Return list of items in list greater than some value
Solution 1:
You can use a list comprehension to filter it:
j2 = [i for i in j if i >= 5]
If you actually want it sorted like your example was, you can use sorted
:
j2 = sorted(i for i in j if i >= 5)
or call sort
on the final list:
j2 = [i for i in j if i >= 5]
j2.sort()
Solution 2:
A list comprehension is a simple approach:
j2 = [x for x in j if x >= 5]
Alternately, you can use filter
for the exact same result:
j2 = filter(lambda x: x >= 5, j)
Note that the original list j
is unmodified.
Solution 3:
You can use a list comprehension:
[x for x in j if x >= 5]
Solution 4:
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
<filter object at 0x000000955D16DC18>
>>> 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!!
Solution 5:
In case you are considering using the numpy
module, it makes this task very simple, as requested:
import numpy as np
j = np.array([4, 5, 6, 7, 1, 3, 7, 5])
j2 = np.sort(j[j >= 5])
The code inside of the brackets, j >= 5
, produces a list of True
or False
values, which then serve as indices to select the desired values in j
. Finally, we sort with the sort
function built into numpy
.
Tested result (a numpy
array):
array([5, 5, 6, 7, 7])