Return list of items in list greater than some value

I have the following list

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

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What's the simplest way to return [5,5,6,7,7] being the elements in j greater or equal to 5?

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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()

---

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.

---

You can use a list comprehension:

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

---

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!!

---

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])

---

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]

---

There is another way,

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

It was tested in Python 3.