基于对一列或多列进行分组来计算日期之间的差异

Question

我的数据集示例如下：

| id |       Date | Buyer    |
|:--:|-----------:|----------|
|  9 | 11/29/2018 | Jenny    |
|  9 | 11/29/2018 | Jenny    |
|  9 | 11/29/2018 | Jenny    |
| 4  | 5/30/2018  | Chang    |
| 4  | 7/4/2018   | Chang    |
| 4  | 8/17/2018  | Chang    |
| 5  | 5/25/2018  | Chunfei  |
| 5  | 2/13/2019  | Chunfei  |
| 5  | 2/16/2019  | Chunfei  |
| 5  | 2/16/2019  | Chunfei  |
| 5  | 2/23/2019  | Chunfei  |
| 5  | 2/25/2019  | Chunfei  |
| 8  | 2/28/2019  | Chunfei  |
| 8  | 2/28/2019  | Chunfei  |

我对这个数据集有两个问题:

< li >我需要计算日期之间的差异，但此差异将基于“买方”和“id”分组来计算，这意味着，买方“Jenny”和Id“9”的日期差异将是一个组，Id为“4”的买方“Chang”将是另一个组，Id为“5”的买方“Chunfei”将是另一个组，Id为“8”的“Chunfei”将是另一个组。因此，输出将是:

| id |       Date | Buyer_id | Diff |
|:--:|-----------:|----------|------|
|  9 | 11/29/2018 | Jenny    | NA   |
|  9 | 11/29/2018 | Jenny    | 0    |
|  9 | 11/29/2018 | Jenny    | 0    |
| 4  | 5/30/2018  | Chang    | NA   |
| 4  | 7/4/2018   | Chang    | 35   |
| 4  | 8/17/2018  | Chang    | 44   |
| 5  | 5/25/2018  | Chunfei  | NA   |
| 5  | 2/13/2019  | Chunfei  | 264  |
| 5  | 2/16/2019  | Chunfei  | 3    |
| 5  | 2/16/2019  | Chunfei  | 0    |
| 5  | 2/23/2019  | Chunfei  | 7    |
| 5  | 2/25/2019  | Chunfei  | 2    |
| 8  | 2/28/2019  | Chunfei  | NA   |
| 8  | 2/28/2019  | Chunfei  | 0    |

问题是我不明白为什么group_by不起作用。下面的代码减去连续的行，而不是为同一个买家和id分组，然后减去。

df=data.frame(id=c("9","9","9","4","4","4","5","5","5","5","5","5","8","8"), 
              Date=c("11/29/2018","11/29/2018","11/29/2018","5/30/2018","7/4/2018", 
                      "8/17/2018","5/25/2018","2/13/2019","2/16/2019","2/16/2019","2/23/2019",
                      "2/25/2019","2/28/2019","2/28/2019"),Buyer=c("Jenny","Jenny","Jenny",
                      "Chang","Chang","Chang","Chunfei","Chunfei","Chunfei","Chunfei","Chunfei",
                      "Chunfei","Chunfei","Chunfei"))
df$id=as.numeric(as.character(df$id))
df$Date=as.Date(df$Date, "%m/%d/%Y")
df$Buyer=as.character(df$Buyer)

df1=df %>% group_by(Buyer,id) %>%
  mutate(diff=as.numeric(difftime(Date,lag(Date),units='days')))

我们可以在最终输出中屏蔽“diff”列，它应该如下所示:

| id |    Date    | Buyer_id |
|----|:----------:|---------:|
| 9  | 11/29/2018 |    Jenny |
| 9  | 11/29/2018 |    Jenny |
| 9  | 11/29/2018 |    Jenny |
| 5  | 2/13/2019  | Chunfei  |
| 5  | 2/16/2019  | Chunfei  |
| 5  | 2/16/2019  | Chunfei  |
| 5  | 2/23/2019  | Chunfei  |
| 5  | 2/25/2019  | Chunfei  |
| 8  | 2/28/2019  | Chunfei  |
| 8  | 2/28/2019  | Chunfei  |

Answer 1

我们可以使用< code>diff减去< code>Date并选择至少有一个值小于5天的组。

library(dplyr)

df %>%
  group_by(id, Buyer) %>%
  filter(any(diff(Date) <= 5))

#      id Date       Buyer  
#   <dbl> <date>     <chr>  
# 1     9 2018-11-29 Jenny  
# 2     9 2018-11-29 Jenny  
# 3     9 2018-11-29 Jenny  
# 4     5 2018-05-25 Chunfei
# 5     5 2019-02-13 Chunfei
# 6     5 2019-02-16 Chunfei
# 7     5 2019-02-16 Chunfei
# 8     5 2019-02-23 Chunfei
# 9     5 2019-02-25 Chunfei
#10     8 2019-02-28 Chunfei
#11     8 2019-02-28 Chunfei

重读这个问题后，我想你可能不会< code>filter整个组，而只是那些有5天差异的行。我们可以得到< code>diff值小于5的索引，并选择它以前的索引。

df %>%
  group_by(id, Buyer) %>%
  mutate(diff = c(NA, diff(Date))) %>%
  slice({i1 <- which(diff <= 5); unique(c(i1, i1-1))}) %>%
  select(-diff)

#      id Date       Buyer  
#   <dbl> <date>     <chr>  
# 1     5 2019-02-16 Chunfei
# 2     5 2019-02-16 Chunfei
# 3     5 2019-02-25 Chunfei
# 4     5 2019-02-13 Chunfei
# 5     5 2019-02-23 Chunfei
# 6     8 2019-02-28 Chunfei
# 7     8 2019-02-28 Chunfei
# 8     9 2018-11-29 Jenny  
# 9     9 2018-11-29 Jenny  
#10     9 2018-11-29 Jenny  

数据

df <- structure(list(id = c(9, 9, 9, 4, 4, 4, 5, 5, 5, 5, 5, 5, 8, 
8), Date = structure(c(17864, 17864, 17864, 17681, 17716, 17760, 
17676, 17940, 17943, 17943, 17950, 17952, 17955, 17955), class = "Date"), 
Buyer = c("Jenny", "Jenny", "Jenny", "Chang", "Chang", "Chang", 
"Chunfei", "Chunfei", "Chunfei", "Chunfei", "Chunfei", "Chunfei", 
"Chunfei", "Chunfei")), row.names = c(NA, -14L), class = "data.frame")