There is a list with numbers. How to find the number as close as possible to the middle of the list?

list_nums = [5, 7, 10, 13, 17]

In this case, this number is 11, but since it is not there the closest to it is 10.

  • The first time not correctly described the task. Average arithmetic I knew how to calculate. - je_inc

2 answers 2

 >>> import statistics >>> avg = statistics.mean(list_nums) # среднее >>> min(list_nums, key=lambda num: abs(num - avg)) # приближённое к среднему 10

lambda num: abs(num - avg) is a function that expresses the concept of "close to average" - the smaller this value is, the closer to the average (at zero it coincides with the average). You can give the name of this function:

 def distance(a, b): return abs(a - b)

If we imagine that the points lie on the line, then the distance returns the Euclidean metric - the length of the segment between the points. That is why the abs() function is used, which returns the absolute value : abs(5 - 11) == abs(17 - 11) == 6 in this case, the numbers 5 and 17 at one distance from the average 11 .

 min(list_nums, key=lambda num: distance(num, avg))

min() returns the number from the list_nums list that is closest to avg by the metric defined by distance() .

statistics.mean(L) is more accurate than the naive sum(L)/len(L) formula if you want floating point numbers to support:

 >>> list_nums = [1e50, 1, -1e50] * 1000 >>> sum(list_nums) / len(list_nums) 0.0 >>> statistics.mean(list_nums) 0.3333333333333333

statistics.mean() does not work this way, but for example it would be enough: math.fsum(list_nums) / len(list_nums) .

  • Class !!! But could you write a noob easier. And it seems like I understand, but not to the end. Or maybe because of abs, in which I am not fully moving - je_inc
  • @je_inc: added explanation - jfs
 In [58]: avg = sum(list_nums)/len(list_nums) In [59]: min([(x,abs(x-avg)) for x in list_nums], key=lambda x: x[1])[0] Out[59]: 10

Step by step analysis:

 In [60]: [(x,abs(x-avg)) for x in list_nums] Out[60]: [(5, 5.4), (7, 3.4000000000000004), (10, 0.40000000000000036), (13, 2.5999999999999996), (17, 6.6)]

further we find a tuple with the minimum difference:

 In [61]: min([(x,abs(x-avg)) for x in list_nums], key=lambda x: x[1]) Out[61]: (10, 0.40000000000000036)

and select the first element of the tuple:

 In [62]: min([(x,abs(x-avg)) for x in list_nums], key=lambda x: x[1])[0] Out[62]: 10

here is a very elegant version suggested by @jfs

 min(list_nums, key=lambda num: abs(num - avg))

Alternative option:

 avg = sum(list_nums)/len(list_nums) x = list_nums[0] for i in list_nums: if abs(i - avg) < abs(x - avg): x = i print(x)
  • You are cool, a separate ATP for a detailed explanation, or else a noob, in an abbreviated form , a bit hard to read the code - je_inc
  • one
    You can simplify the min() call: min(list_nums, key=lambda num: abs(num - avg)) . For float sum / len, the formula may be inaccurate. You can use statistics.mean - jfs
  • @jfs, thanks! min(list_nums, key=lambda num: abs(num - avg)) - very elegant! - MaxU