聚合多个功能答案

【问题标题】：Multiple functions in aggregate聚合多个功能
【发布时间】：2015-12-10 05:36:03
【问题描述】：

有可能从以下数据框df1

 Branch Loan_Amount TAT
      A         100 2.0
      A         120 4.0
      A         300 9.0
      B         150 1.5
      B         200 2.0

我可以使用聚合函数将以下输出作为数据框 df2

 Branch Number_of_loans Loan_Amount Total_TAT
      A               3         520      15.0
      B               2         350       3.5

我知道我可以使用 nrow 来计算 number_of_loans 并合并，但我正在寻找更好的方法。

【问题讨论】：

为什么aggregate对你来说不是一个好方法？

标签： r aggregate

【解决方案1】：

基础包：

df1 <- aggregate(.~ Branch, df, FUN = "sum")
df2 <- setNames(aggregate(Loan_Amount~Branch, df, length)[2], c("Number_of_loans"))
cbind(df1, df2)

输出

  Branch Loan_Amount  TAT Number_of_loans
1      A         520 15.0               3
2      B         350  3.5               2

包sqldf:

library(sqldf)
sqldf("SELECT Branch, COUNT(Loan_Amount) Number_of_loans, SUM(Loan_Amount) Loan_Amount, SUM(TAT) TAT 
      FROM df 
      GROUP BY Branch")

输出

  Branch Number_of_loans Loan_Amount  TAT
1      A               3         520 15.0
2      B               2         350  3.5

数据

df <- structure(list(Branch = structure(c(1L, 1L, 1L, 2L, 2L), .Label = c("A", 
"B"), class = "factor"), Loan_Amount = c(100L, 120L, 300L, 150L, 
200L), TAT = c(2, 4, 9, 1.5, 2)), .Names = c("Branch", "Loan_Amount", 
"TAT"), class = "data.frame", row.names = c(NA, -5L))

【讨论】：

【解决方案2】：

使用 dplyr，您可以这样做：

library(dplyr)
group_by(d,Branch) %>% 
  summarize(Number_of_loans = n(),
            Loan_Amount = sum(Loan_Amount),
            TAT = sum(TAT))

输出

Source: local data frame [2 x 4]

  Branch Number_of_loans Loan_Amount   TAT
  (fctr)           (int)       (int) (dbl)
1      A               3         520  15.0
2      B               2         350   3.5

数据

d <- read.table(text="Branch Loan_Amount TAT
A         100 2.0
A         120 4.0
A         300 9.0
B         150 1.5
B         200 2.0",head=TRUE)

【讨论】：

【解决方案3】：

使用 data.table

library(data.table)
setDT(df)[,list(Number_of_loans=.N, 
                Loan_Amount    =sum(Loan_Amount), 
                Total_TAT      =sum(TAT)), by=Branch]
#    Branch Number_of_loans Loan_Amount Total_TAT
# 1:      A               3         520      15.0
# 2:      B               2         350       3.5

【讨论】：

【解决方案4】：

这很笨拙且效率低下，但它有效且有趣（它使用aggregate()）：

d <- read.table(text="Branch Loan_Amount TAT
A         100 2.0
A         120 4.0
A         300 9.0
B         150 1.5
B         200 2.0",head=TRUE)

library(stringr)
df = aggregate(.~Branch, data=d, FUN=function(x) paste0(length(x), '|',sum(x)))
df_ = cbind(str_split_fixed(df$Loan_Amount, '|', 4)[,c(2,4)], str_split_fixed(df$TAT, '|', 4)[,4])
df_ = apply(df_, 2, as.numeric)
colnames(df_) = c('Number_of_loans','Loan_Amount','Total_TAT')
cbind(df[,'Branch',drop=F], df_)

产生所需的data.frame：

  Branch Number_of_loans Loan_Amount Total_TAT
1      A               3         520      15.0
2      B               2         350       3.5

【讨论】：