Python lambda with if but without else
Question:
I was writing some lambda functions and couldn’t figure this out. Is there a way to have something like lambda x: x if (x<3) in python? As lambda a,b: a if (a > b) else b works ok. So far lambda x: x < 3 and x or None seems to be the closest i have found.
Answers:
What’s wrong with lambda x: x if x < 3 else None?
A lambda, like any function, must have a return value.
lambda x: x if (x<3) does not work because it does not specify what to return if not x<3. By default functions return None, so you could do
lambda x: x if (x<3) else None
But perhaps what you are looking for is a list comprehension with an if condition. For example:
In [21]: data = [1, 2, 5, 10, -1]
In [22]: [x for x in data if x < 3]
Out[22]: [1, 2, -1]
I found that filter provided exactly what I was looking for in python 2:
>>> data = [1, 2, 5, 10, -1]
>>> filter(lambda x: x < 3, data)
[1, 2, -1]
The implementation is different in 2.x and 3.x: while 2.x provides a list, 3.x provides an iterator. Using a list comprehension might make for a cleaner use in 3.x:
>>> data = [1, 2, 5, 10, -1]
>>> [filter(lambda x: x < 3, data)]
[1, 2, -1]
You can always try to invoke ‘filter’ for conditional checks. Fundamentally, map() has to work on every occurrence of the iterables, so it cannot pick and choose. But filter may help narrow down the choices. For example, I create a list from 1 to 19 but want to create a tuple of squares of only even numbers.
x = list(range(1,20))
y = tuple(map(lambda n: n**2, filter(lambda n: n%2==0,x)))
print (y)
You can use ellipsis ... to fill else statement
lambda x: x if (x<3) else ...
Note it does not work with pass.
I was writing some lambda functions and couldn’t figure this out. Is there a way to have something like lambda x: x if (x<3) in python? As lambda a,b: a if (a > b) else b works ok. So far lambda x: x < 3 and x or None seems to be the closest i have found.
What’s wrong with lambda x: x if x < 3 else None?
A lambda, like any function, must have a return value.
lambda x: x if (x<3) does not work because it does not specify what to return if not x<3. By default functions return None, so you could do
lambda x: x if (x<3) else None
But perhaps what you are looking for is a list comprehension with an if condition. For example:
In [21]: data = [1, 2, 5, 10, -1]
In [22]: [x for x in data if x < 3]
Out[22]: [1, 2, -1]
I found that filter provided exactly what I was looking for in python 2:
>>> data = [1, 2, 5, 10, -1]
>>> filter(lambda x: x < 3, data)
[1, 2, -1]
The implementation is different in 2.x and 3.x: while 2.x provides a list, 3.x provides an iterator. Using a list comprehension might make for a cleaner use in 3.x:
>>> data = [1, 2, 5, 10, -1]
>>> [filter(lambda x: x < 3, data)]
[1, 2, -1]
You can always try to invoke ‘filter’ for conditional checks. Fundamentally, map() has to work on every occurrence of the iterables, so it cannot pick and choose. But filter may help narrow down the choices. For example, I create a list from 1 to 19 but want to create a tuple of squares of only even numbers.
x = list(range(1,20))
y = tuple(map(lambda n: n**2, filter(lambda n: n%2==0,x)))
print (y)
You can use ellipsis ... to fill else statement
lambda x: x if (x<3) else ...
Note it does not work with pass.