CODEWITHSUNDEEP
python-3.xlist
Annah Lairy
Annah Lairy

Reputation: 23

Python : element exist in list 1 and list 2

this is my code, with 2 lists. The script check the similar elements and put them in a new list named newlist :

list1 = ['jack sparow' , 'monika hardan', 'hamid tatan', 'california']

list2 = ['california']

newlist = []

if any(x in list1 for x in list2):
    newlist.append(list2)

print(newlist)

Output : [['california']]

What i'm looking for is, even if the element in list 1 contains one word of an element in list 2, should be detected, here is an example :

list1 = ['jack sparow' , 'monika hardan', 'hamid tatan', 'california']

list2 = ['sparow']

newlist = []

if any(x in list1 for x in list2):
    newlist.append(list2)

print(newlist)

Output : []

Wanted Output : [['jack sparow']]

Upvotes: 1

Views: 2038

Answers (3)

Gorisanson
Gorisanson

Reputation: 2322

You can try the following:

list1 = ['jack sparow', 'monika hardan', 'hamid tatan', 'california']

list2 = ['california', 'sparow']

newlist = [[x] for x in list1 for y in list2 if y in x]

print(newlist)

which prints

[['jack sparow'], ['california']]

UPDATE

To achieve the output [['jack sparow'], [None], [None], ['california']], you can try the following:

def is_partially_contained_in(s, t):
    for y in t:
        if y in s:
            return True
    return False


list1 = ['jack sparow', 'monika hardan', 'hamid tatan', 'california']

list2 = ['california', 'sparow']

newlist = [[x] if is_partially_contained_in(x, list2) else [None] for x in list1]

print(newlist)

which prints

[['jack sparow'], [None], [None], ['california']]

Upvotes: 2

Guy Shefer
Guy Shefer

Reputation: 88

list1 = ['jack sparow' , 'monika hardan', 'hamid tatan', 'california']

list2 = ['california']

newlist = []

for a in list2:
    for b in list1:
        if a in b:
            newlist.append(b)

print(newlist)

Upvotes: 2

Daniel Walker
Daniel Walker

Reputation: 6782

Try this:

if any(x in y for y in list1 for x in list2):
    newlist.append(list2)

Upvotes: 1

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