计算两个多维数组之间的相关系数

2 回答

• 28

  Correlation (default 'valid' case) between two 2D arrays:

  您可以像这样使用矩阵乘法np.dot -

  out = np.dot(arr_one,arr_two.T)
  

  与两个输入数组的每个成对行组合（row1，row2）之间的默认 "valid" 情况的相关性将对应于每个（row1，row2）位置处的乘法结果 .

  ---

  Row-wise Correlation Coefficient calculation for two 2D arrays:

  def corr2_coeff(A,B):
      # Rowwise mean of input arrays & subtract from input arrays themeselves
      A_mA = A - A.mean(1)[:,None]
      B_mB = B - B.mean(1)[:,None]
  
      # Sum of squares across rows
      ssA = (A_mA**2).sum(1);
      ssB = (B_mB**2).sum(1);
  
      # Finally get corr coeff
      return np.dot(A_mA,B_mB.T)/np.sqrt(np.dot(ssA[:,None],ssB[None]))
  

  这是基于这个解决方案How to apply corr2 functions in Multidimentional arrays in MATLAB

  Benchmarking

  本节将运行时性能与other answer.中列出的 generate_correlation_map ＆loopy pearsonr 方法的建议方法进行比较（取自函数 test_generate_correlation_map() ，其末尾没有值正确性验证码） . 请注意，建议方法的时间还包括在开始时检查两个输入数组中是否有相同数量的列，这也是在另一个答案中完成的 . 下面列出了运行时 .

  情况1：

  In [106]: A = np.random.rand(1000,100)
  
  In [107]: B = np.random.rand(1000,100)
  
  In [108]: %timeit corr2_coeff(A,B)
  100 loops, best of 3: 15 ms per loop
  
  In [109]: %timeit generate_correlation_map(A, B)
  100 loops, best of 3: 19.6 ms per loop
  

  案例＃2：

  In [110]: A = np.random.rand(5000,100)
  
  In [111]: B = np.random.rand(5000,100)
  
  In [112]: %timeit corr2_coeff(A,B)
  1 loops, best of 3: 368 ms per loop
  
  In [113]: %timeit generate_correlation_map(A, B)
  1 loops, best of 3: 493 ms per loop
  

  案例＃3：

  In [114]: A = np.random.rand(10000,10)
  
  In [115]: B = np.random.rand(10000,10)
  
  In [116]: %timeit corr2_coeff(A,B)
  1 loops, best of 3: 1.29 s per loop
  
  In [117]: %timeit generate_correlation_map(A, B)
  1 loops, best of 3: 1.83 s per loop
  

  另一个循环 pearsonr based 方法似乎太慢，但这里是一个小数据的运行时 -

  In [118]: A = np.random.rand(1000,100)
  
  In [119]: B = np.random.rand(1000,100)
  
  In [120]: %timeit corr2_coeff(A,B)
  100 loops, best of 3: 15.3 ms per loop
  
  In [121]: %timeit generate_correlation_map(A, B)
  100 loops, best of 3: 19.7 ms per loop
  
  In [122]: %timeit pearsonr_based(A,B)
  1 loops, best of 3: 33 s per loop
  

  回复于 2024-04-24T05:24:53+08:00
• 5

  @Divakar为计算未缩放的相关性提供了一个很好的选择，这是我最初要求的 .

  为了计算相关系数，需要更多：

  import numpy as np
  
  def generate_correlation_map(x, y):
      """Correlate each n with each m.
  
      Parameters
      ----------
      x : np.array
        Shape N X T.
  
      y : np.array
        Shape M X T.
  
      Returns
      -------
      np.array
        N X M array in which each element is a correlation coefficient.
  
      """
      mu_x = x.mean(1)
      mu_y = y.mean(1)
      n = x.shape[1]
      if n != y.shape[1]:
          raise ValueError('x and y must ' +
                           'have the same number of timepoints.')
      s_x = x.std(1, ddof=n - 1)
      s_y = y.std(1, ddof=n - 1)
      cov = np.dot(x,
                   y.T) - n * np.dot(mu_x[:, np.newaxis],
                                    mu_y[np.newaxis, :])
      return cov / np.dot(s_x[:, np.newaxis], s_y[np.newaxis, :])
  

  这是对此函数的测试，该函数通过：

  from scipy.stats import pearsonr
  
  def test_generate_correlation_map():
      x = np.random.rand(10, 10)
      y = np.random.rand(20, 10)
      desired = np.empty((10, 20))
      for n in range(x.shape[0]):
          for m in range(y.shape[0]):
              desired[n, m] = pearsonr(x[n, :], y[m, :])[0]
      actual = generate_correlation_map(x, y)
      np.testing.assert_array_almost_equal(actual, desired)
  

  回复于 2024-04-24T05:24:53+08:00