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我正在使用 Bass 扩散模型，通过执行 for 循环，我找到了大约 5000 个产品的参数 m、p 和 q。手动将这些参数放入excel表格中需要做很多工作。有人知道将这些值存储在 R 中的简单方法吗？例如在表中？这是我的代码：

PID=unique(TotalBass4$ProdID)

BassStored=NULL
k=0
pidlistNAs=NULL
for(pid in PID){
  k=k+1
  out <- lm(Sales ~ Cumsaleslag + Cumsalessqrt, data=subset(TotalBass4,subset=ProdID==pid))
  P1a <- out$coef[1]
  P1b <- out$coef[2]
  P1c <- out$coef[3]
  P1mplus <- (-P1b+sqrt(P1b**2-4*P1a*P1c))/(2*P1c)
  P1mminus <- (-P1b-sqrt(P1b**2-4*P1a*P1c))/(2*P1c)
  m <- P1mminus
  p <- 1/m
  q <- P1b+p
  cmsl=subset(TotalBass4,subset=ProdID==pid)$Cumsaleslag
  Spred <- Bassmodel(p, q, m, cmsl,T=30)$Sales
  Spred <- ts(Spred)
  BassStored[[k]]=list(parm=c(m,p,q),ProdID=pid) ## Spred=Spred
  names(BassStored[[k]]$parm)=c("m","p","q")
  if(is.na(P1c))
  pidlistNAs=c(pidlistNAs,pid)
  if((k%%10)==0)
  print(k)
  }

BassStored[1]

4

1 回答 1

1

该示例的一些可重现数据：

set.seed(2017-12-20)

p <- runif(10, min=0.006293, max=0.00689)
m <- runif(10, min=67380.15, max=68980.95)
q <- runif(10, min=0.61809, max=0.65804)

lapply(1:10, function(i) {
  list(parm=c(m=m[i], p=p[i], q=q[i]), ProdID=i)
}) -> BassStored

您已经创建了类似的东西，并且可能仍将其加载到 R 的内存中：

str(BassStored)
## List of 10
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.79e+04 6.69e-03 6.47e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 1
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.81e+04 6.45e-03 6.29e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 2
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.86e+04 6.37e-03 6.18e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 3
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.78e+04 6.54e-03 6.55e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 4
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.88e+04 6.52e-03 6.45e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 5
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.79e+04 6.38e-03 6.19e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 6
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.75e+04 6.63e-03 6.21e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 7
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.76e+04 6.57e-03 6.24e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 8
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.79e+04 6.83e-03 6.55e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 9
##  $ :List of 2
##   ..$ parm  : Named num [1:3] 6.88e+04 6.61e-03 6.29e-01
##   .. ..- attr(*, "names")= chr [1:3] "m" "p" "q"
##   ..$ ProdID: int 10

我们可以把它变成一个数据框：

do.call(
  rbind.data.frame,
  lapply(1:length(BassStored), function(i){
    as.list(unlist(BassStored[i]))
  })
) -> xdf

xdf
##      parm.m      parm.p    parm.q ProdID
## 2  67860.06 0.006689309 0.6468014      1
## 21 68054.35 0.006451261 0.6286121      2
## 3  68640.19 0.006372309 0.6181186      3
## 4  67829.24 0.006541486 0.6551225      4
## 5  68807.85 0.006517481 0.6454875      5
## 6  67886.29 0.006382578 0.6194927      6
## 7  67542.34 0.006625390 0.6212089      7
## 8  67635.12 0.006566107 0.6239669      8
## 9  67878.34 0.006826642 0.6545225      9
## 10 68778.44 0.006609701 0.6287901     10

清理名称：

xdf <- setNames(xdf, c("m", "p", "q", "ProdID"))

xdf
##           m           p         q ProdID
## 2  67860.06 0.006689309 0.6468014      1
## 21 68054.35 0.006451261 0.6286121      2
## 3  68640.19 0.006372309 0.6181186      3
## 4  67829.24 0.006541486 0.6551225      4
## 5  68807.85 0.006517481 0.6454875      5
## 6  67886.29 0.006382578 0.6194927      6
## 7  67542.34 0.006625390 0.6212089      7
## 8  67635.12 0.006566107 0.6239669      8
## 9  67878.34 0.006826642 0.6545225      9
## 10 68778.44 0.006609701 0.6287901     10

并且，写出来：

write.csv(xdf, "bassmodel.csv")
于 2017-12-20T11:14:46.367 回答