比较两个列表的真假

Question

for a program I'm writing, I have two lists at the end that I need to compare. They won't be in order, but here's the tricky part. At the end the first list may have more items than the second list. But as long as the first list has the required items defined in the second list, the function would return a true value. I tried set lists, but that's more for intersection, which is not what I'm looking for. This is the code I have so far:

this is the default player inventory, which is added to throughout the game.

pla_back = ["Combat Knife","Flares","Compass"]

this is the block of code that compares the two lists. I tried to use in, but it doesn't work.

acceptItem = ["Combat Knife","Flares","Compass","silver bar","a cloth","a stick","some rope"]
    if acceptItem in pla_back:
        placeholder
    else:
        placeholder

If someone could help me with this, I would very much appreciate it. I feel like the answer is right on the tip of my toungue, but I just can't figure it out.

Answer 1

if set(pla_back) >= set(acceptItem):
    print 'Player has the required items'
else:
    print "Player doesn't have the required items"

If my interpretation is correct, you want to see if pla_back contains all the elements in acceptItem . So it could have more but as long as it has each element in acceptItem it should return true.

Here >= means 'is superset'

s.issuperset(t)     s >= t      test whether every element in t is in s

So we're checking if every element in acceptItem appears in pla_back

Answer 2

You should make at acceptItem a set. Then you can do

>>> acceptItem = {"Combat Knife","Flares","Compass","silver bar","a cloth","a stick","some rope"}
>>> pla_back = ["Combat Knife","Flares","Compass"]
>>> all(item in acceptItem for item in pla_back)
True

(This also works if acceptItem remains a list, but set lookup is much faster).