如果你指的是质心，你只需要得到所有点的平均值。

x = [p[0] for p in points]
y = [p[1] for p in points]
centroid = (sum(x) / len(points), sum(y) / len(points))

如果这组点是一个大小为N x 2的numpy数组positions，则质心简单地由下式给出：

centroid = positions.mean(axis=0)

它会直接给予你一个numpy数组的两个坐标。
一般来说，numpy数组可以以矢量化的方式用于所有这些度量，与for循环相比，这种方式非常紧凑和 * 非常 * 快速。

我假设一个点是一个像（x，y）这样的元组，所以你可以使用zip来连接x和y，然后使用x和y的最小值和最大值，你可以确定中心点。

x,y=zip(*points)
center=(max(x)+min(x))/2., (max(y)+min(y))/2.

输出示例

Points in an array : [(411, 148), (304, 148), (357, 241)]
x:(411, 304, 357)
y:(148, 148, 241)
center: (357.5, 194.5)

在这种情况下，点的平均值并不是形心。一般来说，面积的中心是面积的一阶矩。因此，你必须计算定义图形形状的多边形的面积，然后计算每个轴的面积的一阶矩：sum((r_i * A_i), for i in range(N))/sum(A_i)。因此，我们可以得到位于图形轮廓上的一组点：
data.dat的内容：

x,y
0.99159,0.00467
0.97822,0.00828
0.96383,0.01237
0.94834,0.01703
0.93166,0.02231
0.91374,0.02816
0.89456,0.03443
0.87415,0.04092
0.85265,0.04755
0.83029,0.05426
0.80736,0.06099
0.78414,0.06766
0.76087,0.07423
0.73768,0.08064
0.71456,0.08687
0.69143,0.09294
0.6681,0.09886
0.64446,0.10469
0.62058,0.11041
0.59684,0.11598
0.57378,0.12127
0.55182,0.12613
0.53101,0.13048
0.51113,0.13432
0.49187,0.13766
0.47287,0.14054
0.4538,0.14301
0.43445,0.14514
0.4148,0.14694
0.39496,0.14844
0.37527,0.14964
0.3561,0.15051
0.33766,0.151
0.31999,0.15111
0.303,0.15081
0.28655,0.15011
0.27048,0.149
0.25467,0.14748
0.23907,0.14558
0.22372,0.14331
0.20869,0.14071
0.19411,0.13782
0.1801,0.1347
0.16677,0.13138
0.1542,0.12788
0.1424,0.12422
0.13136,0.12042
0.12106,0.1165
0.11142,0.11244
0.1024,0.10826
0.09391,0.10393
0.08588,0.09944
0.07824,0.09486
0.07098,0.0903
0.06412,0.08592
0.05772,0.08177
0.05182,0.07782
0.04641,0.07403
0.04142,0.07034
0.03683,0.06673
0.03258,0.0632
0.02864,0.05975
0.025,0.05637
0.02164,0.05307
0.01853,0.04985
0.01567,0.04669
0.01307,0.04357
0.01073,0.04047
0.00864,0.03735
0.00679,0.03423
0.00517,0.0311
0.00377,0.02798
0.00258,0.02487
0.00158,0.02177
0.00078,0.0187
0.00017,0.01565
-0.00025,0.01262
-0.00049,0.00962
-0.00055,0.00663
-0.00042,0.00367
-9.00E-05,0.00073
0.00043,-0.00218
0.00114,-0.00508
0.00206,-0.00793
0.00316,-0.01073
0.00447,-0.01346
0.00599,-0.0161
0.00772,-0.01865
0.00968,-0.02106
0.01188,-0.02333
0.01435,-0.02541
0.01711,-0.02728
0.02016,-0.02894
0.02351,-0.03043
0.02714,-0.03179
0.03101,-0.03309
0.03514,-0.03434
0.03955,-0.03555
0.04429,-0.03675
0.04937,-0.03795
0.05483,-0.03918
0.06069,-0.04048
0.06697,-0.0418
0.07372,-0.04312
0.081,-0.04436
0.0889,-0.0455
0.0976,-0.04653
0.10725,-0.04749
0.11801,-0.0484
0.13002,-0.04926
0.1434,-0.05007
0.15815,-0.05085
0.17419,-0.05161
0.19136,-0.05227
0.20955,-0.05274
0.22884,-0.05293
0.2494,-0.05286
0.27136,-0.05256
0.29473,-0.05206
0.3192,-0.05146
0.3442,-0.05081
0.36921,-0.05008
0.39401,-0.04924
0.41866,-0.04826
0.44342,-0.04713
0.46845,-0.04586
0.49376,-0.04447
0.51919,-0.04299
0.54445,-0.04142
0.56941,-0.03976
0.5941,-0.03796
0.61869,-0.036
0.64343,-0.03387
0.66852,-0.03162
0.69403,-0.0293
0.7199,-0.02698
0.74597,-0.02469
0.77207,-0.02242
0.79809,-0.02016
0.82397,-0.0179
0.84958,-0.01566
0.8746,-0.01347
0.8986,-0.01136
0.92114,-0.00939
0.94183,-0.00758
0.96037,-0.00596
0.97676,-0.00453
0.99124,-0.00326
1,-0.0025

代码：计算截面一阶矩

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt

data = pd.read_table('data.dat',delim_whitespace=True,skiprows=[0],names=['x','y'],index_col=False)
x = data.x.to_numpy() 
y = data.y.to_numpy()

N = range(len(data)-1)
M = np.array([(x[i]-x[i+1])*(y[i]+y[i+1])/2 for i in N]) #Area of each trapezoid
My = np.array([(x[i]+x[i+1])/2 for i in N])*M #Moment of area (area*distance to centroid) with respect to the Y axis of each trapezoid
Mx = np.array([(y[i]+y[i+1])/4 for i in N])*M #Moment of area (area*distance to centroid) with respect to the X axis of each trapezoid
X = sum(My)/sum(M)
Y = sum(Mx)/sum(M)

centroid = [X , Y]

points_ave = data.mean(axis=0)

plt.plot(data.x, data.y, 'r',marker='.',markeredgecolor='black', markersize=3)
plt.plot(*centroid, 'blue', marker='o',markeredgecolor='black', markersize=7)
plt.plot(*points_ave, 'green', marker='o',markeredgecolor='black', markersize=7)
plt.axis('equal')
plt.xlim((-0.05, 1.05))
plt.legend(['GOE 383 AIRFOIL','Centroid','Average of points'])

在下图中，您可以非常清楚地看到非均匀点采样是如何使结果偏斜的。点的平均值仅对点质量或集中属性有用。