根据另一个列表从子列表中删除元素

Question

I have a list filled with sublist:

a = [['12345/0111','57894/0311','45698/2333'],['12345/0600','87456/1234']]

Then I have another list:

b = ['0111','0600','0311']

I would like to delete all elements from list a which doesn't contain elements from list b . I thought that it should be like this for the firts element in b (in this case 45698/2333 and 87456/1234):

for x in a:
    for y in x:
        if b[0] not in y:
            x.remove(y)

But it doesn't work even for the first element and I really don't know how to do it for all the elements in b .

EDIT: I am sorry I didn't specified that in the output I need to have the same nested list structure.

Answer 1

Give this a try:

# create a new list to store values
c = []

for x in a:
   for y in x:
      if str.split(y, '/')[1] in b:
         c.append(y)

My Results:

a
[['12345/0111', '57894/0311', '45698/2333'], ['12345/0600', '87456/1234']]
b
['0111', '0600', '0311']
c = []
for x in a:
    for y in x:
       if str.split(y, '/')[1] in b:
          c.append(y)

c
['12345/0111', '57894/0311', '12345/0600']

EDITED AFTER OP CLARIFICATION:

for x in a:
  for y in x:
    if str.split(y, '/')[1] in b:
      x.remove(y)

EDITED RESULTS:

for x in a:
   for y in x:
     if str.split(y, '/')[1] in b:
       x.remove(y)

a
[['57894/0311', '45698/2333'], ['87456/1234']]

Answer 2

You're saying to remove anything that's in b, which doesn't appear to be what you mean. In any case if you remove the 'not' from either of the following two samples they will give you the ones that are in b.

a = [['12345/0111','57894/0311','45698/2333'],['12345/0600','87456/1234']]
b = ['0111','0600','0311']

Not Nested

output = []
for l in a:
    for e in l:
        if not any([x in e for x in b]):
            output.append(e)

Nested

output = []
for l in a:
    output.append([x for x in l if not any(z in x for z in b)])