我想使用data.table来简洁地计算一些百分比变化,但我在完全理解.SD操作方式方面遇到了一些问题。假设我有以下表格:
dt = structure(list(type = c("A", "A", "A", "B", "B", "B"), Year = c(2000L,
2005L, 2010L, 2000L, 2005L, 2010L), alpha = c(0.0364325563237498,
0.0401968159729988, 0.0357395587861466, 0.0317236054181487, 0.0328213742235379,
0.0294694430578336), beta = c(0.0364325563237498, 0.0401968159729988,
0.0357395587861466, 0.0317236054181487, 0.0328213742235379, 0.0294694430578336
)), .Names = c("type", "Year", "alpha", "beta"), row.names = c(NA,
-6L), class = c("data.table", "data.frame"))
> dt
## type Year alpha beta
## 1: A 2000 0.03643256 0.03643256
## 2: A 2005 0.04019682 0.04019682
## 3: A 2010 0.03573956 0.03573956
## 4: B 2000 0.03172361 0.03172361
## 5: B 2005 0.03282137 0.03282137
## 6: B 2010 0.02946944 0.02946944
为了计算每个类别中alpha的百分比变化,我想出了以下代码:
dt[,change:=list(lapply(3:2,function(x)(.SD[x,alpha]/.SD[
(x-1),alpha]))),by=list(type)][][Year==2000,change:=NA]
但是我觉得可能有更简洁的方法来完成它。特别是如果想要对两列执行百分比变化,下面的方法将不起作用。
dt[,c("changeAlpha","changeBeta"):=list(lapply(3:2,
function(x)(.SD[x]/.SD[(x-1)]))),by=list(type)][Year==2000,change:=NA][]
所以我采取了以下措施:
dt[,c("changeAlpha","changeBeta"):=list(
lapply(3:2,function(x)(.SD[x,alpha]/.SD[(x-1),alpha])),
lapply(3:2,function(x)(.SD[x,beta]/.SD[(x-1),beta]))),by=list(type)][
Year==2000,c("changeAlpha","changeBeta"):=list(NA,NA)][]
## type Year alpha beta changeAlpha changeBeta
## 1: A 2000 0.03643256 0.03643256 NA NA
## 2: A 2005 0.04019682 0.04019682 1.10332131557826 1.10332131557826
## 3: A 2010 0.03573956 0.03573956 0.889114172877617 0.889114172877617
## 4: B 2000 0.03172361 0.03172361 NA NA
## 5: B 2005 0.03282137 0.03282137 1.03460416276522 1.03460416276522
## 6: B 2010 0.02946944 0.02946944 0.897873527693412 0.897873527693412
但操作似乎是正确的,但出现了很多警告,导致我来到这里。
- 我的思考方法完全错误还是这种操作的正确方式?
cols <- names(dt)[-(1:2)]
,然后运行代码。 - David Arenburg