Question

我有一个如下数据集：

date,       time,product,shop_id

20140104    900 Banana  18  
20140104    900 Banana  19  
20140104    924 Banana  18  
20140104    929 Banana  18  
20140104    932 Banana  20  
20140104    948 Banana  18

我需要使用不同的product和不同的shop_id

来提取观察结果

所以，我需要按product+shop_id

对观察进行分组

这是我的代码：

library(plyr)
  d_ply( shop, .( product,shop_id  ),table   )
 print(p)

不幸的是，它会打印null

数据集：

date=c(20140104,20140104,20140104,20140104,20140104)
time=c(924  ,900,854,700,1450)
product=c(Banana    ,Banana ,Banana ,Banana ,Banana)
shop_id=c(18,18,18,19,20)
shop<-data.frame(date=date,time=time,product=product,shop_id=shop_id)

输出应为

         date,    time, product, shop_id


        20140104    900 Banana  19  
        20140104    932 Banana  20  
        20140104    948 Banana  18

Answer 1

我们可以做到

library(tidyverse)
shop %>%
   group_by(product, shop_id) %>% 
   mutate(n = n()) %>% 
   group_by(time) %>% 
   arrange(n) %>%
   slice(1) %>%
   group_by(product, shop_id) %>% 
   arrange(-time) %>%
   slice(1) %>% 
   select(-n) %>%
   arrange(time)
#      date  time product shop_id
#      <int> <int>   <chr>   <int>
#1 20140104   900  Banana      19
#2 20140104   932  Banana      20
#3 20140104   948  Banana      18

Answer 2

要仅采用第一个独特组合，只需使用包stats中的aggregate：

> aggregate(shop, by=list(shop$product, shop$shop_id), FUN=function(x){x[1]})

Group.1 Group.2     date time product shop_id
1  Banana      18 20140104  924  Banana      18
2  Banana      19 20140104  700  Banana      19
3  Banana      20 20140104 1450  Banana      20

说明：我的FUN=function(x){x[1]}在碰撞时仅采用第一个元素

删除“Group.1”，“Group.2”或其他列：

> res <- aggregate(shop, by=list(shop$product, shop$shop_id), FUN=function(x){x[1]})
> res[ , !(names(res) %in% c("Group.1", "Group.2"))]
      date time product shop_id
1 20140104  924  Banana      18
2 20140104  700  Banana      19
3 20140104 1450  Banana      20

P.S。您提供的数据集与您需要的示例不一致，因此这就是数字存在差异的原因。

P.S.2如果您想在发生碰撞时获取所有数据：

> aggregate(shop, by=list(shop$product, shop$shop_id), FUN="identity")
  Group.1 Group.2                         date          time product    shop_id
1  Banana      18 20140104, 20140104, 20140104 924, 900, 854 1, 1, 1 18, 18, 18
2  Banana      19                     20140104           700       1         19
3  Banana      20                     20140104          1450       1         20

如果您想标记碰撞：

> aggregate(shop, by=list(shop$product, shop$shop_id), FUN=function(x){if (length(x) > 1) NA else x})
  Group.1 Group.2     date time product shop_id
1  Banana      18       NA   NA      NA      NA
2  Banana      19 20140104  700       1      19
3  Banana      20 20140104 1450       1      20

如果要排除非唯一行：

> res <- aggregate(shop, by=list(shop$product, shop$shop_id), FUN=function(x){if (length(x) > 1) NULL else x})

> res[res$product != "NULL", !(names(res) %in% c("Group.1", "Group.2"))]
      date time product shop_id
2 20140104  700       1      19
3 20140104 1450       1      20

如果你想避免从String到Int（对于产品）的强制，使用“”/“NULL”/“NA”而不是NULL / NA。

Answer 3

可以使用dplyr完成，如下所示：

# create the sample dataset
date=c(20140104,20140104,20140104,20140104,20140104)
time=c(924  ,900,854,700,1450)
product=c("Banana","Banana","Banana","Banana","Banana")
shop_id=c(18,18,18,19,20)
shop<-data.frame(date=date,time=time,product=product,shop_id=shop_id)

# load a dplyr library
library(dplyr)

# take shop data
shop %>%
        # group by product, shop id, date
        group_by(product, shop_id, date) %>%
        # for each such combination, find the earliest time
        summarise(time = min(time)) %>%
        # group by product, shop id
        group_by(product, shop_id) %>%
        # for each combination of product & shop id
        # return the earliest date and time recorded on the earliest date
        summarise(date = min(date), time = time[date == min(date)])

通过对某些特征进行分组来进行子集观察

3 个答案: