我有以下数据框(df),尽管我认为我无法解决如何执行以下操作:
输入:
id business_id type date1 date2 date3
1 A1 Month 13/10/13 13/09/13 13/08/13
1 A1 Total Net Deposits 1500 951 190
1 A1 Month end Bank Balance 729 650 164
预期产出:
id business_id Month Total Net Deposits Month end Bank Balance
1 A1 13/10/13 1500 729
1 A1 13/09/13 951 650
1 A1 13/09/13 190 164
答案 0 :(得分:5)
这是一个tidyr选项:
.so
如果您的列存储为因子,则会出现警告,但您可以忽略它。
您可以使用基础R(统计信息)和其他一些库中的library(tidyr)
df %>%
gather(date, val, date1:date3) %>%
spread(key = type, val = val)
# id business_id date Month Month end Bank Balance Total Net Deposits
#1 1 A1 date1 13/10/13 729 1500
#2 1 A1 date2 13/09/13 650 951
#3 1 A1 date3 13/08/13 164 190
#Warning:
#attributes are not identical across measure variables; they will be dropped
,reshape2
,data.table
执行相同操作。
答案 1 :(得分:4)
我们可以使用base R
cbind(df1[1:2],setNames(as.data.frame(t(df1[-(1:3)])), df1$type))
# id business_id Month Total Net Deposits Month end Bank Balance
#date1 1 A1 13/10/13 1500 729
#date2 1 A1 13/09/13 951 650
#date3 1 A1 13/08/13 190 164
答案 2 :(得分:3)
library(reshape2)
df.m<-melt(x,id.var=c("id", "business_id","type"))
dcast(df.m,id+business_id+variable~type)
如果你愿意的话,你可以摆脱变量“变量”。
id business_id variable Month MonthendBankBalance TotalNetDeposits
1 1 A1 date1 13/10/13 729 1500
2 1 A1 date2 13/09/13 650 951
3 1 A1 date3 13/08/13 164 190