finding the element in a list closest to the mean of elements in python?-CodePudding

This is my array a= [5, 25, 50, 100, 250, 500] . The mean value of a is 155 (i calculated using sum(a)/len(a)) but i have to store 100 in a variable instead of 155.

Is there any easy way to solve this problem.

CodePudding user response：

IIUC, use numpy.argmin to find the the index of the value closest to the mean by computing the absolute difference to the mean:

a = np.array([5, 25, 50, 100, 250, 500])

out = a[np.argmin(np.abs(a-a.mean()))]

output: 100

CodePudding user response：

If you want to keep it pure python, you can use a custom key for sorted to find the lowest difference element:

a= [5, 25, 50, 100, 250, 500]
a_mean = sum(a)/len(a)
out = sorted(a, key=lambda val:abs(val-a_mean))[0]
# 100

CodePudding user response：

If we want to go with pure python, I would write a function like this:

from typing import List
numb_list = [1, 4, 10, 20, 55, 102, 77, 89]


def find_closest_to_mean(num_list: List[int])->int:
    mean = sum(numb_list)/len(num_list)
    distance_list = [abs(mean - num) for num in numb_list]
    return num_list[distance_list.index(min(distance_list))]


print(find_closest_to_mean(numb_list))

[Out]

mean = 44.75
closest number = 55

here I create a function called find_closest_to_mean that expects a num_list argument that is a list of integers. It then first calculates the mean of the list, creates a distance_list in which each element corresponds to the distance of the num_list in that position with the mean (as an absolute value). lastly it returns an integer from the num_list that has the least distance to the mean.