Question

我创建了以下函数来确定两个变量的滞后。

但是，这个函数只需要两个参数，我想在整个数据集上运行它：

datSel <- structure(list(stat.resProp.Dwell.4 = c(0.000887705, 0.007954085, 
-0.025859667, 0.024097552, 0.114052787, 0.023329207, 0.042143181, 
-0.092587287, -0.004050228, -0.001624696, 0.020121403, -0.100502922, 
0.057354185, 0.025463388, 0.037409854, 0.001561281, -0.028482938, 
-0.004827041, 0.014411779, -0.029034298, 0.021053409, -0.067963182, 
0.032070259, -0.038091783, 0.039751534, 0.027802281, -0.027802281, 
-0.013355791, 0.009201236, -0.073403679, 0.021277398, -0.033901552, 
0.012624153, -0.065733979, 0.032017801, -0.072042665, 0.041936911, 
0.002861232, 0.017933468, -0.01698154, 0.006638242, -0.08375153, 
-0.007220248, 0.0255507, 0.019980685, 0.013752673, 0.026000502, 
-0.021134312, -0.019608471, 0.0166916, -0.021654389, 0.066402455, 
0.024828862, -0.083302632, 0.042518482, -0.052439198, 0.037186281, 
-0.056311172, -0.012270093), stat.lohn = c(0, -0.007558004, -0.015289567, 
0, 0, -0.009609384, -0.019500305, 0, 0, -0.012458015, -0.025391532, 
-0.000983501, 0, -0.00165265, -0.003313516, 0.000204576, 0, -0.004898564, 
-0.009869709, 0, 0, -0.010574012, -0.021489482, 0, 0, -0.011534651, 
-0.023476287, 0, 0, -0.00814845, -0.016498838, 0, 0, -0.0099856, 
-0.020275409, -0.002818337, 0, -0.007212389, -0.014582736, 0, 
0, -0.004121565, -0.008294445, 0, 0, -0.010766386, -0.021886884, 
0, 0, -0.010179741, -0.02067574, 0, 0, -0.011797067, -0.024020039, 
-0.002017983, -0.007343864, -0.007398196, -0.014962644), stat.resProp.Dwell.1 = c(0.012777325, 
-0.002991775, -0.057819571, -0.00796817, -0.019386714, 0, 0.009740337, 
0.005638356, -0.035148694, 0, 0.027084134, -0.160377856, 0.101169235, 
-0.043007944, 0.043007944, -0.002580647, -0.015625318, 0.023347364, 
0.007662873, -0.09607383, -0.024575906, 0.056733018, -0.000904568, 
-0.058703392, 0.011450507, 0.007561473, 0.037879817, -0.032246, 
0.042169401, -0.001796946, -0.024580209, -0.148788737, 0.082097362, 
-0.000985707, -0.00098668, 0.003940892, -0.049380309, 0.005151995, 
0.027371197, -0.025317808, 0.019299736, -0.047382704, -0.010604553, 
0.082827084, -0.04516573, 0.003075348, 0.007139245, 0.022111454, 
-0.004982571, -0.038701368, 0.018519048, -0.049096021, 0.061254226, 
-0.020346582, 0.023363175, -0.00402415, -0.014213437, 0.023245109, 
0.027587957), stat.carReg = c(0.022775414, 0.008073857, 0.002624717, 
0.169431097, -0.144595366, 0.066716837, -0.086971929, 0.037928208, 
0.071752161, -0.046824102, 0.106085873, 0.049965928, -0.057984255, 
-0.091650262, 0.090732857, -0.082282389, 0.053376121, -0.044203971, 
-0.022855425, 0.025856271, 0.000136493, 0.05579193, -0.293966656, 
0.013645739, 0.059732986, 0.187020956, -0.145234848, 0.11041385, 
-0.126539687, -0.000949877, 0.031473389, 0.020267816, -0.02180532, 
-0.07175183, 0.147500145, -0.040559138, 0.008394819, 0.049045337, 
-0.043050615, 0.094358754, -0.058408438, -0.005018402, -0.061717889, 
0.100150837, -0.071100417, -0.084393865, 0.002854733, 0.002141389, 
-0.026538398, 0.013480513, -0.046002189, -0.030495611, 0.052899746, 
0.012842017, 0.064086498, 0.020757573, -0.043441298, -0.009563043, 
0.048033848)), .Names = c("stat.resProp.Dwell.4", "stat.lohn", 
"stat.resProp.Dwell.1", "stat.carReg"), row.names = c(NA, -59L
), class = "data.frame")

函数和我的函数调用是：

select.lags<-function(x,y,max.lag=8) {
  y<-as.numeric(y)
  y.lag<-embed(y,max.lag+1)[,-1,drop=FALSE]
  x.lag<-embed(x,max.lag+1)[,-1,drop=FALSE]

  t<-tail(seq_along(y),nrow(y.lag))

  ms=lapply(1:max.lag,function(i) lm(y[t]~y.lag[,1:i]+x.lag[,1:i]))

  pvals<-mapply(function(i) anova(ms[[i]],ms[[i-1]])[2,"Pr(>F)"],max.lag:2)
  ind<-which(pvals<0.05)[1]
  ftest<-ifelse(is.na(ind),1,max.lag-ind+1)

  aic<-as.numeric(lapply(ms,AIC))
  bic<-as.numeric(lapply(ms,BIC))
  structure(list(ic=cbind(aic=aic,bic=bic),pvals=pvals,
                 selection=list(aic=which.min(aic),bic=which.min(bic),ftest=ftest)))
}

for (i in length(datSel) ) {
  for (y in length(datSel) ) {
    d1<-ts(datSel[i])
    d2<-ts(datSel[y])
    lag <- select.lags(d1,d2,5)
  } 
}

作为延迟的输出我得到：

> lag
$ic
           aic        bic
[1,] -115.3623 -109.56679
[2,] -114.3370 -106.60972
[3,] -116.2026 -106.54350
[4,] -114.7030 -103.11210
[5,] -112.7153  -99.19253
[6,] -110.8018  -95.34721
[7,] -110.0812  -92.69477
[8,] -110.1427  -90.82446

$pvals
[1] 0.1952302 0.3017934 0.7858944 0.9176337 0.5040079 0.0604511 0.3406657

$selection
$selection$aic
[1] 3

$selection$bic
[1] 1

$selection$ftest
[1] 1

正如您所看到的，我只得到8个结果，但是，我的data.frame有20个变量。

任何建议我做错了什么？

感谢您的回复！

Answer 1

如果你想要，例如存储AIC标准的结果：

lag.aic.store = matrix(NA, 4, 4)
for (i in 1:length(datSel) ) {
  for (y in 1:length(datSel) ) {
    d1<-ts(datSel[,i])
    d2<-ts(datSel[,y])
    lag <- select.lags(d1,d2,5)
    lag.store.aic[i,y] = lag$selection$aic
  } 
}

$ic中有8个值，因为max.lag为8，它与您的变量数量无关。还请注意，为了清晰起见，我在变量索引时添加了逗号，并且必须循环遍历1:length(datSel)，否则您将只捕获最后一个变量。

迭代函数并获取每个结果值

1 个答案: