Question

我大致有以下data.frame：

df <- structure(list(WK = c("W212015", "W212015", "W222015", "W222015", 
"W232015", "W232015", "W372016", "W372016", "W382016", "W382016", 
"W392016", "W392016"), YEAR = c(2015, 2015, 2015, 2015, 2015, 
2015, 2016, 2016, 2016, 2016, 2016, 2016), SEM = c("W21", "W21", 
"W22", "W22", "W23", "W23", "W37", "W37", "W38", "W38", "W39", 
"W39"), DEPTO = c("FARMA", "NO FARMA", "FARMA", "NO FARMA", "FARMA", 
"NO FARMA", "FARMA", "NO FARMA", "FARMA", "NO FARMA", "FARMA", 
"NO FARMA"), UNIDADES = c(1010, 1252, 1354, 1275, 1411, 1171, 
1057, 1362, 1435, 1120, 1451, 1140), VENTAS = c(128270, 172776, 
173312, 204000, 189074, 165111, 130011, 226092, 231035, 181440, 
220552, 197220)), .Names = c("WK", "YEAR", "SEM", "DEPTO", "UNIDADES", 
"VENTAS"), class = "data.frame", row.names = c(NA, -12L))

head(df)
  WK      YEAR SEM    DEPTO UNIDADES VENTAS
1 W212015 2015 W21    FARMA     1010 128270
2 W212015 2015 W21 NO FARMA     1252 172776
3 W222015 2015 W22    FARMA     1354 173312
4 W222015 2015 W22 NO FARMA     1275 204000
5 W232015 2015 W23    FARMA     1411 189074
6 W232015 2015 W23 NO FARMA     1171 165111

数据采用整洁格式，通常建议使用。但是，当我试图计算年度变化时，我会不断遇到一些问题（计算方法为：t1 / t0-1）。

期望的结果将是这样的：

res <- structure(list(DEPTO = c("FARMA", "FARMA", "FARMA", "FARMA", 
"FARMA", "FARMA", "FARMA", "FARMA", "FARMA", "FARMA", "FARMA", 
"FARMA", "FARMA", "FARMA", "FARMA", "FARMA", "FARMA", "FARMA", 
"FARMA", "NO FARMA", "NO FARMA", "NO FARMA", "NO FARMA", "NO FARMA", 
"NO FARMA", "NO FARMA", "NO FARMA", "NO FARMA", "NO FARMA", "NO FARMA", 
"NO FARMA", "NO FARMA", "NO FARMA", "NO FARMA", "NO FARMA", "NO FARMA", 
"NO FARMA", "NO FARMA"), SEM = c("W21", "W22", "W23", "W24", 
"W25", "W26", "W27", "W28", "W29", "W30", "W31", "W32", "W33", 
"W34", "W35", "W36", "W37", "W38", "W39", "W21", "W22", "W23", 
"W24", "W25", "W26", "W27", "W28", "W29", "W30", "W31", "W32", 
"W33", "W34", "W35", "W36", "W37", "W38", "W39"), YEAR = c(2016, 
2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 
2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 
2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 2016, 
2016, 2016, 2016, 2016), YOYDIFF = c(0.103960396039604, -0.231166912850812, 
0.0276399716513112, -0.0818120351588911, 0.0528169014084507, 
0.128721541155867, -0.105511811023622, -0.227333333333333, -0.103495544893763, 
0.017304189435337, -0.0495652173913044, -0.239972807613868, -0.258199581297976, 
-0.204038997214485, 0.281081081081081, -0.255927475592748, -0.0130718954248366, 
-0.0231449965963241, 0.4169921875, 0.0511182108626198, -0.167058823529412, 
-0.0341588385994877, 0.00649953574744666, 0.05859375, 0.10594512195122, 
0.0839469808541973, 0.250877192982456, 0.233502538071066, 0.0282738095238095, 
-0.0850077279752705, 0.103135313531353, 0.121684867394696, -0.0892448512585813, 
0.137738853503185, -0.489478114478115, NA, -0.206236711552091, 
-0.105180533751962)), .Names = c("DEPTO", "SEM", "YEAR", "YOYDIFF"
), row.names = c(NA, 38L), class = "data.frame")

head(res)
  DEPTO SEM YEAR     YOYDIFF
1 FARMA W21 2016  0.10396040
2 FARMA W22 2016 -0.23116691
3 FARMA W23 2016  0.02763997
4 FARMA W24 2016 -0.08181204
5 FARMA W25 2016  0.05281690
6 FARMA W26 2016  0.12872154

所以，我按周（SEM）和分组变量（DEPTO）做了一年一年。

通常情况下，我会使用dcast或某些过滤和inner_joins与dplyr的组合执行此操作，但我遇到速度问题，看起来真的很像为简单的计算而做这一切的效率太高。

有没有＆＃34;通用＆＃34;功能或包装，以加快速度，使其更直观？我有兴趣看看是否有人使用超快data.table或任何类似的没有进行各种连接和操作来解决这个问题。

更新

结果（YOYDIFF）将通过例如除以变量来获得 UNIDADES DEPTO =＆＃34; FARMA＆＃34;和YEAR =＆＃34; 2016＆＃34;和SEM =＆＃34; W21＆＃34;由UNIDADES划分DEPTO =＆＃34; FARMA＆＃34;和YEAR =＆＃34; 2015＆＃34;和SEM =＆＃34; W21＆＃34;最后减去1。

更新2

df是一个非常小的示例，仅用于显示结构，而res是在Excel上执行某些计算的结果。但是df

中的原始数据不

Answer 1

也许这并不完全是你所追求的，但这是我的猜测：

library(data.table)
setDT(df)
df[,list(YOYDIFF = UNIDADES[YEAR==2016] / UNIDADES[YEAR==2015]), by=c('SEM','DEPTO')]

给出了：

   SEM    DEPTO   YOYDIFF
1: W21    FARMA 1.0465347
2: W21 NO FARMA 1.0878594
3: W22    FARMA 1.0598227
4: W22 NO FARMA 0.8784314
5: W23    FARMA 1.0283487
6: W23 NO FARMA 0.9735269

启动数据集（已修改为可用）：

df <- structure(list(WK = c("W212015", "W212015", "W222015", "W222015", 
"W232015", "W232015", "W372016", "W372016", "W382016", "W382016", 
"W392016", "W392016"), YEAR = c(2015, 2015, 2015, 2015, 2015, 
2015, 2016, 2016, 2016, 2016, 2016, 2016), SEM = c("W21", "W21", 
"W22", "W22", "W23", "W23", "W21", "W21", "W22", "W22", "W23", 
"W23"), DEPTO = c("FARMA", "NO FARMA", "FARMA", "NO FARMA", "FARMA", 
"NO FARMA", "FARMA", "NO FARMA", "FARMA", "NO FARMA", "FARMA", 
"NO FARMA"), UNIDADES = c(1010, 1252, 1354, 1275, 1411, 1171, 
1057, 1362, 1435, 1120, 1451, 1140), VENTAS = c(128270, 172776, 
173312, 204000, 189074, 165111, 130011, 226092, 231035, 181440, 
220552, 197220)), .Names = c("WK", "YEAR", "SEM", "DEPTO", "UNIDADES", 
"VENTAS"), row.names = c(NA, -12L), class = "data.frame")

整齐数据中行之间差异的通用功能？

1 个答案: