Select n values from a list, spanning across the range

Question

I have a list of numbers and I would like to select n values evenly distributed across the list.

For example:

vals = list(range(10))
select_n(vals, 4)
select_n(vals, 5)

should give

[0, 3, 6, 9]
[0, 2, 5, 7, 9]

My current hack is to iterate as such:

[vals[round((len(vals) - 1)/(n-1) * i)] for i in range(n)]

Is there a Python or NumPy function to do this? If not, is there a more efficient way to write this?

You can translate your hack to numpy using `vals[((len(vals)-1)*np.arange(4)/(4-1)).astype(int)]` assuming `vals` is a numpy array. — user545424, Jun 19 '19 at 21:59
@wwii, that would not choose the end for n=5. Perhaps I should have added another example. — Jonathon Vandezande, Jun 19 '19 at 22:35
Possible duplicate of [(Nearly) Evenly select items from a list](https://stackoverflow.com/questions/46494029/nearly-evenly-select-items-from-a-list) — LiuXiMin, Jun 19 '19 at 22:54
`Perhaps I should have added another example` - maybe the behaviour you need when the list cannot be evenly divided by `n`, and any other specs like `must include endpoints`... — wwii, Jun 19 '19 at 23:14
@LiuXiMin, that appears to create a similar mask, but doesn't include the endpoints. — Jonathon Vandezande, Jun 19 '19 at 23:53

score 1 · Accepted Answer · answered Jun 19 '19 at 22:50

you could use np.linspace for the "heavy" lifting:

from operator import itemgetter

a = [*range(10)]

N = 5

# if tuple ok 
itemgetter(*np.linspace(0.5,len(a)-0.5,N,dtype=int))(a)
# (0, 2, 5, 7, 9)

# if must be list
[a[i] for i in np.linspace(0.5,len(a)-0.5,N,dtype=int)]
# [0, 2, 5, 7, 9]

score 0 · Answer 2 · answered Jun 19 '19 at 21:59

You can do something like this:

def select_n(vals,cnt):
    inc = int(len(vals)/cnt)
    # print(inc)
    res = [vals[i] for i in range(0,len(vals),inc)]
    # print(res)
    return res

vals = list(range(10))
# print(vals)
res = select_n(vals,4)
print(res)

Select n values from a list, spanning across the range

2 Answers2