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 or equal to 5?

Answer 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()

Answer 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.

Answer 3

您可以使用列表理解：

[x for x in j if x >= 5]

Answer 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!!

Answer 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])

Answer 6

Since your desired output is sorted, you also need to sort it:

>>> j=[4, 5, 6, 7, 1, 3, 7, 5]
>>> sorted(x for x in j if x >= 5)
[5, 5, 6, 7, 7]

Answer 7

There is another way,

j3 = j2 > 4; print(j2[j3])

tested in 3.x