R Dataframe过滤：使用基于时间因素的唯一或重复功能

Posted 2023-03-29

【中文标题】R Dataframe过滤：使用基于时间因素的唯一或重复功能

【英文标题】：R Dataframe filtering: Using unique or duplicate function based on time factor

【发布时间】：2021-11-17 21:21:47

【问题描述】：

我正在尝试过滤贷款数据的数据框，但如果每个月度报告仍然未偿还贷款，则会重复贷款，或者如果已付款则放弃贷款（不能只使用最新的月度报告）。我想通过贷方过滤贷款的唯一到期日期，并删除重复项并仅保留报告日期的最新数据。以下是数据示例：

df <- data.frame(Reporting.date=c("6/30/2020","6/30/2020","6/30/2020","8/31/2021","8/31/2021"
                                  ,"8/31/2021","6/30/2020","7/31/2021","5/31/2020","12/31/2020")
                 , Lender.name=c("Lender1","Lender1","Lender1","Lender1","Lender1","Lender1"
                                 ,"Lender1","Lender1","Lender2","Lender2")
                 , Date.of.maturity=c("6/20/2025","6/20/2025","6/20/2025","6/20/2025","6/20/2025"
                                      ,"6/20/2025","6/30/2022","6/30/2022","5/15/2024","5/15/2024")
                 , Loan.amount=c(13129474,14643881,44935677,13129474,14643881,44935677
                                 ,150000,150000,2750000,2750000))

从示例数据中可以看出，Lender1 有 2 个唯一的到期日。第一个日期有 3 笔贷款在 2 个报告日期重复，第二个到期日有 1 笔贷款重复。我想删除重复项以保留最新的报告数据。我希望之后得到一个看起来像这样的数据框：

Reporting.date	Lender.name	Date.of.maturity	Loan.amount
8/31/2021	Lender1	6/20/2025	13129474
8/31/2021	Lender1	6/20/2025	14643881
8/31/2021	Lender1	6/20/2025	44935677
7/31/2021	Lender1	6/30/2022	150000
12/31/2020	Lender2	5/15/2024	2750000

【参考方案1】：

您需要将Reporting.date 转换为日期格式，可以是mutate（和我一样），也可以直接转换为filter。

library(tidyverse)

df %>%
  mutate(Reporting.date = as.Date(Reporting.date, format = '%m/%d/%Y')) %>%
  group_by(Lender.name, Date.of.maturity, Loan.amount) %>%
  filter(Reporting.date == max(Reporting.date)) %>%
  ungroup()

【参考方案2】：

我们也可以通过arrange 来实现这一点

library(dplyr)
library(lubridate)
df %>%
  arrange(Lender.name, Date.of.maturity, Loan.amount, 
         desc(mdy(Reporting.date))) %>%
  group_by(Lender.name, Date.of.maturity, Loan.amount) %>%
  slice_head(n = 1) %>%
  ungroup

-输出

# A tibble: 5 x 4
  Reporting.date Lender.name Date.of.maturity Loan.amount
  <chr>          <chr>       <chr>                  <dbl>
1 8/31/2021      Lender1     6/20/2025           13129474
2 8/31/2021      Lender1     6/20/2025           14643881
3 8/31/2021      Lender1     6/20/2025           44935677
4 7/31/2021      Lender1     6/30/2022             150000
5 12/31/2020     Lender2     5/15/2024            2750000

【参考方案3】：

使用 subset、transform 和 ave 的基本 R 选项 -

subset(transform(df, Reporting.date = as.Date(Reporting.date, format = '%m/%d/%Y')), 
       Reporting.date == ave(Reporting.date, Lender.name, Date.of.maturity, FUN = max))

#   Reporting.date Lender.name Date.of.maturity Loan.amount
#4      2021-08-31     Lender1        6/20/2025    13129474
#5      2021-08-31     Lender1        6/20/2025    14643881
#6      2021-08-31     Lender1        6/20/2025    44935677
#8      2021-07-31     Lender1        6/30/2022      150000
#10     2020-12-31     Lender2        5/15/2024     2750000