在 R 中执行时间序列的 fft

我想使用 FFT 将波拟合到时间序列。 目标是绘制具有不同谐波的图，并用它来预测 n 个数据点。

我正在使用的代码基于此answer https://stackoverflow.com/questions/41435777/perform-fourier-analysis-to-a-time-series-in-r from @灾难性故障 https://stackoverflow.com/users/2874779/catastrophic-failure

 nff = function(y = NULL, n = NULL, up = 10L, plot = TRUE, add = FALSE, main = NULL, ...){
  #The direct transformation
  #The first frequency is DC, the rest are duplicated
  dff = fft(y)
  #The time
  t = seq(from = 1, to = length(y))
  #Upsampled time
  nt = seq(from = 1, to = length(y)+1-1/up, by = 1/up)
  #New spectrum
  ndff = array(data = 0, dim = c(length(nt), 1L))
  ndff[1] = dff[1] #Always, it's the DC component
  if(n != 0){
    ndff[2:(n+1)] = dff[2:(n+1)] #The positive frequencies always come first
    #The negative ones are trickier
    ndff[length(ndff):(length(ndff) - n + 1)] = dff[length(y):(length(y) - n + 1)]
  }
  #The inverses
  indff = fft(ndff/as.integer(length(y)), inverse = TRUE)
  idff = fft(dff/as.integer(length(y)), inverse = TRUE)
  if(plot){
    if(!add){
      plot(x = t, y = y, xlab = "Time", ylab = "Data",
           main = ifelse(is.null(main), paste(n, "harmonics"), main), type="l", col="green")
      lines(y = Mod(idff), x = t, col = "red")
    }
    lines(y = Mod(indff), x = nt, ...)
  }
  ret = data.frame(time = nt, y = Mod(indff))
  return(ret)
}

对我来说，问题是，因为我的数据集中也有负值，所以我无法弄清楚为什么包含正值。

这是原著的剧情data https://drive.google.com/drive/folders/1P_oRxJW82OrdDOB8ND5Xd6I-COgGZveM?usp=sharing

与 fft 后的图相比

我如何调整代码，使谐波也涵盖缺失的负值，以及如何使用它来计算（预测）接下来的 n 个时间点？

当您尝试使用以下命令绘制结果时会出现问题Mod(idff) and Mod(indff)如下：

...
  lines(y = Mod(idff), x = t, col = "red")
}
lines(y = Mod(indff), x = nt, ...)

Mod将始终返回与复数大小相对应的正数。

由于您正在计算具有 Hermitian 对称性的序列（通过构造）的逆 FFT，因此您应该期望得到实值结果。但实际上，由于舍入误差，可能存在较小的虚部。通过仅提取实部，可以安全地忽略这些Re(idff) and Re(indff)， 如下：

...
  lines(y = Re(idff), x = t, col = "red")
}
lines(y = Re(indff), x = nt, ...)

请注意，通常最好首先确认虚部与实部相比确实是非常小的数字，因为相反的情况表明频域值不具有预期的埃尔米特对称性。