How to find item in list - list

Question

I am new and have been working on this for a week now but can't find any solution. I hope someone can help me figure this out.
How can I find items in the list - listitems and output their items individually?
listitems = ['Beer, Chicken', 'Cake, Chocolate', 'Lemon with ci, Chicken', 'Beer, Beer, Cake, Chocolate']
Is there anyway i can compute the related food in the receipt list?
So far I am only able to find the foods for ONE item. My code is as follows:
I also computed another list of each item is for comparison.
eachitems = ['Beer', 'Cake', 'Chocolate', 'Lemon with ci', 'Chicken']

Answer 1

I would personally use a dictionary based on keys for each item with associated items as their values, would also be much easier for you to get the results you want, not exactly sure right know how I would accomplish it from the list you made only.
From your original code, add " print (combi) ", " print (checklist) " and " print (correlatedlist) " at the end and you will see it doesn't really append it the way you want.

Answer 2

In Python 3.5:
import itertools
listOfStrings = ['Beer, Chicken', 'Cake, Chocolate', 'Lemon with ci, Chicken', 'Beer, Beer, Cake, Chocolate']
print ('Original list:', listOfStrings)
print ()
listOfLists = [group.replace (', ', ',') .split (',') for group in listOfStrings]
print ('Turned into list of lists:', listOfLists)
print ()
allFoods = set (itertools.chain (*listOfLists))
print ('All individual foods:', allFoods)
print ()
listOfSets = [set (group) for group in listOfLists]
print ('A list of sets is handy, since sets contain no duplicates:', listOfSets)
print ()
dictOfFoods = dict ([[food, set ()] for food in allFoods])
print ('Prepare a dictionary, where we can put the associated foods:', dictOfFoods)
print ()
for food in dictOfFoods:
for foodSet in listOfSets:
if food in foodSet:
dictOfFoods [food] .update (foodSet)
dictOfFoods [food] .remove (food)
print ('The dictionary is now filled:', dictOfFoods)
print ()
for food in dictOfFoods:
print ('People who buy', food, 'also buy:')
for otherFood in dictOfFoods [food]:
print (otherFood)
print ()
Will print:
Original list: ['Beer, Chicken', 'Cake, Chocolate', 'Lemon with ci, Chicken', 'Beer, Beer, Cake, Chocolate']
Turned into list of lists: [['Beer', 'Chicken'], ['Cake', 'Chocolate'], ['Lemon with ci', 'Chicken'], ['Beer', 'Beer', 'Cake', 'Chocolate']]
All individual foods: {'Chocolate', 'Lemon with ci', 'Chicken', 'Cake', 'Beer'}
A list of sets is handy, since sets contain no duplicates: [{'Chicken', 'Beer'}, {'Chocolate', 'Cake'}, {'Chicken', 'Lemon with ci'}, {'Chocolate', 'Cake', 'Beer'}]
Prepare a dictionary, where we can put the associated foods: {'Chocolate': set(), 'Lemon with ci': set(), 'Cake': set(), 'Beer': set(), 'Chicken': set()}
The dictionary is now filled: {'Chocolate': {'Cake', 'Beer'}, 'Lemon with ci': {'Chicken'}, 'Cake': {'Chocolate', 'Beer'}, 'Beer': {'Chocolate', 'Chicken', 'Cake'}, 'Chicken': {'Lemon with ci', 'Beer'}}
People who buy Chocolate also buy:
Cake
Beer
People who buy Lemon with ci also buy:
Chicken
People who buy Cake also buy:
Chocolate
Beer
People who buy Beer also buy:
Chocolate
Chicken
Cake
People who buy Chicken also buy:
Lemon with ci
Beer
If you don't want to use itertools and *, you can also make a loop in a loop to traverse all elements of the listOfLists and add them to allFoods, which you initially make empty.

Answer 3

It took me sometime to understand exactly what you wanted, but this is a working solution.
listitems = ['Beer, Chicken', 'Cake, Chocolate', 'Lemon with ci, Chicken', 'Beer, Beer, Cake, Chocolate']
eachitems = ['Beer', 'Cake', 'Chocolate', 'Lemon with ci', 'Chicken']
for item in eachitems:
assoc = [associated for associated in listitems if item in associated]
result = set()
for itemlist in assoc:
itemlist = itemlist.replace(', ', ',').split(',')
itemlist = set(itemlist)
itemlist.remove(item)
result = result | itemlist
print('People who buy {} also buy: '.format(item), ', '.join(sorted(result)))
Output
People who buy Beer also buy: Cake, Chicken, Chocolate
People who buy Cake also buy: Beer, Chocolate
People who buy Chocolate also buy: Beer, Cake
People who buy Lemon with ci also buy: Chicken
People who buy Chicken also buy: Beer, Lemon with ci
They key part to this solution is the use of Sets to remove the duplicate items and the |(union) operator.
As a side note, instead of using | like this
result = result | itemlist
you can modify the set in place with
result.update(itemlist)