Remove items in an array that are contained in other items [Python]

I've tried looping through the array and looking to see if one item contains another item as followed.

you were almost there:

list_ = []
Array = ['orangutan', 'sun', 'tan']
for i in Array:
    list_.extend([x for x in Array if x in i and x != i])
print(list_)

output:

['tan']

The remove() method removes the first matching element (which is passed as an argument) from the list.

I assume that you have another array with the strings that you want to keep. If this is the case, you can loop and use .pop():

array=['orangutan', 'sun', 'tan']
other_array=['orangutan', 'sun']
for i in range(0,len(array)):
    if array[i] not in other_array:
        array.pop(i)
        
print(array)

array=['orangutan', 'sun']

You can do:

lst = ['orangutan', 'sun', 'tan']

print([x for i, x in enumerate(lst)
       if not any(x in y for j, y in enumerate(lst) if i != j)])

Update:

It won't escape notice that I have arrived at essentially the same solution as Riccardo Bucco. But the tests are different here. For the equivalence of these two tests, see De Morgan's laws. The any (i.e. a reduction of or over the boolean inputs) gives the inverse of all (i.e. reduction of and) on the inverse of those inputs -- hence the not any(... in ...) is equivalent to all(... not in ...).

In either case, iteration will stop at the first match: all will stop iterating and return False if it sees a false-like value, and any will stop iterating and return True if it sees a true-like value.

Here is a possible solution:

result = [x
          for i, x in enumerate(Array)
          if all(x not in y for j, y in enumerate(Array) if i != j)]