Possible Duplicate:
Get difference from two lists in Python

What is a simplified way of doing this? I have been trying on my own, and I can't figure it out. list a and list b, the new list should have items that are only in list a. So:

a = apple, carrot, lemon b = pineapple, apple, tomato new_list = carrot, lemon

I tried writing code, but every time it always returns the whole list a to me.

0

5 Answers

You can write this using a list comprehension which tells us quite literally which elements need to end up in new_list:

a = ['apple', 'carrot', 'lemon'] b = ['pineapple', 'apple', 'tomato'] # This gives us: new_list = ['carrot' , 'lemon'] new_list = [fruit for fruit in a if fruit not in b]

Or, using a for loop:

new_list = [] for fruit in a: if fruit not in b: new_list.append(fruit)

As you can see these approaches are quite similar which is why Python also has list comprehensions to easily construct lists.

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You can use a set:

# Assume a, b are Python lists # Create sets of a,b setA = set(a) setB = set(b) # Get new set with elements that are only in a but not in b onlyInA = setA.difference(b)

UPDATE
As iurisilvio and mgilson pointed out, this approach only works if a and b do not contain duplicates, and if the order of the elements does not matter.

5

You may want this:

a = ["apple", "carrot", "lemon"] b = ["pineapple", "apple", "tomato"] new_list = [x for x in a if (x not in b)] print new_list

Would this work for you?

a = ["apple", "carrot", "lemon"] b = ["pineapple", "apple", "tomato"] new_list = [] for v in a: if v not in b: new_list.append(v) print new_list

Or, more concisely:

new_list = filter(lambda v: v not in b, a)
2

How about using sets (or the built in set since Sets was deprecated in 2.6)?

from sets import Set a = Set(['apple', 'carrot', 'lemon']) b = Set(['pineapple','apple','tomato']) new_set = a.difference(b) print new_set

gives the output

Set(['carrot', 'lemon'])
4