Question

我想根据条件length计算两个函数。当lst自变量为<=4时，当scale自变量> 4时，一个函数为lm，第二个函数为length模型，我尝试使用：

lapply(lst, function (x) ifelse((sapply(lst,length)<=4)==T,scale,
                                (fit <- lm(x ~ poly(seq(1:length(x)), 5, raw=TRUE)))
                                (d <- resid(fit)-mean(resid(fit)))/sd(resid(fit))))

但是有一个错误：

ifelse（（sapply（lst，length）<= 4）== T，scale，（fit <-lm（x〜：未使用的参数（d <-resid（fit）-Mean（resid（fit））/ sd（resid（fit）））

当长度大于4时，如何计算同一函数中残差的拟合和z.score？

数据

lst <- list(a=c(2.5,9.8,5.0,6.7,6.5,5.2,34.4, 4.2,39.5, 1.3,0.0,0.0,4.1,0.0,0.0,25.5,196.5, 0.0,104.2,0.0,0.0,0.0,0.0,0.0),
            b=c(147.4,122.9,110.2,142.3))

Answer 1

您需要if和else

out <- lapply(lst, function(x) {
  if (length(x) <= 4) {
    scale(x)
  } else {
    fit <- lm(x ~ poly(
      x = 1:length(x),
      degree = 5,
      raw = TRUE
    ))
    return(resid(fit) - mean(resid(fit)) / sd(resid(fit)))
  }
})

out
#$a
#         1          2          3          4          5          6          7          8          9 
# 12.047467  -1.783343 -14.687917 -12.902068  -8.652886  -4.053071  30.404730   3.453593  39.257979 
#        10         11         12         13         14         15         16         17         18 
# -1.380593  -7.817887 -15.061277 -19.464085 -32.319915 -40.256981 -20.832448 146.873239 -49.437978 
#        19         20         21         22         23         24 
# 58.823894 -37.457437 -26.198873 -12.712270   1.201236  12.956892 

#$b
#           [,1]
#[1,]  0.9671130
#[2,] -0.4517055
#[3,] -1.1871746
#[4,]  0.6717671
#attr(,"scaled:center")
#[1] 130.7
#attr(,"scaled:scale")
#[1] 17.26789

在R中具有多边形拟合和比例的lapply和ifelse条件

1 个答案: