我的数据框看起来像这样,我想要两个单独的累积列,一个用于基金A,另一个用于基金B
free
但这完全不是我想要的,因为它向我展示了两个基金的运行总额,尽管只是在资金匹配的时候。 请帮助。
答案 0 :(得分:9)
怎么样:
library(dplyr)
d %>%
group_by(Name) %>%
mutate(cA=cumsum(ifelse(!is.na(Fund) & Fund=="A",SalesAmount,0))) %>%
mutate(cB=cumsum(ifelse(!is.na(Fund) & Fund=="B",SalesAmount,0)))
输出:
Source: local data frame [10 x 8]
Groups: Name
Name Event SalesAmount Fund Cum.A.desired. Cum.B.desired. cA cB
1 John Webinar NA NA NA NA 0 0
2 John Sale 1000 A 1000 NA 1000 0
3 John Sale 2000 B 1000 2000 1000 2000
4 John Sale 3000 A 4000 2000 4000 2000
5 John Email NA NA 4000 2000 4000 2000
6 Tom Webinar NA NA NA NA 0 0
7 Tom Sale 1000 A 1000 NA 1000 0
8 Tom Sale 2000 B 1000 2000 1000 2000
9 Tom Sale 3000 A 4000 2000 4000 2000
10 Tom Email NA NA 4000 2000 4000 2000
如果需要,之后可以用NA
替换结果列中的零:
result$cA[result$cA==0] <- NA
result$cB[result$cB==0] <- NA
您的输入数据集:
d <- structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L), .Label = c("John", "Tom"), class = "factor"), Event = structure(c(3L, 2L, 2L, 2L, 1L, 3L, 2L, 2L, 2L, 1L), .Label = c("Email", "Sale", "Webinar"), class = "factor"), SalesAmount = c(NA, 1000L, 2000L, 3000L, NA, NA, 1000L, 2000L, 3000L, NA), Fund = structure(c(NA, 1L, 2L, 1L, NA, NA, 1L, 2L, 1L, NA), .Label = c("A", "B"), class = "factor"), Cum.A.desired. = c(NA, 1000L, 1000L, 4000L, 4000L, NA, 1000L, 1000L, 4000L, 4000L), Cum.B.desired. = c(NA, NA, 2000L, 2000L, 2000L, NA, NA, 2000L, 2000L, 2000L)), .Names = c("Name", "Event", "SalesAmount", "Fund", "Cum.A.desired.", "Cum.B.desired." ), class = "data.frame", row.names = c(NA, -10L))
答案 1 :(得分:3)
这是一种使用zoo
和data.table
来推广更多资金的方法:
# prep
require(data.table)
require(zoo)
setDT(d)
d[,Fund:=as.character(Fund)] # because factors are the worst
uf <- unique(d[Event=="Sale"]$Fund) # collect set of funds
首先,在相关的观察子集中分配累计销售额:
for (f in uf) d[(Event=="Sale"&Fund==f),paste0('c',f):=cumsum(SalesAmount),by=Name]
然后,进行最后一次观察:
d[,paste0('c',uf):=lapply(.SD,na.locf,na.rm=FALSE),.SDcols=paste0('c',uf),by=Name]
答案 2 :(得分:2)
您可以将@ Marat的答案略微缩短为一个targetSdkVersion
来缩短@ Marat的答案:
mutate