Question

我有一个这样的数据框：

    data.frame(home=c("A","B","C","A","C"),weight=c(0.1,0.25,0.36,0.14,0.2),region=c("north","south","east","North","south"))

Home Weigth  region  
A     0.1     North      
B     0.25    South    
C     0.36    East   
A     0.14    North
C     0.2     South

我希望聚合我的data.frame跟随两个因子变量并总和第三个。结果将给出：

    data.frame(home=c("A","B","C"),north=c(0.24,0,0),south=c(0,0.25,0.2),east=c(0.36,0,0))

Home North  South  East
A     0.24   0      0
B     0      0.25    0
C     0      0.2    0.36

我正在尝试使用像聚合这样的快速简单的功能，但也许唯一的解决方案是使用我想要的东西手动制作data.frame

Answer 1

数据

df <- data.frame( home = c("A", "B", "C", "A", "C"), weight = c(0.1, 0.25, 0.36, 0.14, 0.2), region = c("north", "south", "east", "North", "south") )

的 tidyr

library(tidyr) spread(df, region, weight, fill = 0)

的 reshape2

library(reshape2) dcast(df, home ~ region, value.var = "weight", fill = 0)

的碱

# xtabs xtabs(weight ~ home + region, data = df) # reshape df_wide <-reshape(df, idvar ='home', timevar ='region', direction ='wide') df_wide[is.na(df_wide)] <- 0

<强>输出

home east north North south 1 A 0.00 0.1 0.14 0.00 2 B 0.00 0.0 0.00 0.25 3 C 0.36 0.0 0.00 0.20

Answer 2

基本上有两个步骤，（1）总和; （2）将结果转化为双向表

library(dplyr)
df <-  data.frame(home=c("A","B","C","A","C"),weight=c(0.1,0.25,0.36,0.14,0.2),region=c("north","south","east","North","south"))
df$region <- Hmisc::capitalize(as.character(df$region))

df_sum <- df %>% group_by(home, region) %>% summarize(weight_sum = sum(weight, na.rm=TRUE))

reshape2::dcast(df_sum, home ~ region, function(V) sum(V, na.rm=TRUE))

第二笔金额是无关的，并非真的有必要，我在这里只是为了避免将NA转换为0的额外步骤。

Answer 3

我想这样做，h01是你想要的结果

x00<-data.frame(home=c("A","B","C","A","C"),weight=c(0.1,0.25,0.36,0.14,0.2),
                region=c("north","South","East","North","South"),stringsAsFactors = F)
x00$region<-tolower(x00$region)
x01<-ddply(x00,.(region,home),summarize,result=sum(weight))
h01<-data.frame(north=c(0,0,0),south=c(0,0,0),east=c(0,0,0),row.names = c("A","B","C"))
for (x in 1:nrow(x01)){
  h01[x01$home[x],x01$region[x]]=x01$result[x]
}

h01$Home=row.names(h01)
row.names(h01)<-c()

r聚合多因子变量

3 个答案: