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?

Asked By: Carlton

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Answers:

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()
Answered By: Michael Mrozek

You can use a list comprehension:

[x for x in j if x >= 5]
Answered By: sepp2k

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.

Answered By: Justin Ardini

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]
Answered By: Lennart Regebro

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 the 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!!

Answered By: U12-Forward

There is another way,

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

It was tested in Python 3.

Answered By: Harsha VK

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])
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