如何在Python中比较列表元素

Question

Here i am having two lists :

list1 = ['2C535EB58F19B58' , '7B89D9071EB531B143594FF909BAC846' , '0509']

list2 = ['1641AB0C9C5B8867' , '0098968C' , '509']

I need to compare the elements inside list2 with list1 elements.

I want the output to be , after comparing :

509

since 509 is present in 0509.

How can i achieve so ? Does regex can help me in this ?

Answer 1

Try this: Here we are checking whether element in list2 is a substring of an element in list1.

list1 = ['2C535EB58F19B58' , '7B89D9071EB531B143594FF909BAC846' , '0509']

list2 = ['1641AB0C9C5B8867' , '0098968C' , '509']

for i, j in zip(list1, list2):
    if j in i:
        print(j)

One-liner which will append to a list:

print( [j for i, j in zip(list1, list2) if j in i])

There could be more simple and better answers. If it helps, you can opt it.

Answer 2

You can do something like this:

common_elements = []
for x in list1:
    for y in list2:
        if y in x:
            common_elements.append(y)

common_elements will contain the elements you need.

Answer 3

As already proposed by BearBrown in a comment, there is an easy way to achieve your goal. There is no reason to think about regular expressions at all (the in -operator is powerful enough).

[x for x in list2 if any(y for y in list1 if x in y)]

Here you are looking for each string x in list2 if it is a substring of any string y in list1 and finally save each matching substring x in a new list.

Answer 4

for x in list2:
    for y in list1:
        if x in y:
            print x #or whatever

nested for-loops, I think this is a simple way, but I'm sure there is a better one

Answer 5

You can simply use in operation in python to check whether one string is in another string.

The easiest way to solve your problem would be

[y for y in list2 if any(y in x for x in list1)]