Return list of items in list greater than some value

Question

I have the following list

j=[4,5,6,7,1,3,7,5]

What\'s the simplest way to return [5,5,6,7,7] being the elements in j greater o

Answer 1

Use filter (short version without doing a function with lambda, using __le__):

j2 = filter((5).__le__, j)

Example (python 3):

>>> j=[4,5,6,7,1,3,7,5]
>>> j2 = filter((5).__le__, j)
>>> j2

>>> list(j2)
[5, 6, 7, 7, 5]
>>> 

Example (python 2):

>>> j=[4,5,6,7,1,3,7,5]
>>> j2 = filter((5).__le__, j)
>>> j2
[5, 6, 7, 7, 5]
>>>

Use __le__ i recommend this, it's very easy, __le__ is your friend

If want to sort it to desired output (both versions):

>>> j=[4,5,6,7,1,3,7,5]
>>> j2 = filter((5).__le__, j)
>>> sorted(j2)
[5, 5, 6, 7, 7]
>>> 

Use sorted

Timings:

>>> from timeit import timeit
>>> timeit(lambda: [i for i in j if i >= 5]) # Michael Mrozek
1.4558496298222325
>>> timeit(lambda: filter(lambda x: x >= 5, j)) # Justin Ardini
0.693048732089828
>>> timeit(lambda: filter((5).__le__, j)) # Mine
0.714461565831428
>>> 

So Justin wins!!

With number=1:

>>> from timeit import timeit
>>> timeit(lambda: [i for i in j if i >= 5],number=1) # Michael Mrozek
1.642193421957927e-05
>>> timeit(lambda: filter(lambda x: x >= 5, j),number=1) # Justin Ardini
3.421236300482633e-06
>>> timeit(lambda: filter((5).__le__, j),number=1) # Mine
1.8474676011237534e-05
>>> 

So Michael wins!!

>>> from timeit import timeit
>>> timeit(lambda: [i for i in j if i >= 5],number=10) # Michael Mrozek
4.721306089550126e-05
>>> timeit(lambda: filter(lambda x: x >= 5, j),number=10) # Justin Ardini
1.0947956184281793e-05
>>> timeit(lambda: filter((5).__le__, j),number=10) # Mine
1.5053439710754901e-05
>>> 

So Justin wins again!!