CODEWITHSUNDEEP
pythonlistlist-comprehension
badatprog
badatprog

Reputation: 19

Python 3 List Comprehension remove tuples from list

I have to solve a task for university. I have 4 given lists and I have to count the possible variations of burgers that can be made under some restrictions.

breads = ["Weissbrot", "Vollkorn", "Dinkel", "Speckbrot"]
patties = ["Wildschwein", "Rind", "Halloumi", "Aubergine"]
souces = ["Kaese", "Knoblauch", "Curry"]
toppings = ["Kopfsalat", "Bacon", "Tomate"]

My code so far:

i = "bottom, patty, souce, topping, top"

burger = [i for bottom in breads
          for top in breads
          for patty in patties
          for souce in souces
          for topping in toppings
          if bottom != top
          if i != ("Speckbrot", "Aubergine", "Kaese", "Bacon", "Weissbrot")]

print(len(burger))

The restrictions:

The finished burger needs to have the structure (bottom, patty, souce, topping, top). I safed this under the variable 'i'. Bottom and top must have different bread. I solved this with if bottom != top.

Mixing Aubergine with Speckbrot, Kaese or Bacon and Halloumi with Speckbrot or Bacon is not allowed. I tried to sovle this with if i != ("Speckbrot", "Aubergine", "Kaese", "Bacon", "Weissbrot") but its obviously not correct.

Furthermore, if bottom and top are swapped and the rest stays the same this counts as 1 burger and not 2. I have no plan how to solve this yet.

Sorry for the german words, i can translate it if needed.

Many thanks in advance

EDIT: The correct answer is 138 variations.

Upvotes: 0

Views: 62

Answers (3)

Blckknght
Blckknght

Reputation: 104712

Your issue is with the variable i, which doesn't work the way you seem to want it to. It is not a substitute for naming five variables. It's just a string that happens to have the variable names in it.

This would be a valid comprehension, though I suspect the restriction from the last if clause is a lot more narrow than you want (it forbids exactly one combination).

burger = [(bottom, patty, souce, topping, top)
          for bottom in breads
          for top in breads
          for patty in patties
          for souce in souces
          for topping in toppings
          if bottom != top
          if (bottom, patty, souce, topping, top) !=
             ("Speckbrot", "Aubergine", "Kaese", "Bacon", "Weissbrot")]

Upvotes: 1

code_vader
code_vader

Reputation: 256

Using list comprehension, this works

choices = [[b, p, s, t] for b in breads for p in patties for s in souces for t in toppings] #add restrections (if [b,p, s, t] !=)

print (choices)

Upvotes: 0

code_vader
code_vader

Reputation: 256

First, in this case you probably shouldn't be using list comprehension. List comprehension is for smaller things, and this is far too messy. This is probably the best way to do it (using for loops). In the code below, count will be the total number and choices are all possible burgers.

choices = []
count = 0
for b in breads:
    for p in patties:
        for s in souces:
            for t in toppings:
                temp = [b, p, s, t]
                #add restrictions here
                choices.append(temp)
                count += 1

print (choices, count)

Upvotes: 0

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