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?
b = ('q')
doesn't create a tuple. Tuples with one element need an explicit ,
, that is b = ('q',)
tuples
into iterables
.
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]
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.
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 ]
else
in a list comprehension, not where you put one at least. Don't confuse a list comprehension (filtering), with a conditional expression (which must have a value, making the else expression mandatory).
else
be used in list comprehension though as shown in the code.
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]
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'}
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