命名水果的两列上的自定义聚合

Question

我希望按名称聚合数据框的两列，具体如下：

• 通过专门汇总两列parts和fruits
来删除结果中的parts列
• 虽然Apple，Banana和Strawberry的parts值并不重要，所有内容都已汇总，但Grape和Kiwi的parts值应该成为新的fruits名称< / LI>
• 结果（在底部）应该有8个聚合行而不是20个

第一眼看上去听起来很简单，但经过数小时的反复试验后，我找不到任何有用的解决方案。这是一个例子：

theDF <- data.frame(dates = as.Date(c(today()+20)),
    fruits = c("Apple","Apple","Apple","Apple","Banana","Banana","Banana","Banana",
      "Strawberry","Strawberry","Strawberry","Strawberry","Grape", "Grape",
      "Grape","Grape", "Kiwi","Kiwi","Kiwi","Kiwi"),
    parts = c("Big Green Apple","Apple2","Blue Apple","XYZ Apple4",
      "Yellow Banana1","Small Banana","Banana3","Banana4",
      "Red Small Strawberry","Red StrawberryY","Big Strawberry",
       "StrawberryZ","Green Grape", "Blue Grape", "Blue Grape",
       "Blue Grape","Big Kiwi","Small Kiwi","Big Kiwi","Middle Kiwi"),
    stock = as.vector(sample(1:20)) )      

当前数据框：

所需的输出：

Answer 1

我们可以使用data.table。如果有一些模式，比如结尾字符是大写字母或数字中的数字＆＃39;要删除的列，我们可以使用sub来执行该操作，并将其用作分组变量以及“日期”。并获得＆＃39; stock＆＃39;的sum。

library(data.table)
setDT(theDF)[,.(stock = sum(stock)) , .(dates, fruits = sub("([0-9]|[A-Z])$", "", parts))]
#        dates      fruits stock
#1: 2016-06-19       Apple    46
#2: 2016-06-19      Banana    35
#3: 2016-06-19  Strawberry    38
#4: 2016-06-19 Green Grape    12
#5: 2016-06-19  Blue Grape    21
#6: 2016-06-19    Big Kiwi    37
#7: 2016-06-19  Small Kiwi    14 
#8: 2016-06-19 Middle Kiwi     7

或者使用dplyr，我们可以类似地实施相同的方法。

library(dplyr)
theDF %>%
    group_by(dates, fruits = sub('([0-9]|[A-Z])$', '', parts)) %>% 
    summarise(stock = sum(stock))

更新

如果没有模式且仅基于手动识别“水果”中的元素，则创建vector元素，使用%chin%获取＆＃中的逻辑索引39;我＆＃39;，分配（:=）＆＃39;部分＆＃39;对应于＆＃39; i＆＃39;到了水果＆＃39;然后通过＆＃39; date＆＃39;，＆＃39; fruit＆＃39;来做小组。并获得＆＃39; stock＆＃39;的sum。

setDT(theDF)[as.character(fruits) %chin% c("Grape", "Kiwi"),
          fruits := parts][, .(stock = sum(stock)), .(dates, fruits)]

数据

theDF <- structure(list(dates = structure(c(16971, 16971, 16971, 16971, 
16971, 16971, 16971, 16971, 16971, 16971, 16971, 16971, 16971, 
16971, 16971, 16971, 16971, 16971, 16971, 16971), class = "Date"), 
    fruits = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 5L, 
    5L, 5L, 5L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L), .Label = c("Apple", 
    "Banana", "Grape", "Kiwi", "Strawberry"), class = "factor"), 
    parts = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 14L, 
    15L, 16L, 16L, 11L, 10L, 10L, 10L, 9L, 13L, 9L, 12L), .Label = c("Apple1", 
    "Apple2", "Apple3", "Apple4", "Banana1", "Banana2", "Banana3", 
    "Banana4", "Big Kiwi", "Blue Grape", "Green Grape", "Middle Kiwi", 
    "Small Kiwi", "StrawberryX", "StrawberryY", "StrawberryZ"
    ), class = "factor"), stock = c(8, 19, 15, 4, 6, 18, 1, 10, 
    9, 16, 11, 2, 12, 13, 5, 3, 17, 14, 20, 7)), .Names = c("dates", 
"fruits", "parts", "stock"), row.names = c(NA, -20L), class = "data.frame")

Answer 2

另一种方法是在第一步中创建适当的分组变量，然后使用您希望按组汇总的任何方法。在这里，我使用dplyr，您可以使用其他人（data.table等）。

library(dplyr)
theDF <- data.frame(fruits, parts, stock, stringsAsFactors = F)
theDF$fruits <- with(theDF, ifelse(fruits=="Kiwi" | fruits=="Grape", parts, fruits))

theDF %>% group_by(fruits) %>% summarise(stock = sum(stock))

Source: local data frame [8 x 2]

       fruits stock
        (chr) (int)
1       Apple    34
2      Banana    35
3    Big Kiwi    26
4  Blue Grape    32
5 Green Grape     7
6 Middle Kiwi    12
7  Small Kiwi    19
8  Strawberry    45

我没有找到函数today()，所以我跳过了日期列。您可以将date插入到分组中，然后将其添加回group_by(fruits, date)，以便将其保留在所需的输出中。