Lambda with if condition in map()

Question

I have this bit of a practice code:

even = []
odd = []

for x in range(1000):
    if x % 2 != 0:
        odd.append(x)
    else:
        even.append(x)

print map(lambda x: x if str(x)[-1] == '2' else pass, even)

print even
print odd

In my mind, I should get in the end full list of odd numbers in 0 - 999 range and a list of even numbers from the same range which do not end with "2". In practice however, I keep on getting syntax error pointing to the "pass" in lambda expression.

What am I doing wrong here?

Cheers, Greem

You're doing extra work by making a string. Try x % 10 == 2 — OneCricketeer
– OneCricketeer, Commented Apr 9, 2017 at 3:02

Ry- · Accepted Answer · 2017-04-09 02:56:27Z

2

pass is a statement, but inline if, being an operator, needs its operands to be expressions. map can’t actually remove elements from the sequence, but filter (returns a new list with only the values for which the function returns True) can:

print filter(lambda x: str(x)[-1] == '2', even)

answered Apr 9, 2017 at 2:56

community wiki

Ry-

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Comments

cjbrog · Accepted Answer · 2017-04-09 03:01:37Z

2

If you're like me and don't like filters and lambda, you can accomplish this with Python list comprehension:

print [x for x in even if str(x)[-1] == '2']

answered Apr 9, 2017 at 3:01

cjbrog

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Comments

Banee Ishaque K · Accepted Answer · 2017-04-09 12:52:24Z

1

even = []
odd = []

for x in range(1000):
    if x % 2 != 0:
        odd.append(x)
    else:
        even.append(x)

print (filter(lambda x: str(x)[-1] == '2', even))

print (even)
print (odd)

Will work on python 3 too..

answered Apr 9, 2017 at 12:52

Banee Ishaque K

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