Checking if any elements in one list are in another

Question

I’m trying to compare two lists and simply print a message if any value from the first list is in the second list.

def listCompare():
  list1 = [1, 2, 3, 4, 5]
  list2 = [5, 6, 7, 8, 9]
  if list1 in list2:
    print("Number was found")
  else:
    print("Number not in list")

In this example, I want the if to evaluate to True because 5 is in both lists. This doesn’t work, and I’m not sure of the simplest way to compare the two lists.

Asked By: h1h1

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Source

Answer 1

You could solve this many ways. One that is pretty simple to understand is to just use a loop.

def comp(list1, list2):
    for val in list1:
        if val in list2:
            return True
    return False

A more compact way you can do it is to use map and reduce:

reduce(lambda v1,v2: v1 or v2, map(lambda v: v in list2, list1))

Even better, the reduce can be replaced with any:

any(map(lambda v: v in list2, list1))

You could also use sets:

len(set(list1).intersection(list2)) > 0

Answered By: David Alber

Answer 2

There are different ways. If you just want to check if one list contains any element from the other list, you can do this..

not set(list1).isdisjoint(list2)

I believe using isdisjoint is better than intersection for Python 2.6 and above.

Answered By: vidit

Answer 3

Your original approach can work with a list comprehension:

def listCompare():
  list1 = [1, 2, 3, 4, 5]
  list2 = [5, 6, 7, 8, 9]
  if [item for item in list1 if item in list2]:
    print("Number was found")
  else:
    print("Number not in list")

Answered By: dansalmo

Answer 4

There is a built in function to compare lists:

Following is the syntax for cmp() method −

cmp(list1, list2)

#!/usr/bin/python

list1, list2 = [123, 'xyz'], [123, 'xyz']

print cmp(list1,list2)

When we run above program, it produces following result −

If the result is a tie, meaning that 0 is returned

Answered By: user1463110

Answer 5

You could change the lists to sets and then compare both sets using the & function. eg:

list1 = [1, 2, 3, 4, 5]
list2 = [5, 6, 7, 8, 9]

if set(list1) & set(list2):
    print "Number was found"
else:
    print "Number not in list"

The “&” operator gives the intersection point between the two sets. If there is an intersection, a set with the intersecting points will be returned. If there is no intersecting points then an empty set will be returned.

When you evaluate an empty set/list/dict/tuple with the “if” operator in Python the boolean False is returned.

Answered By: colins44

Answer 6

I wrote the following code in one of my projects. It basically compares each individual element of the list. Feel free to use it, if it works for your requirement.

def reachedGoal(a,b):
    if(len(a)!=len(b)):
        raise ValueError("Wrong lists provided")

    for val1 in range(0,len(a)):
        temp1=a[val1]
        temp2=b[val1]
        for val2 in range(0,len(b)):
            if(temp1[val2]!=temp2[val2]):
                return False
    return True

Answered By: Aman Madan

Answer 7

What if I want to to return the list of occurrence of the values in list1 as compared to list2 as below

list1 = [1, 2, 3, 4, 5]

list2 = [5, 6, 7, 8, 9]

I expect to 0 occurrence of 1 ,0 occurrence of 2, 0 occurrence of 3, 0 occurrence of 4, and 1 occurrence of 5 to have a new list as below,

new_list=[0,0,0,0,1]

Answered By: Francis kinyuru

Checking if any elements in one list are in another

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