Scipy - 当矩阵稀疏时避免手动循环的更好方法

Question

Logistic回归的目标函数是

，渐变是

其中w是一个scipy的csr稀疏矩阵，其中d-by为1。

我的问题是，当我有一个scipy的csr稀疏矩阵和一个numpy数组时，分别是X_train和y_train。 （X_train的每一行都是x_i，y_train的每个元素都是y_i） 有没有更好的方法来计算梯度而不使用manully for loop？

有关详细信息，请执行large scale logistic regression。因此，表现很重要。

感谢。

更新5/19（添加我当前的代码）

感谢@ Jaime的提醒，这是我的代码。我基本上想看看是否有更好的方法来实现gradient(X, y, w)。

import numpy as np
import scipy as sp
from sklearn import datasets
from numpy.linalg import norm
from scipy import sparse

eta = 0.01
xi  = 0.1
C   = 1

X_train, y_train = datasets.load_svmlight_file('lr/datasets/a9a')
X_test, y_test   = datasets.load_svmlight_file('lr/datasets/a9a.t', n_features=X_train.shape[1])

def gradient(X, y, w):
  # w should be a col vector
  summation = w
  for i in range(X.shape[0]):
    exp_i = np.exp( y[i] *  X.getrow(i).dot(w)[0, 0] )
    summation = summation - (y[i] / (1 + exp_i)) * X.getrow(i).T

  return summation

def hes_mul(X, D, s):
  # w and s should be a col vector
  # should return a col vector
  return s + C * X.T.dot( D.dot( X.dot(s) ) )


def cg(X, y, w):
  # gradF is col vector, so all of these are col vectors
  gradF = gradient(X, y, w)
  s = sparse.csr_matrix( np.zeros(X_train.shape[1]) ).T
  r = -1 * gradF
  d = r

  D = []
  for i in range(X.shape[0]):
    exp_i = np.exp( (-1) * y[i] * w.T.dot(X.getrow(i).T)[0, 0] )
    D.append(exp_i / ((1 + exp_i) ** 2))
  D = sparse.diags(D, 0)

  while True:

    r_norm = np.sqrt((r.data ** 2).sum())
    print r_norm
    print np.sqrt((gradF.data ** 2).sum())

    if r_norm <= xi * np.sqrt((gradF.data ** 2).sum()):
      return s

    hes_mul_d = hes_mul(X, D, d)

    alpha = (r_norm ** 2) / d.T.dot( hes_mul_d )[0, 0]

    s = s + alpha * d

    r = r - alpha * hes_mul_d

    beta = (r.data ** 2).sum() / (r_norm ** 2)

    d = r + beta * d


w = sparse.csr_matrix( np.zeros(X_train.shape[1]) ).T
s = cg(X_train, y_train, w)