Finding element closest to mean in python list

Question

I have a list of some random numbers (integers and floats)

In[]:  list1
Out[]: [8.6, 9, 15, 20, 1]

I have found out the average/ mean value of the list:

m = reduce(lambda x, y: x + y, list1) / len(list1)

This outputs 10.72 .

Now, I have to find the closest number to avg from the list list1 .

In the above eg, from list1 the closes number to 10.72 is 9.

One way is to find the difference between the avg and each element in the list, and then look for the smallest difference and then output that element producing smallest difference, But I was looking for more crisp and efficient way, please suggest me one.

I guess : `a = np.asarray(list1)` and then : `a[np.abs(a - a.mean()).argmin()]`? — Divakar, Jul 02 '17 at 22:28
All the answers below in effect do what OP doesn't want to, but at least they do it crisply. — cs95, Jul 02 '17 at 22:29
Use an array version : `a = np.asarray(list1)` and use `a.mean()` to get that mean value and then the linked dup for the rest of the work. — Divakar, Jul 02 '17 at 22:33
If you want things to be more crisp and efficient, use `statistics.mean` (as of Python 3.4) rather than your own flavor of `reduce`. If you have to use an older version, at least use `sum` for the first part. — Thijs van Dien, Jul 02 '17 at 22:34
Use `sum` instead of implementing it yourself using `reduce` — juanpa.arrivillaga, Jul 02 '17 at 22:36
The question referred in duplicate, doesn't really answer the question about efficiency, yet it provide nice answer using syntactic sugar of numpy, while has nothing to do with complexity efficient. — Artem Barger, Jul 02 '17 at 22:52
@ArtemBarger NumPy isn't just sugar, its underlying code is in C. So, unless you consider C as sugar too, I would have to disagree and should be as efficient as one could get staying in Python/NumPy. — Divakar, Jul 02 '17 at 23:02
answer in the link hides details, it makes two passes on list, while there is solution which can do it in one pass only, with some math trick — Artem Barger, Jul 02 '17 at 23:20

score 3 · Accepted Answer · answered Jul 02 '17 at 22:30

3

You can use built-in min() function and find minimum distance from your number

min(list1, key=lambda x:abs(x-m))

answered Jul 02 '17 at 22:30

Logovskii Dmitrii

2,629
4
27
44

Finding element closest to mean in python list

1 Answers1