Question

我试图在R中实现以下卷积，但没有得到预期的结果：

$$ C _ {\ sigma} [i] = \ sum \ limits_ {k = -P} ^ P SDL _ {\ sigma} [i-k，i] \ centerdot S [i] $$

其中$ S [i] $是光谱强度的矢量（洛伦兹信号/核磁共振光谱），而$ i \ in [1，N] $其中$ N $是数据点的数量（在实际例子中），也许32K值）。这是Jacob，Deborde和Moing的等式1， Analytical Bioanalytical Chemistry （2013）405：5049-5061（DOI 10.1007 / s00216-013-6852-y）。

$ SDL _ {\ sigma} $是计算洛伦兹曲线的二阶导数的函数，我已经实现如下（基于论文中的等式2）：

SDL <- function(x, x0, sigma = 0.0005){
    if (!sigma > 0) stop("sigma must be greater than zero.")
    num <- 16 * sigma * ((12 * (x-x0)^2) - sigma^2)
    denom <- pi * ((4 * (x - x0)^2) + sigma^2)^3
    sdl <-  num/denom
    return(sdl)
    }

sigma是半峰宽，x0是洛伦兹信号的中心。

我相信SDL正常工作（因为返回的值的形状类似于经验的Savitzky-Golay二阶导数）。我的问题是实现$ C _ {\ sigma} $，我写的是：

CP <- function(S = NULL, X = NULL, method = "SDL", W = 2000, sigma = 0.0005) {
    # S is the spectrum, X is the frequencies, W is the window size (2*P in the eqn above)
    # Compute the requested 2nd derivative
    if (method == "SDL") {

        P <- floor(W/2)
        sdl <- rep(NA_real_, length(X)) # initialize a vector to store the final answer

        for(i in 1:length(X)) {
            # Shrink window if necessary at each extreme
            if ((i + P) > length(X)) P <- (length(X) - i + 1)
            if (i < P) P <- i
            # Assemble the indices corresponding to the window
            idx <- seq(i - P + 1, i + P - 1, 1)
            # Now compute the sdl
            sdl[i] <- sum(SDL(X[idx], X[i], sigma = sigma))
            P <- floor(W/2) # need to reset at the end of each iteration
            }
        }

    if (method == "SG") {
        sdl <- sgolayfilt(S, m = 2)     
        }

    # Now convolve!  There is a built-in function for this!
    cp <- convolve(S, sdl, type = "open")
    # The convolution has length 2*(length(S)) - 1 due to zero padding
    # so we need rescale back to the scale of S
    # Not sure if this is the right approach, but it doesn't affect the shape
    cp <- c(cp, 0.0)
    cp <- colMeans(matrix(cp, ncol = length(cp)/2)) # stackoverflow.com/q/32746842/633251
    return(cp)
    }

根据参考，二阶导数的计算限于约2000个数据点的窗口以节省时间。我认为这部分工作正常。它应该只产生微不足道的扭曲。

以下是整个过程和问题的演示：

require("SpecHelpers")
require("signal")
# Create a Lorentzian curve
loren <- data.frame(x0 = 0, area = 1, gamma = 0.5)
lorentz1 <- makeSpec(loren, plot = FALSE, type = "lorentz", dd = 100, x.range = c(-10, 10))
#
# Compute convolution
x <- lorentz1[1,] # Frequency values
y <- lorentz1[2,] # Intensity values
sig <- 100 * 0.0005 # per the reference
cpSDL <- CP(S = y, X = x, sigma = sig)
sdl <- sgolayfilt(y, m = 2)
cpSG <- CP(S = y, method = "SG")
#
# Plot the original data, compare to convolution product
ylabel <- "data (black), Conv. Prod. SDL (blue), Conv. Prod. SG (red)"
plot(x, y, type = "l", ylab = ylabel, ylim = c(-0.75, 0.75))
lines(x, cpSG*100, col = "red")
lines(x, cpSDL/2e5, col = "blue")

正如您所看到的，CP使用SDL（蓝色）的卷积产品与CP使用SG方法的卷积积相似（红色，这是正确的，除了比例）。我希望使用SDL方法的结果应具有相似的形状，但规模不同。

如果你到目前为止一直困扰我，a）谢谢，b）你能看出错误吗？毫无疑问，我有一个根本的误解。

Answer 1

您正在执行的手动卷积存在一些问题。如果您查看维基百科页面上为“Savitzky-Golay过滤器”here定义的卷积函数，您会看到总和中的y[j+i]项与S[i]项冲突在您引用的等式中。我相信您引用的等式可能不正确/错误。

我按照以下方式修改了你的功能，它似乎现在可以产生与sgolayfilt()版本相同的形状，虽然我不确定我的实现是完全正确的。请注意，sigma的选择很重要，确实会影响最终的形状。如果您最初没有获得相同的形状，请尝试显着调整sigma参数。

CP <- function(S = NULL, X = NULL, method = "SDL", W = 2000, sigma = 0.0005) {
    # S is the spectrum, X is the frequencies, W is the window size (2*P in the eqn above)
    # Compute the requested 2nd derivative
    if (method == "SDL") {
        sdl <- rep(NA_real_, length(X)) # initialize a vector to store the final answer

        for(i in 1:length(X)) {
            bound1 <- 2*i - 1
            bound2 <- 2*length(X) - 2*i + 1
            P <- min(bound1, bound2)
            # Assemble the indices corresponding to the window
            idx <- seq(i-(P-1)/2, i+(P-1)/2, 1)
            # Now compute the sdl
            sdl[i] <- sum(SDL(X[idx], X[i], sigma = sigma) * S[idx])
            }
        }

    if (method == "SG") {
        sdl <- sgolayfilt(S, m = 2)     
        }

    # Now convolve!  There is a built-in function for this!
    cp <- convolve(S, sdl, type = "open")
    # The convolution has length 2*(length(S)) - 1 due to zero padding
    # so we need rescale back to the scale of S
    # Not sure if this is the right approach, but it doesn't affect the shape
    cp <- c(cp, 0.0)
    cp <- colMeans(matrix(cp, ncol = length(cp)/2)) # stackoverflow.com/q/32746842/633251
    return(cp)
}

Answer 2

您的代码存在一些问题。在计算SDL的CP中，看起来你试图在$ C _ {\ sigma} $的等式中求和，但这个求和是卷积的定义。

当你实际计算SDL时，你正在改变x0的值，但这个值是洛伦兹的平均值，应该是常数（在这种情况下它是0）。

最后，您可以计算卷积的边界并使用原始边界拉出信号

CP <- function(S = NULL, X = NULL, method = "SDL", W = 2000, sigma = 0.0005) {
# S is the spectrum, X is the frequencies, W is the window size (2*P in the eqn above)
# Compute the requested 2nd derivative
if (method == "SDL") {


    sdl <- rep(NA_real_, length(X)) # initialize a vector to store the final answer

    for(i in 1:length(X)) {
        sdl[i] <- SDL(X[i], 0, sigma = sigma)
        }
    }

if (method == "SG") {
    sdl <- sgolayfilt(S, m = 2)     
    }

# Now convolve!  There is a built-in function for this!
cp <- convolve(S, sdl, type = "open")
shift <- floor(length(S)/2) #both signals are the same length and symmetric around zero
                             #so the fist point of the convolution is half the signal 
                             #before the first point of the signal   
print(shift)      
cp <- cp[shift:(length(cp)-shift)]
return (cp)
}

运行此测试。

require("SpecHelpers")
require("signal")
# Create a Lorentzian curve
loren <- data.frame(x0 = 0, area = 1, gamma = 0.5)
lorentz1 <- makeSpec(loren, plot = FALSE, type = "lorentz", dd = 100,    x.range = c(-10, 10))
#
# Compute convolution
x <- lorentz1[1,] # Frequency values
y <- lorentz1[2,] # Intensity values
sig <- 100 * 0.0005 # per the reference
cpSDL <- CP(S = y, X = x, sigma = sig)

#
# Plot the original data, compare to convolution product

plot(x, cpSDL)

产生预期的形状：

naming conventions

我也不完全确定您的SDL定义是否正确。 cpSDL对于洛伦兹的二阶导数有一个更复杂的公式。

意外的卷积结果

2 个答案: