我对许多国家的样本进行了几年的抽样调查，其中包含有关产出（GDP）的信息。我想使用我在R-Bloggers here中找到的函数来计算“输出差距”，但希望考虑到年份，它可以遍历样本中的所有国家/地区，并将结果存储在矩阵（跨行绑定）。
该函数如下所示：
hp <- function(data,l=1600){
#h-p filter code from Farnsworth
hpfilterq <- function(x=data,lambda=l){
eye <- diag(length(x))
result <- solve(eye+lambda*crossprod(diff(eye,lag=1,d=2)),x)
return(result)
}
hpfiltered<-hpfilterq(data)
hpgap <- data - hpfiltered
#
t1<-1:length(data)
t2<-t1^2
t3<-t1^3
t1<-ts(t1)
t2<-ts(t2)
t3<-ts(t3)
#
datats<-ts(data)
myseries<-ts.union(datats,t1,t2,t3)
#
polynomial1 <- lm(datats ~ t1,data=myseries)
polynomial2 <- lm(datats ~ t1 + t2,data=myseries)
polynomial3 <- lm(datats ~ t1 + t2 + t3,data=myseries)
#
returndata<-data.frame(hpgap,polynomial1$residuals,polynomial2$residuals,polynomial3$residuals)
colnames(returndata) <- c("H-P Gap", "Poly1","Poly2","Poly3")
return(returndata)
}
假设我的样本如下：
country year output
1 AUS 2000 49709.21
2 AUS 2001 59805.90
3 AUS 2002 46501.57
4 AUS 2003 53521.78
5 AUS 2004 53824.41
6 AUS 2005 55001.43
7 AUS 2006 48356.12
8 AUS 2007 55125.00
9 AUS 2008 58551.84
10 AUS 2009 57805.95
11 AUS 2010 64858.86
12 AUS 2011 67395.81
13 AUS 2012 69043.00
14 AUS 2013 73789.00
15 AUS 2014 77869.09
16 BEL 2000 7110.00
17 BEL 2001 7235.10
18 BEL 2002 7204.10
19 BEL 2003 7327.60
20 BEL 2004 7558.70
21 BEL 2005 7123.10
22 BEL 2006 7539.00
23 BEL 2007 7943.40
24 BEL 2008 8052.50
25 BEL 2009 7509.60
26 BEL 2010 8455.50
27 BEL 2011 8749.40
28 BEL 2012 9694.10
29 BEL 2013 9614.40
30 BEL 2014 8707.50
我想将功能“ hp”应用于“ AUS”，
H-P Gap Poly1 Poly2 Poly3
1 2393.2324 2751.8407 -3684.68922 -3536.276248
2 10838.8666 11069.4941 7391.47697 7412.678826
3 -4118.0018 -4013.8596 -5357.75042 -5414.832333
4 1239.8878 1227.3108 1793.15960 1698.566717
5 -135.2596 -249.0938 1802.10804 1702.622415
6 -657.9475 -851.1054 2261.06288 2181.148204
7 -9031.2784 -9275.4400 -5526.69186 -5570.726477
8 -4024.7279 -4285.5977 -324.65619 -324.656186
9 -2394.8369 -2637.7883 1110.95990 1154.994518
10 -4970.1209 -5162.7119 -2050.54360 -1970.628924
11 224.6324 111.1625 2162.36431 2261.849928
12 881.4586 869.0828 1434.93163 1529.524516
13 633.1949 737.2392 -606.65163 -549.569720
14 3474.5127 3704.2066 26.18952 4.987662
15 5646.3875 6005.2600 -431.26992 -579.682899
“ BEL”，
H-P Gap Poly1 Poly2 Poly3
1 291.55895 311.04333 -4.1179412 -188.99755
2 253.89032 266.24190 86.1497479 59.73838
3 59.93946 65.34048 -0.4624273 70.64511
4 19.96547 18.93905 46.6455333 164.48089
5 86.68999 80.13762 180.5736296 304.50392
6 -514.57784 -525.36381 -372.9781383 -273.42758
7 -266.08305 -279.36524 -95.8097705 -40.95538
8 -30.94904 -44.86667 149.0787330 149.07873
9 -92.93293 -105.66810 77.8873723 23.03298
10 -808.67250 -818.46952 -666.0838526 -765.63441
11 -37.24743 -42.47095 57.9650582 -65.96523
12 81.16801 81.52762 109.2341047 -8.60125
13 850.02282 856.32619 790.5232870 719.41575
14 594.71530 606.72476 426.6326050 453.04398
15 -487.48754 -470.07667 -785.2379412 -600.35833
和所有其他国家/地区，并将它们存储在矩阵（或某种形式的列表）中：
H-P Gap Poly1 Poly2 Poly3
1 2393.23236 2751.84068 -3684.6892235 -3536.276248
2 10838.86665 11069.49406 7391.4769723 7412.678826
3 -4118.00184 -4013.85956 -5357.7504192 -5414.832333
4 1239.88784 1227.31082 1793.1596021 1698.566717
5 -135.25961 -249.09380 1802.1080361 1702.622415
6 -657.94746 -851.10542 2261.0628828 2181.148204
7 -9031.27839 -9275.44005 -5526.6918577 -5570.726477
8 -4024.72789 -4285.59767 -324.6561855 -324.656186
9 -2394.83690 -2637.78829 1110.9598994 1154.994518
10 -4970.12091 -5162.71191 -2050.5436029 -1970.628924
11 224.63238 111.16247 2162.3643075 2261.849928
12 881.45859 869.08285 1434.9316306 1529.524516
13 633.19493 737.23923 -606.6516335 -549.569720
14 3474.51272 3704.20660 26.1895151 4.987662
15 5646.38752 6005.25998 -431.2699235 -579.682899
16 291.55895 311.04333 -4.1179412 -188.997549
17 253.89032 266.24190 86.1497479 59.738375
18 59.93946 65.34048 -0.4624273 70.645114
19 19.96547 18.93905 46.6455333 164.480888
20 86.68999 80.13762 180.5736296 304.503916
21 -514.57784 -525.36381 -372.9781383 -273.427580
22 -266.08305 -279.36524 -95.8097705 -40.955381
23 -30.94904 -44.86667 149.0787330 149.078733
24 -92.93293 -105.66810 77.8873723 23.032983
25 -808.67250 -818.46952 -666.0838526 -765.634411
26 -37.24743 -42.47095 57.9650582 -65.965228
27 81.16801 81.52762 109.2341047 -8.601250
28 850.02282 856.32619 790.5232870 719.415746
29 594.71530 606.72476 426.6326050 453.043978
30 -487.48754 -470.07667 -785.2379412 -600.358333
无需使用效率极低的hp(data$output[c(1:15)])和hp(data$output[c(16:30)])
答案 0 :(得分：2)
您可以结合使用split和lapply在新函数中应用hp函数，这将自动将结果导出为list中的data.frames按国家
hp_per_country <- function(x) {
data_list <- split(x$output, x$country)
result <- lapply(data_list, FUN = hp)
} #note that `return` is assumed in R for the last object computed by a function
overall_hp_results <- hp_per_country(input_data)
如果这对计算很重要，则假定input_data按年份排序。