Python 3.6。获得所有相同X坐标的平均Y.

Question

I have a list of coordinates that looks like this:

my_list = [[1, 1], [1, 3], [1, 5], [2, 1], [2, 3]]

As we see, there are same X values for first three coordinates with different Y and same situation for another two coordiantes. I want to make new list which will look like this:

new_list = [[1, 3], [2, 2]]

where y1 = 3 = (1+3+5)/3 and y2 = 2 = (1+3)/2 . I have written my code which is below, but it works slowly.

I work with hundreds of thousands coordinates so the question is: How to make this code work faster? Is there any optimization or special open source libraty, which can speed up my code?

Thank you in advance.

x_mass = []

for m in mass:
  x_mass.append(m[0])

set_x_mass = set(x_mass) 
list_x_mass = list(set_x_mass) 

performance_points = [] 

def function(i):
  unique_x_mass = []
  for m in mass:
    if m[0] == i:
      unique_x_mass.append(m)

  summ_y = 0
  for m in unique_x_mass:
    summ_y += m[1]
  point = [float(m[0]), float(summ_y/len(unique_x_mass))] 
  performance_points.append(point)
  return performance_points

for x in list_x_mass:
  function(x)

Answer 1

Create DataFrame and aggregate mean :

L = [[1, 1], [1, 3], [1, 5], [2, 1], [2, 3]]

L1 = pd.DataFrame(L).groupby(0, as_index=False)[1].mean().values.tolist()
print (L1)
[[1, 3], [2, 2]]

Answer 2

The pandas solution offered by @jezrael is elegant but slow (like almost everything pandas). I would suggest using modules itertools and statistics :

from statistics import mean
from itertools import groupby

grouper = groupby(L, key=lambda x: x[0])
#The next line is again more elegant, but slower:
#grouper = groupby(L, key=operator.itemgetter(0))
[[x, mean(yi[1] for yi in y)] for x,y in grouper]

The result is, of course, the same. The execution time for the sample list is two orders of magnitude faster.