Question

我正在重写一些旧代码，以使用新型的参数表示形式。旧版本使用矩阵表示参数，而新版本使用List和arma::fcube。我只有在从另一个函数中多次调用该函数时才能观察到性能下降：

旧功能ConditionalProbs比新功能conditional_probabilities慢2倍。另一方面，multiple_times_old（多次调用ConditionalProbs）比multiple_times_new快4倍，即使此功能之间唯一的区别就是多次调用。

我为长代码表示歉意：这是我的旧代码（请注意，我使用的是NumericMatrix，但确实通过在新代码上更改IntegerMatrix来提高了速度）

// [[Rcpp::depends(RcppArmadillo)]]
// [[Rcpp::export]]
NumericVector ConditionalProbs(NumericMatrix X, IntegerVector position,  int C, NumericMatrix cMat, NumericMatrix vMat, NumericVector V) {

  int n = X.nrow(); int m = X.ncol();
  int n_neighbors = cMat.nrow();
  int x = position[0] -1; int y = position[1] -1;
  int neix, neiy;
  NumericVector p(C + 1);
  IntegerVector vals = seq_len(C+1) - 1;
  double U;
  int dif;

  for(int value = 0; value <= C; value++){
    U = V[value];
    for(int ne=0; ne < n_neighbors; ne++){
      neix = x + cMat(ne,0); neiy = y + cMat(ne,1);
      if(neix < n && neix >=0 && neiy < m && neiy>=0){
        dif = X(neix,neiy) - vals[value] ;
        U = U + vMat(ne,dif+C);
      }
    }
    p[value] = exp(U);
  }
  p = p/sum(p);
  return(p);
}


// [[Rcpp::depends(RcppArmadillo)]]
// [[Rcpp::export]]
NumericMatrix multiple_times_old(NumericMatrix X, NumericMatrix cMat, NumericMatrix vMat, NumericVector V, int C, int n_times){
  NumericVector probability;
  int N = X.nrow(); int M = X.ncol();
  int x,y;
  IntegerVector position(2);
  for(int i = 0; i < n_times; i++){
    x = 2;
    y = 2;
    position[0] = x; position[1] = y;
    probability = ConditionalProbs(X, position, C, cMat, vMat, V);
  }
  return(X);
}

现在的新版本是：

// [[Rcpp::depends(RcppArmadillo)]]
// [[Rcpp::export]]
NumericVector conditional_probabilities(IntegerMatrix X, IntegerVector position, List R, arma::fcube theta){
  int n_neighbors = theta.n_rows;
  int C = theta.n_cols - 1;
  int x = position[0] - 1; int y = position[1] - 1;
  int N = X.nrow(); int M = X.ncol();

  IntegerVector this_pos;
  NumericVector probs(C+1);
  float this_prob;
  int dx, dy;

  for(int value = 0; value <= C; value++){
    this_prob = 0;
    for(int i = 0; i < n_neighbors; i++){
      this_pos = as<IntegerVector>(R[i]);
      dx = this_pos[0]; dy = this_pos[1];
      if(0 <= x+dx && x+dx < N && 0 <= y+dy && y+dy < M){
        this_prob = this_prob + theta(i, value, X(x+dx, y+dy));}
      //if(0 <= x-dx && x-dx < N && 0 <= y-dy && y-dy < M){
      // this_prob = this_prob + theta(i, X(x-dx, y-dy), value);}
    }
    probs[value] = exp(this_prob);
  }
  return(probs/sum(probs));
}

// [[Rcpp::depends(RcppArmadillo)]]
// [[Rcpp::export]]
IntegerMatrix multiple_times_new(IntegerMatrix X, List R, arma::fcube theta, int n_times){
  NumericVector probability;
  int N = X.nrow(); int M = X.ncol();
  int x,y;
  IntegerVector position(2);
  for(int i = 0; i < n_times; i++){
    x = 2;
    y = 2;
    position[0] = x; position[1] = y;
    probability = conditional_probabilities(X, position, R, theta);
  }
  return(X);
}

我编写了基准测试功能：

library(Rcpp)
sourceCpp("bench.cpp") #All the c++ functions are on this file

# parameters for old functions
cMat <- matrix(c(1,0,0,1), nrow = 2, byrow = TRUE)
vMat <- matrix(c(-1,-1,-1,-1,1,-1,-1,-1,-1,
                 -1,-1,-1,-1,1,-1,-1,-1,-1), nrow = 2, byrow = TRUE)
vMat <- rbind(vMat, vMat) + 0.0
cMat <- rbind(cMat, -cMat)
V <- rep(0.0,5)

# parameters for new functions
R <- list(c(1L,0L), c(0L,1L), c(-1L,0L), c(0L, -1L))
theta <- array(0, c(4,5,5))
theta[1,,] <- theta[2,,] <- theta[3,,] <- theta[4,,] <- diag(rep(2,5)) - 1.0

X <- matrix(sample(0:4,64*64, replace = TRUE), nrow = 64)

library(rbenchmark)

benchmark(ConditionalProbs = ConditionalProbs(X, c(30,30), 4, cMat, vMat, V),
          conditional_probabilities = conditional_probabilities(X, c(30,30), R, theta), 
          replications = 50000)[ ,c("test", "relative","elapsed", "replications")]

                       test relative elapsed replications
2 conditional_probabilities    1.000   0.314        50000
1          ConditionalProbs    1.538   0.483        50000

benchmark(multiple_times_old = multiple_times_old(X, cMat2, vMat2, V, 4, 50000),
          multiple_times_new = multiple_times_new(X, R, theta, 50000),
          replications = 10)[ ,c("test", "relative", "elapsed", "replications")]

                test relative elapsed replications
2 multiple_times_new    4.359   0.959           10
1 multiple_times_old    1.000   0.220           10

如果第一个调用的函数比第二个调用的函数快，为什么multiple_times_new比multiple_times_old慢？

从另一个函数中调用时，RCPP函数性能降低

0 个答案: