PROSAGA码农传奇-ECN-计算2D Numpy数组中数字对的频率的最有效方法

<div class =“post-text”itemprop =“text”>
  <P>
    我有一个想法，代码如下。
我的代码的最大缺点是它随着列的增加而运行得非常慢，并且比@Paul Panzer的代码慢。我向Paul Panzer道歉。
  </p>
  <P>
    如果你想更快，只需忽略num_to_items的功能。因为
     <code>
      (1, 1)
    </code>
     等于
     <code>
      1*2**20 + 1
    </code>
    。
  </p>
   <pre>
    <code>
      import numpy as np
from random import choice
from itertools import izip
from scipy.sparse import csr_matrix, csc_matrix
from scipy import sparse as sp

c_10 = np.array([[choice(range(1, 10)) for _ in range(3)] for _ in range(1000)])
c_1000 = np.array([[choice(range(1, 1000)) for _ in range(3)] for _ in range(1000)])

def _bit_to_items(num):
    return (num >> 20, num & 0b1111111111111111111)

def unique_bit_shit(c):
    cc = c << 20 # suppose that: 2**20 > max(c)

dialog_mtx_1 = np.array([[1, 0, 0],
                             [1, 0, 0],
                             [0, 1, 0]])

dialog_mtx_2 = np.array([[0, 1, 0],
                             [0, 0, 1],
                             [0, 0, 1]])

dialog_mtx_1 = dialog_mtx_1.T
    dialog_mtx_2 = dialog_mtx_2.T

pairs = cc.dot(dialog_mtx_1) + c.dot(dialog_mtx_2)
    pairs_num, count = np.unique(pairs, return_counts=True)
    return [(_bit_to_items(num), v) for num, v in izip(pairs_num, count)]

def _dot_to_items(num):
    # 2**20 is 1048576
    return (num / 1048576, num % 1048576)

def unique_dot(c):
    dialog_mtx_3 = np.array([[2**20, 2**20, 0],
                             [1,     0,     2**20],
                             [0,     1,     1]])

pairs = c.dot(dialog_mtx_3)

pairs_num, count = np.unique(pairs, return_counts=True)
    return [(_dot_to_items(num), v) for num, v in izip(pairs_num, count)]

</code>
  </pre>
</DIV>