python在3 x n数组中最常见的索引对

import numpy as np

array = np.array([[   0,    0,    0, 2879, 2879, 2879],
       [  40,   40,   40, 162,  165,  168],
       [ 249,  250,  251, 195,  196,  198]])

# Zip together only the second and third rows
only_coords = zip(array[1,:], array[2,:])

from collections import Counter

Counter(only_coords).most_common()

生产:

Out[11]: 
[((40, 249), 1),
 ((165, 196), 1),
 ((162, 195), 1),
 ((168, 198), 1),
 ((40, 251), 1),
 ((40, 250), 1)]

这是一种矢量化方法-

IDs = a[1].max()+1 + a[2]
unq, idx, count = np.unique(IDs, return_index=1,return_counts=1)
out = a[1:,idx[count.argmax()]]

如果可能存在负坐标,请使用 a[1].max()-a[1].min()+1 + a[2] IDs .

In [44]: a
Out[44]: 
array([[8, 3, 6, 6, 8, 5, 1, 6, 6, 5],
       [5, 2, 1, 1, 5, 1, 5, 1, 1, 4],
       [8, 2, 3, 3, 8, 1, 7, 3, 3, 3]])

In [47]: IDs = a[1].max()+1 + a[2]

In [48]: unq, idx, count = np.unique(IDs, return_index=1,return_counts=1)

In [49]: a[1:,idx[count.argmax()]]
Out[49]: array([1, 3])

这可能看起来有点抽象,但您可以尝试将每个坐标保存为一个数字,例如[2,1]=2.1。并将数据放入这些坐标的列表中。例如,[1,1,2]的第二行和[2,2,1]的第三行将是[1.2,1.2,2.1],然后您可以使用代码:

from collections import Counter
list1=[1.2,1.2,2.1]
data = Counter(list1)
print (data.most_common(1))  # Returns the highest occurring item

打印最常见的数字,以及它出现的次数,然后如果需要在代码中使用它,您可以简单地将数字转换回坐标。

import numpy as np
import collections

a = np.array([[0, 1, 2, 3], [10, 10, 30 ,40], [25, 25, 10, 50]])
# You don't care about time
b = np.transpose(a[1:])

# convert list items to tuples
c = map(lambda v:tuple(v), b)
collections.Counter(c)

Counter({(10, 25): 2, (30, 10): 1, (40, 50): 1})