &" + word
  word = word.split(" ")
  word.pop()
  wordF = ""
  for i in word:
    wordF += str(i)
  if len(eval(wordF)) == 0:
    return None
  else:
    return eval (wordF)

Could you give a little more detail on the problem you're trying to solve? You almost definitely do not need eval() for this. — Tillson, Apr 30 '19 at 02:36
Or turning something into a weird string only to parse it back again — Mad Physicist, Apr 30 '19 at 02:39
Yes it works. The solution that @isaactfa is what i was looking for — Daniel, Apr 30 '19 at 15:31

isaactfa · Accepted Answer · 2019-04-30T02:54:22.880

0

def comunes(*lista):
    commons = set(lista[0])
    for i in lista[1:]:
        commons &= set(i)
    return commons

This function creates a set commons with all the elements of the first list and then for each next list i it'll replace commons with the intersection (& operator) of commons with the set of i. The intersection of two sets is the set with only the elements that are both in the first and the second set. This'll ensure that commons has only elements that are in all the lists you passed to comunes.

edited Apr 30 '19 at 02:54

answered Apr 30 '19 at 02:49

isaactfa

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As a rule of thumb, if you're using `eval` or `exec` you should rethink your approach. It's almost never a good idea to use them for 'everyday' problems. – isaactfa Apr 30 '19 at 16:09

Is there a more "logic" form of programming this function?

1 Answers1