Question

我有数组np.array二维数组

[[8, 12, 5, 2], [12,15, 6,10], [15, 8, 12, 5], [12,15,8,6]]

我想创建另一个二维数组，（数组中的每个数字，重复的次数，位置）

(2,1,[1,4]), (5,2,[1,3],[3,4]) ,(6,2,[2,3],[4,4]) , (8,3,[1,1],[3,1],[4,3]) (12,4,[1,2],[2,1],[3,3],[4,1]) ,(15,3,[2,2],[3,1],[4,2])
I'd like to generate comparisons between rows and columns.

进行解释（以15号为例）

重复：3

位置：[2,2]，[3,1]，[4,2]

Answer 1

这是使用np.unqiue和np.where的一种方法，请注意numpy array中的索引是从0开始而不是1

x,y=np.unique(a.ravel(), return_counts=True)
l=[]
for v,c in zip(x,y):
    l.append((v,c,np.column_stack(np.where(a==v)).tolist()))


l
Out[344]: 
[(2, 1, [[0, 3]]),
 (5, 2, [[0, 2], [2, 3]]),
 (6, 2, [[1, 2], [3, 3]]),
 (8, 3, [[0, 0], [2, 1], [3, 2]]),
 (10, 1, [[1, 3]]),
 (12, 4, [[0, 1], [1, 0], [2, 2], [3, 0]]),
 (15, 3, [[1, 1], [2, 0], [3, 1]])]

Answer 2

使用帖子Most efficient way to sort an array into bins specified by an index array?中的代码作为模块stb，我们可以做到

import numpy as  np
from stb import sort_to_bins_sparse as sort_to_bins
from pprint import pprint

X = np.array([[8, 12, 5, 2], [12,15, 6,10], [15, 8, 12, 5], [12,15,8,6]])

unq, inv, cnt = np.unique(X, return_inverse=True, return_counts=True)
sidx = sort_to_bins(inv, np.arange(X.size))
# or (slower but avoids dependency on stb module)
# sidx = np.argsort(inv, kind='stable')

pprint(list(zip(unq, cnt, np.split(np.transpose(np.unravel_index(sidx, X.shape)) + 1, cnt[:-1].cumsum()))))[(2, 1, array([[1, 4]])),
#  (5, 2, array([[1, 3],
#        [3, 4]])),
#  (6, 2, array([[2, 3],
#        [4, 4]])),
#  (8, 3, array([[1, 1],
#        [3, 2],
#        [4, 3]])),
#  (10, 1, array([[2, 4]])),
#  (12, 4, array([[1, 2],
#        [2, 1],
#        [3, 3],
#        [4, 1]])),
#  (15, 3, array([[2, 2],
#        [3, 1],
#        [4, 2]]))]

python搜索并查找指定数字位置在数组2d中

2 个答案: