根据分类变量分配日期

Question

>我有一组以下形式的Payment Date：

ID  Payment Date   
1   18-01-01
1   18-02-03
2   18-04-03
2   18-05-08
2   18-06-06
3   17-12-23
3   18-01-22
3   18-02-24
4   17-11-09
4   18-12-06

我想添加一列，Activation Date，作为每个ID的最早付款日期，例如：

ID  Payment Date   Activation Date
1   18-01-01       18-01-01
1   18-02-03       18-01-01
2   18-04-03       18-04-03
2   18-05-08       18-04-03
2   18-06-06       18-04-03
3   17-12-23       17-12-23
3   18-01-22       17-12-23
3   18-02-24       17-12-23
4   17-11-09       17-11-09
4   18-12-06       17-11-09

我想知道与其进入循环并逐个处理每个 ID，不如有一种更聪明的方法可以做到这一点。

Answer 1

df = read.table(text = "
ID  PaymentDate   
1   18-01-01
1   18-02-03
2   18-04-03
2   18-05-08
2   18-06-06
3   17-12-23
3   18-01-22
3   18-02-24
4   17-11-09
4   18-12-06
", header=T)
library(dplyr)
library(lubridate)
df %>%
group_by(ID) %>%
mutate(ActivationDate = min(ymd(PaymentDate))) %>%
ungroup()
# # A tibble: 10 x 3
#     ID PaymentDate ActivationDate
#   <int> <fct>       <date>        
# 1     1 18-01-01    2018-01-01    
# 2     1 18-02-03    2018-01-01    
# 3     2 18-04-03    2018-04-03    
# 4     2 18-05-08    2018-04-03    
# 5     2 18-06-06    2018-04-03    
# 6     3 17-12-23    2017-12-23    
# 7     3 18-01-22    2017-12-23    
# 8     3 18-02-24    2017-12-23    
# 9     4 17-11-09    2017-11-09    
#10     4 18-12-06    2017-11-09 

假设您的数据集已经订购，并且您不想使用Date格式，则可以使用

df %>%
group_by(ID) %>%
mutate(ActivationDate = first(PaymentDate)) %>%
ungroup()

Answer 2

使用data.table的解决方案

数据：

df1<-
fread("ID  Payment
1   18-01-01
1   18-02-03
2   18-04-03
2   18-05-08
2   18-06-06
3   17-12-23
3   18-01-22
3   18-02-24
4   17-11-09
4   18-12-06") %>% setDF

法典：

data.table::setDT(df1)[,Activation := Payment[1],by="ID"][]

结果：

#   ID  Payment Activation
#1:  1 18-01-01   18-01-01
#2:  1 18-02-03   18-01-01
#3:  2 18-04-03   18-04-03
#4:  2 18-05-08   18-04-03
#5:  2 18-06-06   18-04-03
#6:  3 17-12-23   17-12-23
#7:  3 18-01-22   17-12-23
#8:  3 18-02-24   17-12-23
#9:  4 17-11-09   17-11-09
#10:  4 18-12-06   17-11-09

---

快速提示：

• 切勿再在列名中使用"空格">
• 使用下划线或驼峰大小写。 例如：payment_date、paymentDate

Answer 3

使用sqldf：

您的数据集：

df=read.table(text="ID  PaymentDate   
1   18-01-01
1   18-02-03
2   18-04-03
2   18-05-08
2   18-06-06
3   17-12-23
3   18-01-22
3   18-02-24
4   17-11-09
4   18-12-06",header=T)

法典

# we can first find the minimum PaymentDate using the inner query and then
# populate the data.frame using the inner query
sqldf("select a.ID,a.PaymentDate, b.ActivationDate from df as a JOIN 
(select ID,min(PaymentDate) as ActivationDate from df group by ID) as b where a.ID=b.ID")

输出：

ID PaymentDate ActivationDate
1   1    18-01-01       18-01-01
2   1    18-02-03       18-01-01
3   2    18-04-03       18-04-03
4   2    18-05-08       18-04-03
5   2    18-06-06       18-04-03
6   3    17-12-23       17-12-23
7   3    18-01-22       17-12-23
8   3    18-02-24       17-12-23
9   4    17-11-09       17-11-09
10  4    18-12-06       17-11-09