我想按组计算差异。虽然我在 SO 上提到了 R: Function “diff” over various groups 线程，但由于未知原因，我无法找到区别。我尝试了三种方法:a) spread b) dplyr::mutate with base::diff() c) data.表与base::diff()。虽然 a) 有效，但我不确定如何使用 b) 和 c) 解决此问题。
数据说明:
我有该产品每年的收入数据。我将 >= 2013 年分类为周期 2(称为 P2)，将 <2013 年分类为周期 1(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 [有效]
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:NA 在 Diff 列中生成。]
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:它在 Diff 列中生成所有零。]
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？我是 R 的新手，所以如果我的工作听起来太基础，我深表歉意。我仍在学习这门语言。
最佳答案
我们可以用 data.table 来做到这一点。将 'data.frame' 转换为 'data.table' (setDT(Test_File))，按 'Name' 和 'Name' 的 run-length-id 分组，得到 sum 的 'Revenue'，使用 dcast 将其 reshape 为 'wide' 格式，获取 'P2' 和 'P1' 之间的差异并分配 (:=)它到“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'Name'中的相邻元素是否相同('grp')创建一个分组列，然后聚合 'Revenue' 通过 'Name' 和 'grp' 找到 sum，创建一个序列列，reshape 到 'wide' 和 转换数据集以创建“差异”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
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
根据 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
或者用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)
关于r - 使用 dplyr 和 data.table 按组区分，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43627015/