Question

我想按组计算差异。虽然我在SO上提到R: Function “diff” over various groups帖子，但由于未知原因，我无法找到差异。我尝试了三种方法：a）spread b）dplyr::mutate base::diff() c）data.table base::diff()。虽然a）有效，但我不确定如何使用b）和c）来解决这个问题。

有关数据的说明： 我有按年收入的产品数据。我将年份＆gt; = 2013分类为期间2（称为P2），年份＆lt; 2013年为第1期（称为P1）。

示例数据：

dput(Test_File)
structure(list(Ship_Date = c(2010, 2010, 2012, 2012, 2012, 2012, 
2017, 2017, 2017, 2016, 2016, 2016, 2011, 2017), Name = c("Apple", 
"Apple", "Banana", "Banana", "Banana", "Banana", "Apple", "Apple", 
"Apple", "Banana", "Banana", "Banana", "Mango", "Pineapple"), 
    Revenue = c(5, 10, 13, 14, 15, 16, 25, 25, 25, 1, 2, 4, 5, 
    7)), .Names = c("Ship_Date", "Name", "Revenue"), row.names = c(NA, 
14L), class = "data.frame")

预期输出

dput(Diff_Table)
structure(list(Name = c("Apple", "Banana", "Mango", "Pineapple"
), P1 = c(15, 58, 5, NA), P2 = c(75, 7, NA, 7), Diff = c(60, 
-51, NA, NA)), .Names = c("Name", "P1", "P2", "Diff"), class = "data.frame", row.names = c(NA, 
-4L))

这是我的代码：

方法1：使用spread [作品]

data.table::setDT(Test_File)
cutoff<-2013
Test_File[Test_File$Ship_Date>=cutoff,"Ship_Period"]<-"P2"
Test_File[Test_File$Ship_Date<cutoff,"Ship_Period"]<-"P1"

Diff_Table<- Test_File %>%
  dplyr::group_by(Ship_Period,Name) %>%
  dplyr::mutate(Revenue = sum(Revenue)) %>%
  dplyr::select(Ship_Period, Name,Revenue) %>%
  dplyr::ungroup() %>%
  dplyr::distinct() %>%
  tidyr::spread(key = Ship_Period,value = Revenue) %>% 
  dplyr::mutate(Diff = `P2` - `P1`)

方法2：使用dplyr [不起作用：在Diff列中生成新内容。]

Diff_Table<- Test_File %>%
  dplyr::group_by(Ship_Period,Name) %>%
  dplyr::mutate(Revenue = sum(Revenue)) %>%
  dplyr::select(Ship_Period, Name,Revenue) %>%
  dplyr::ungroup() %>%
  dplyr::distinct() %>%
  dplyr::arrange(Name,Ship_Period, Revenue) %>%
  dplyr::group_by(Ship_Period,Name) %>%
  dplyr::mutate(Diff = diff(Revenue))

方法3：使用data.table [不起作用：它会在Diff列中生成全部零。]

Test_File[,Revenue1 := sum(Revenue),by=c("Ship_Period","Name")]
Diff_Table<-Test_File[,.(Diff = diff(Revenue1)),by=c("Ship_Period","Name")]

问题：有人可以帮我解决上面的方法2和方法3吗？我对R很新，所以如果我的工作听起来太基础，我会道歉。我还在学习这门语言。

Answer 1

我们可以使用data.table执行此操作。将'data.frame'转换为'data.table'（setDT(Test_File)），按“名称”和“名称”的游程长度ID分组，获取“收入”的sum，使用dcast将其重新整理为“宽”格式，获取“P2”和“P1”之间的差异，并将其分配（:=）为“差异”

library(data.table)
dcast(setDT(Test_File)[, .(Revenue = sum(Revenue)),
   .(grp=rleid(Name), Name)], Name~ paste0("P", rowid(Name)), 
        value.var = "Revenue")[, Diff := P2 - P1][]
#        Name P1 P2 Diff
#1:     Apple 15 75   60
#2:    Banana 58  7  -51
#3:     Mango  5 NA   NA
#4: Pineapple  7 NA   NA

或者对于第三种情况，即base R，我们根据“姓名”中相邻元素是否相同（'grp'）创建分组列，然后aggregate'收入' 'by'Name'和'grp'找到sum，创建一个序列列，reshape将其设置为'wide'并transform数据集以创建'Diff'列

Test_File$grp <- with(Test_File, cumsum(c(TRUE, Name[-1]!=Name[-length(Name)])))
d1 <- aggregate(Revenue~Name +grp, Test_File, sum)
d1$Seq <- with(d1, ave(seq_along(Name), Name, FUN = seq_along))
transform(reshape(d1[-2], idvar = "Name", timevar = "Seq", 
            direction = "wide"), Diff = Revenue.2- Revenue.1)

tidyverse方法也可以使用

完成

library(dplyr)
library(tidyr)
Test_File %>% 
       group_by(grp = cumsum(c(TRUE, Name[-1]!=Name[-length(Name)])), Name)  %>%
       summarise(Revenue = sum(Revenue)) %>%
       group_by(Name) %>% 
       mutate(Seq = paste0("P", row_number()))  %>% 
       select(-grp) %>% 
       spread(Seq, Revenue) %>% 
       mutate(Diff = P2-P1)
 #Source: local data frame [4 x 4]
 #Groups: Name [4]

#      Name    P1    P2  Diff
#      <chr> <dbl> <dbl> <dbl>
#1     Apple    15    75    60
#2    Banana    58     7   -51
#3     Mango     5    NA    NA
#4 Pineapple     7    NA    NA

更新

基于OP的评论仅使用diff函数

library(data.table)
setDT(Test_File)[, .(Revenue = sum(Revenue)), .(Name, grp = rleid(Name))
  ][, .(P1 = Revenue[1L], P2 = Revenue[2L], Diff = diff(Revenue)) , Name]
#        Name P1 P2 Diff
#1:     Apple 15 75   60
#2:    Banana 58  7  -51
#3:     Mango  5 NA   NA
#4: Pineapple  7 NA   NA

或dplyr

Test_File %>% 
   group_by(grp = cumsum(c(TRUE, Name[-1]!=Name[-length(Name)])), Name)  %>%
   summarise(Revenue = sum(Revenue)) %>%
   group_by(Name) %>% 
   summarise(P1 = first(Revenue), P2 = last(Revenue)) %>%
   mutate(Diff = P2-P1)

Answer 2

这样做：

library("data.table")
setDT(Test_File)
T <- Test_File[, .(P=sum(Revenue)),by=.(Ship_Date, Name)]
T[Ship_Date>=2013][T[Ship_Date<2013][CJ(Name=T$Name, unique=TRUE), on="Name"], on="Name"][,`:=`(P1=i.P, P2=P, Diff=P-i.P)][] 
#    Ship_Date      Name  P i.Ship_Date i.P P1 P2 Diff
# 1:      2017     Apple 75        2010  15 15 75   60
# 2:      2016    Banana  7        2012  58 58  7  -51
# 3:        NA     Mango NA        2011   5  5 NA   NA
# 4:      2017 Pineapple  7          NA  NA NA  7   NA

或只有想要的列：

T[Ship_Date>=2013][T[Ship_Date<2013][CJ(Name=T$Name, unique=TRUE), on="Name"], on="Name"][,`:=`(P1=i.P, P2=P, Diff=P-i.P)][,.(Name, P1, P2, Diff)]
#         Name P1 P2 Diff
# 1:     Apple 15 75   60
# 2:    Banana 58  7  -51
# 3:     Mango  5 NA   NA
# 4: Pineapple NA  7   NA

以下是使用setnames()的变体：

setnames(T[Ship_Date>=2013][T[Ship_Date<2013][CJ(Name=T$Name, unique=TRUE), on="Name"], on="Name"], 
         c("P", "i.P"), c("P2", "P1"))[, Diff:=P2-P1][]

使用dplyr和data.table按组进行区分

2 个答案:

更新