我正在尝试应用函数embed(x=time series, dimension=n)，它返回原始时间序列（ts）并且ts滞后n-1次。我的时间序列（8）是20年，我使用的是n = 5。这是将embed()应用于我的一个ts之后的输出：
L4 L3 L2 L1 L0
NA NA NA NA NA
NA NA NA NA NA
NA NA NA NA NA
495.7 NA NA NA NA
495.7 495.7 NA NA NA
577 495.7 495.7 NA NA
577 577 495.7 495.7 NA
577 577 577 495.7 495.7
577 577 577 577 495.7
577 577 577 577 577
577 577 577 577 577
577 577 577 577 577
577 577 577 577 577
577 577 577 577 577
577 577 577 577 577
266 577 577 577 577
我想将这个功能应用到我的8 ts并按行排出。这是我写的那个函数：
lagging<-function(df,col){#df= data frame, col=column of df to be lagged
dfL<-data.frame(rep(0,128),rep(0,128),rep(0,128),rep(0,128),rep(0,128))#128= 8 categories*16 rows
#after application of embed*
for(i in c('nsw','tas',"sasz","sanz","vicwz","vicez","nzC8","nzC4")){
for(j in seq(1,128,16)){
dfL[j:(j+15),]<-rbind(embed(df[df$fishery==i,col],5))#* allow lag a time series
}
}
colnames(dfL)<-c("L4","L3","L2","L1","L0")
return(dfL)
}
lagging(lob,3)
我的功能运行不正常，它返回第一个ts的输出（重复8次）。我想问题是索引。任何建议将不胜感激。以下是我的数据。谢谢！
数据：
season fishery tac cpue
1990 nsw NA NA
1991 nsw NA NA
1992 nsw NA NA
1993 nsw NA NA
1994 nsw NA 1.83
1995 nsw NA 2.22
1996 nsw 106 2.23
1997 nsw 117 2.3
1998 nsw 125 2.51
1999 nsw 140 2.34
2000 nsw 150 1.62
2001 nsw 150 1.73
2002 nsw 135 2.22
2003 nsw 135 2.38
2004 nsw 102 2.47
2005 nsw 102 2.89
2006 nsw 112 3.43
2007 nsw 124 4.29
2008 nsw 128 3.94
2009 nsw 128 3.15
1990 tas NA 0.97
1991 tas NA 0.97
1992 tas NA 0.89
1993 tas NA 0.85
1994 tas NA 0.82
1995 tas NA 0.94
1996 tas NA 0.95
1997 tas NA 0.88
1998 tas 1502.5 0.94
1999 tas 1502.5 1.01
2000 tas 1502.5 1.02
2001 tas 1502.5 1.04
2002 tas 1523.5 1.12
2003 tas 1523.5 1.09
2004 tas 1523.5 1.16
2005 tas 1523.5 1.2
2006 tas 1523.5 1.18
2007 tas 1523.5 1.17
2008 tas 1523.5 1.01
2009 tas 1470.98 0.89
1990 sasz NA 0.83
1991 sasz NA 0.96
1992 sasz NA 1.01
1993 sasz 1720 1.03
1994 sasz 1740 1.15
1995 sasz 1720 1.07
1996 sasz 1720 0.95
1997 sasz 1720 0.97
1998 sasz 1720 1.13
1999 sasz 1720 1.49
2000 sasz 1770 1.66
2001 sasz 1770 1.90
2002 sasz 1770 2.08
2003 sasz 1900 1.83
2004 sasz 1900 1.81
2005 sasz 1900 1.60
2006 sasz 1900 1.41
2007 sasz 1900 1.12
2008 sasz 1770 0.74
2009 sasz 1400 0.61
1990 sanz NA 1.51
1991 sanz NA 1.52
1992 sanz NA 1.43
1993 sanz NA 1.29
1994 sanz NA 1.27
1995 sanz NA 1.25
1996 sanz NA 1.25
1997 sanz NA 1.31
1998 sanz NA 1.41
1999 sanz NA 1.43
2000 sanz NA 1.24
2001 sanz NA 1.08
2002 sanz NA 1.04
2003 sanz 625 0.84
2004 sanz 520 0.8
2005 sanz 520 0.81
2006 sanz 520 0.86
2007 sanz 520 0.76
2008 sanz 470 0.88
2009 sanz 310 1.07
1990 vicwz NA 0.5
1991 vicwz NA 0.56
1992 vicwz NA 0.55
1993 vicwz NA 0.59
1994 vicwz NA 0.58
1995 vicwz NA 0.56
1996 vicwz NA 0.52
1997 vicwz NA 0.55
1998 vicwz NA 0.59
1999 vicwz NA 0.6
2000 vicwz NA 0.6
2001 vicwz 320 0.58
2002 vicwz 450 0.65
2003 vicwz 450 0.71
2004 vicwz 450 0.66
2005 vicwz 450 0.55
2006 vicwz 450 0.47
2007 vicwz 380 0.47
2008 vicwz 381 0.41
2009 vicwz 240 0.38
1990 vicez NA 0.42
1991 vicez NA 0.39
1992 vicez NA 0.34
1993 vicez NA 0.32
1994 vicez NA 0.28
1995 vicez NA 0.27
1996 vicez NA 0.26
1997 vicez NA 0.3
1998 vicez NA 0.3
1999 vicez NA 0.32
2000 vicez NA 0.32
2001 vicez 42 0.34
2002 vicez 60 0.36
2003 vicez 60 0.43
2004 vicez 60 0.42
2005 vicez 60 0.4
2006 vicez 60 0.43
2007 vicez 66 0.37
2008 vicez 67 0.39
2009 vicez 66 0.37
1990 nzC8 1152.4 0.87
1991 nzC8 1054.6 0.82
1992 nzC8 986.8 0.68
1993 nzC8 888.1 0.92
1994 nzC8 888.1 0.88
1995 nzC8 888.1 0.9
1996 nzC8 888.1 0.87
1997 nzC8 888.1 0.72
1998 nzC8 888.1 0.79
1999 nzC8 711 0.84
2000 nzC8 711 0.98
2001 nzC8 568 0.92
2002 nzC8 568 1.11
2003 nzC8 568 1.67
2004 nzC8 603.4 1.58
2005 nzC8 603.4 1.75
2006 nzC8 755.2 2.19
2007 nzC8 755.2 2.47
2008 nzC8 966 3.22
2009 nzC8 1019 3.13
1990 nzC4 NA NA
1991 nzC4 NA NA
1992 nzC4 NA NA
1993 nzC4 NA NA
1994 nzC4 NA NA
1995 nzC4 NA NA
1996 nzC4 NA NA
1997 nzC4 495.7 NA
1998 nzC4 495.7 NA
1999 nzC4 577 1.27
2000 nzC4 577 1.26
2001 nzC4 577 1.06
2002 nzC4 577 1.09
2003 nzC4 577 1.14
2004 nzC4 577 1
2005 nzC4 577 0.88
2006 nzC4 577 0.65
2007 nzC4 577 0.6
2008 nzC4 577 0.52
2009 nzC4 266 0.57
答案 0 :(得分：1)
重复最后fishery索引数据的结果的问题是由变量j引起的。对于i的每个值，使用j的所有值。因此，在外部for循环的每次运行中，前一次运行的数据将被与当前fishery索引关联的新数据替换。
以下是使用tapply的更有效方法：
lagging <- function(df, col, dimension = 5) {
setNames(as.data.frame(do.call(rbind, tapply(df[[col]], df$fishery,
embed, dimension))),
paste0("L", seq(dimension - 1, 0)))
}
lagging(lob, 3)