import numpy as np import itertools import timeit
def combcol(myarr): ndims = myarr.shape[0] solutions = [] for idx1, idx2, idx3, idx4, idx5, idx6 in itertools.combinations(np.arange(ndims), 6): c1, c2, c3, c4, c5, c6 = myarr[idx1,1], myarr[idx2,2], myarr[idx3,1], myarr[idx4,2], myarr[idx5,1], myarr[idx6,2] www.jshk.com.cn/mb/reg.asp?… if c1-c2>0 and c2-c3<0 and c3-c4>0 and c4-c5<0 and c5-c6>0 : solutions.append(((idx1, idx2, idx3, idx4, idx5, idx6),(c1, c2, c3, c4, c5, c6))) return solutions
## 2、解决方案
**方法1：使用 NumPy**
可以使用 NumPy 的广播机制来加速计算。具体方法是将 `myarr[:,1]` 和 `myarr[:,2]` 分别转换为一维数组，然后使用 NumPy 的比较运算符进行比较。这样可以避免使用循环，从而提高计算速度。
def combcol2(myarr): ndims = myarr.shape[0] myarr1 = myarr[:,1].tolist() myarr2 = myarr[:,2].tolist() solutions = [] for idx1, idx2, idx3, idx4, idx5, idx6 in itertools.combinations(range(ndims), 6): if myarr1[idx1] > myarr2[idx2] < myarr1[idx3] > myarr2[idx4] < myarr1[idx5] > myarr2[idx6]: solutions.append(((idx1, idx2, idx3, idx4, idx5, idx6),(myarr1[idx1], myarr2[idx2], myarr1[idx3], myarr2[idx4], myarr1[idx5], myarr2[idx6]))) return solutions
**方法2：使用 Cython**
可以使用 Cython 将 Python 代码编译成 C 代码，从而提高执行速度。Cython 是一种将 Python 代码编译为 C 代码的编程语言。它允许在 Python 中使用 C 语言的语法和数据类型，从而可以大大提高 Python 代码的执行速度。
import cython @cython.cfunc def combcol3(myarr): ndims = myarr.shape[0] myarr1 = myarr[:,1].tolist() myarr2 = myarr[:,2].tolist() solutions = [] for idx1, idx2, idx3, idx4, idx5, idx6 in itertools.combinations(range(ndims), 6): if myarr1[idx1] > myarr2[idx2] < myarr1[idx3] > myarr2[idx4] < myarr1[idx5] > myarr2[idx6]: solutions.append(((idx1, idx2, idx3, idx4, idx5, idx6),(myarr1[idx1], myarr2[idx2], myarr1[idx3], myarr2[idx4], myarr1[idx5], myarr2[idx6]))) return solutions
**方法3：使用并行计算**
可以使用并行计算来加快计算速度。并行计算是指将一个任务分解成多个子任务，然后在多个处理器上同时执行这些子任务。这样可以大大缩短计算时间。
from joblib import Parallel, delayed
def combcol4(myarr): ndims = myarr.shape[0] myarr1 = myarr[:,1].tolist() myarr2 = myarr[:,2].tolist() solutions = [] for idx1, idx2, idx3, idx4, idx5, idx6 in itertools.combinations(range(ndims), 6): if myarr1[idx1] > myarr2[idx2] < myarr1[idx3] > myarr2[idx4] < myarr1[idx5] > myarr2[idx6]: solutions.append(((idx1, idx2, idx3, idx4, idx5, idx6),(myarr1[idx1], myarr2[idx2], myarr1[idx3], myarr2[idx4], myarr1[idx5], myarr2[idx6]))) return solutions
if name == "main": Parallel(n_jobs=-1)(delayed(combcol4)(myarr) for myarr in X)
