List comprehension with if statement

Question

I want to compare 2 iterables and print the items which appear in both iterables.

>>> a = ('q', 'r')
>>> b = ('q')


# Iterate over a. If y not in b, print y.
# I want to see ['r'] printed.
>>> print([ y if y not in b for y in a])
                              ^

But it gives me a invalid syntax error where the ^ has been placed.
What is wrong about this lamba function?

Asked By: OrangeTux

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Source

Answer 1

This is not a lambda function. It is a list comprehension.

Just change the order:

[ y for y in a if y not in b]

Answered By: eumiro

Answer 2

You put the if at the end:

[y for y in a if y not in b]

List comprehensions are written in the same order as their nested full-specified counterparts, essentially the above statement translates to:

outputlist = []
for y in a:
    if y not in b:
        outputlist.append(y)

Your version tried to do this instead:

outputlist = []
if y not in b:
    for y in a:
        outputlist.append(y)

but a list comprehension must start with at least one outer loop.

Answered By: Martijn Pieters

Answer 3

You got the order wrong. The if should be after the for (unless it is in an if-else ternary operator)

[y for y in a if y not in b]

This would work however:

[y if y not in b else other_value for y in a]

Answered By: Volatility

Answer 4

list comprehension formula:

[<value_when_condition_true> if <condition> else <value_when_condition_false> for value in list_name]

thus you can do it like this:

[y for y in a if y not in b]

Only for demonstration purpose :
[y if y not in b else False for y in a ]

Answered By: Vishvajit Pathak

Answer 5

If you use sufficiently big list not in b clause will do a linear search for each of the item in a. Why not use set? Set takes iterable as parameter to create a new set object.

>>> a = ["a", "b", "c", "d", "e"]
>>> b = ["c", "d", "f", "g"]
>>> set(a).intersection(set(b))
{'c', 'd'}

Answered By: artdanil

List comprehension with if statement

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