如何扫描一个大列表以查找所有元素是否与第二个较小列表的元素匹配？

Question

I'm hvaing a hard time with a problem which I thought was pretty simple. I have two lists:

1. A small one, that looks like this:

list1 = ['A', 'B', 'C', 'D','E']

2. The second list is much bigger, going on for about 800 elements. It looks like this:

list2 = ['E', 'B', 'F', 'A', 'C', 'N'...]

I want to scan list2 and see if all of its elements are the same as the ones in list1. If they are different, I want to see which are the elements that differ and cancel them from list2. In this example, I want to print "F" and "N" from list2 and cancel them.

I tried:

found = False
lenght2 = len(list2)
i = 0
for j in list1:
  for i in range(0, lenght2):
     if i != j:
        found = True
        #I don't know how to cancel i
        print(i)
        i = i + 1
     break

However, the whole thing does not work. Is there anyone who could help me?

Answer 1

You could go through all of list2 then check if it is in list one, like this:

for i in range(len(list2)):
  if list2[i] in list1:
    pass
  else:
    #Cancel list2[i] ? Or whatever.

Answer 2

Keep it simple:

There's no need to use nested loops here -- you are removing elements from list2 , based on information from list1 , so loop through list2 .

list1 = ['A', 'B', 'C', 'D', 'E']
list2 = ['E', 'B', 'F', 'A', 'C', 'N']

for e in list2[:]:
    if e not in list1:
        list2.remove(e)
print(list2)

This code loops through list2 and checks for each element ( e ) if it occurs in list1 . If not, it removes it from list2 .

Note: the use of list2[:] here rather than list2 is due to this .

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Or use a list comprehension:

list1 = ['A', 'B', 'C', 'D', 'E']
list2 = ['E', 'B', 'F', 'A', 'C', 'N']

list2 = [e for e in list2 if e in list1]
print(list2)

See more about how list comprehensions work here .

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In both cases, you should get:

['E', 'B', 'A', 'C']