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RCPP函数未返回所需的NumericVector

[英]Rcpp function not returning desired NumericVector

I am trying to rewrite an R function (Fourier smoothing) in Rcpp for faster computation. 我试图在Rcpp中重写R函数(傅立叶平滑),以加快计算速度。 My Rcpp function is not returning the desired values. 我的Rcpp函数未返回所需的值。

I have a vector 我有一个向量

    x = c(6262, 5862.5, 5463, 5408, 5353, 5687, 5901, 6245, 5864, 5483, 5692, 5708.5, 5054.75, 5072.375, 5090, 5462, 4939, 5248.5, 5558, 5226, 5125, 5006, 4887, 5334.5, 5782, 5501, 5524.5, 5548)

My Rcpp function 我的Rcpp功能

cppFunction("
     NumericVector smo(NumericVector x){
        int n = x.size();
        NumericVector realpart1(5);
        NumericVector imagpart1(5);
        NumericVector sm1(n);
        for (int i = 0; i<5; i++){
            double realpart = 0;
            double imagpart = 0;
            for (int j = 0; j<n; j++) {
                realpart = realpart + 0.07142857*x[j]*cos(2 * 3.142857 * (i+1-1) * (j+2)/28);
                imagpart = imagpart + 0.07142857 * x[j] * sin(2 * 3.142857 * (i+1 - 1) * (j+2) /28);
            }
            realpart1[i]=realpart;
            imagpart1[i] = imagpart;
        }

        for (int j = 0; j<n; j++){
         double sm = realpart1[0]/2;

        for (int i=0; i<5; i++){
            sm = sm + realpart1[i]*cos(2 * 3.142857 * (i+1 - 1) * (j+2) / 28) + imagpart1[i]*sin(2 * 3.142857 * (i+1-1) * (j+2) / 28);
        }
         sm1[j] = sm;

         }
        return sm1; 
}
")

Output of the function smo is coming like below 功能smo的输出如下

16804.81 16674.97 16518.58 16425.55 16453.36 16594.95 16780.77 16914.47
16922.49 16789.76 16563.30 16324.47 16147.96 16070.53 16083.19 16145.65
16210.29 16241.81 16226.19 16170.64 16099.70 16049.52 16058.45 16152.36
16328.20 16545.64 16736.58 16833.36

If I subtract value 10949.12 from the output of function(smo) I am getting the desired result like below 如果我从function(smo)的输出中减去值10949.12function(smo)得到如下所示的期望结果

Desired output 所需的输出

5855.689 5725.846 5569.459 5476.428 5504.237 5645.833 5831.647 5965.351
5973.369 5840.640 5614.181 5375.346 5198.844 5121.412 5134.069 5196.534
5261.174 5292.694 5277.066 5221.517 5150.584 5100.398 5109.330 5203.243
5379.080 5596.524 5787.462 5884.235

The value 10949.12 is the first value of NumericVector realpart1 10949.12NumericVector realpart1的第一个值

I am not able to resolve this issue as I am trying Rcpp for the first time. 我第一次尝试Rcpp时,无法解决此问题。 I have checked loops various times, up to the calculation of realpart1 and imagpart1 loop is working fine... There is some problem with the second loop but I am not able to figure out why the value 10949.12 is being added in the output. 我已经多次检查循环,直到realpart1imagpart1循环的计算都可以正常工作...第二个循环存在一些问题,但是我无法弄清楚为什么在输出中添加了10949.12值。

I will really appreciate any help in this regard. 在这方面的任何帮助,我将非常感谢。

equivalent R Code 等效的R代码

har = 4
pi = 22/7
realpart1 = c()
imagpart1 = c()
for (p in 1:(har+1)){
    realpart = 0
    imagpart = 0
    for (i in 1:length(x)){
        realpart = realpart + (2 /length(x)) * x[i] * cos(2 * pi * (p - 1) * (i+1) / length(x))
        imagpart = imagpart + (2 / length(x)) * x[i] * sin(2 * pi * (p - 1) * (i+1) / length(x))
    }
    realpart1 = c(realpart1,realpart)
    imagpart1 = c(imagpart1,imagpart)
    #print(realpart)
    #print(imagpart)
}   
sm1 = c()
for (i in 1:length(x)){

    sm = realpart1[1]/2

    for (p in 2:(har+1)){
        sm = sm + realpart1[p]*cos(2 * pi * (p - 1) * (i+1) / length(x))+ imagpart1[p]*sin(2 * pi * (p - 1) * (i+1) / length(x))
    }
    sm1 = c(sm1,sm)
}   

There is a difference in the limits of the nested for loop in the second for loop. 第二个for循环中嵌套的for循环的限制有所不同。 In R it goes from 2 to 5, while in C++ it goes from 0 to 4. It should go from 1 to 4 in C++ to be comparable with R. 在R中,它从2到5,而在C ++中,它从0到4。在C ++中,它应该从1到4,以便与R相当。

However, you can probably make the R code faster by avoiding dynamically growing vectors inside the loop. 但是,通过避免在循环内动态增加向量,可以使R代码更快。 In a for loop that is almost never necessary, since you know the size of the resulting vector beforehand and can use eg realpart <- numeric(length = har + 1) and realpart[p] <- ... . 在几乎不需要的for循环中,因为您事先知道了结果向量的大小,因此可以使用realpart <- numeric(length = har + 1)realpart[p] <- ...

However, in this case one can go even further and formulate the problem in terms of matrices and linear algebra, avoiding the (explicit) loops altogether: 但是,在这种情况下,您可以走得更远,用矩阵和线性代数来表达问题,而完全避免(显式)循环:

x <- c(6262, 5862.5, 5463, 5408, 5353, 5687, 5901, 6245, 5864, 5483, 5692, 5708.5,
       5054.75, 5072.375, 5090, 5462, 4939, 5248.5, 5558, 5226, 5125, 5006, 4887,
       5334.5, 5782, 5501, 5524.5, 5548)
fourier_smooth <- function(x, har) {
    pi <- 22 / 7 # this should be removed!
    phase <- 2 * pi * outer(seq_len(har + 1) - 1, seq_along(x) + 1) / length(x) 
    real <- 2 / length(x) * cos(phase) %*% x
    imag <- 2 / length(x) * sin(phase) %*% x
    y <- t(cos(phase)) %*% real + t(sin(phase)) %*% imag
    as.numeric(y - real[1]/2)
}
fourier_smooth(x, 4)
#>  [1] 5855.695 5725.852 5569.463 5476.432 5504.240 5645.837 5831.651
#>  [8] 5965.355 5973.373 5840.644 5614.185 5375.350 5198.848 5121.417
#> [15] 5134.073 5196.538 5261.177 5292.697 5277.070 5221.522 5150.588
#> [22] 5100.402 5109.334 5203.247 5379.084 5596.528 5787.468 5884.242

Created on 2019-08-13 by the reprex package (v0.3.0) reprex软件包 (v0.3.0)创建于2019-08-13

Note that I am including the redefinition of pi only to reproduce your desired result. 请注意,我包含pi的重新定义只是为了重现您想要的结果。 For correct results, the real value of pi should be used. 为了获得正确的结果,应使用pi的实际值。

However, it is even faster to use R's build in FFT: 但是,使用R的FFT内置甚至更快:

x <- c(6262, 5862.5, 5463, 5408, 5353, 5687, 5901, 6245, 5864, 5483, 5692, 5708.5,
       5054.75, 5072.375, 5090, 5462, 4939, 5248.5, 5558, 5226, 5125, 5006, 4887,
       5334.5, 5782, 5501, 5524.5, 5548)
fourier_smooth <- function(x, har) {
    phase <- 2 * pi * outer(seq_len(har + 1) - 1, seq_along(x) - 1) / length(x) 
    real <- 2 / length(x) * cos(phase) %*% x
    imag <- 2 / length(x) * sin(phase) %*% x
    y <- t(cos(phase)) %*% real + t(sin(phase)) %*% imag
    as.numeric(y - real[1]/2)
}

fourier_smooth2 <- function(x, har) {
    y <- fft(x, inverse = TRUE) / length(x)
    y[(har+2):(length(x)-har)] <- 0 # filter higher harmonics while keeping the symmetry for real input
    Re(fft(y)) # result is already real
}

bench::mark(fourier_smooth(x, 4), fourier_smooth2(x, 4))[1:5]
#> # A tibble: 2 x 5
#>   expression                 min   median `itr/sec` mem_alloc
#>   <bch:expr>            <bch:tm> <bch:tm>     <dbl> <bch:byt>
#> 1 fourier_smooth(x, 4)   31.66µs  34.97µs    26342.    4.13MB
#> 2 fourier_smooth2(x, 4)   4.82µs   5.49µs   152845.    3.98KB

Created on 2019-08-13 by the reprex package (v0.3.0) reprex软件包 (v0.3.0)创建于2019-08-13

  • The redefinition of pi was removed to make sure that results are equal. 删除pi的重定义以确保结果相等。
  • The filtering is a bit tricky, but I don't know of any function that's specifically tailored for real time-series. 过滤有些棘手,但我不知道专门针对实时序列量身定制的任何功能。

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