根据日期按组比较值，并在值下降的位置创建新数据帧 [英] Comparing a value by group based on date and creating new dataframe where value dropped

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本文介绍了根据日期按组比较值，并在值下降的位置创建新数据帧的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个DF As Under

+-----+------+--------+--------------------+------+---------+
| ID1 | ID2  | DOC_NO |        DATE        | COST | CLIENT  |
+-----+------+--------+--------------------+------+---------+
| ABC | A123 |      1 | 2021-01-01 0:10:00 |   11 | ABC123  |
| DEF | B456 |      2 | 2021-01-01 0:10:00 |   12 | DEF256  |
| GHI | C789 |      3 | 2021-01-01 0:10:00 |   13 | GHI389  |
| JKL | D890 |      4 | 2021-01-01 0:10:00 |   14 | JKL490  |
| MNO | E012 |      5 | 2021-01-01 0:10:00 |   15 | MNO512  |
| ABC | A123 |      6 | 2021-01-01 0:15:00 |   11 | ABC623  |
| DEF | B456 |      7 | 2021-01-01 0:15:00 |   12 | DEF756  |
| GHI | C789 |      8 | 2021-01-01 0:15:00 |   13 | GHI889  |
| JKL | D890 |      9 | 2021-01-02 0:15:00 |   14 | JKL990  |
| MNO | E012 |     10 | 2021-01-03 0:15:00 |   15 | MNO1012 |
| ABC | A123 |     11 | 2021-01-03 0:20:00 |   10 | GHI890  |
| DEF | B456 |     12 | 2021-01-03 0:20:00 |   11 | JKL991  |
| GHI | C789 |     13 | 2021-01-03 0:20:00 |   12 | MNO1013 |
| JKL | D890 |     14 | 2021-01-03 0:20:00 |   13 | GHI891  |
| MNO | E012 |     15 | 2021-01-03 0:20:00 |   14 | JKL992  |
| ABC | A123 |     16 | 2021-01-03 0:20:00 |   12 | MNO1014 |
| DEF | B456 |     17 | 2021-01-03 0:20:00 |   13 | GHI892  |
| GHI | C789 |     18 | 2021-01-03 0:20:00 |   14 | JKL993  |
| JKL | D890 |     19 | 2021-01-03 0:20:00 |   15 | MNO1015 |
| MNO | E012 |     20 | 2021-01-03 0:20:00 |   16 | GHI893  |
| ABC | A123 |     21 | 2021-01-03 0:25:00 |   11 | ABC124  |
| DEF | B456 |     22 | 2021-01-03 0:25:00 |   12 | DEF257  |
| GHI | C789 |     23 | 2021-01-03 0:25:00 |   13 | GHI390  |
| JKL | D890 |     24 | 2021-01-03 0:25:00 |   14 | JKL491  |
| MNO | E012 |     25 | 2021-01-03 0:25:00 |   15 | MNO513  |
+-----+------+--------+--------------------+------+---------+
我要将ID1和ID2分组，并按DOC_NO和日期排列DF 我要创建一个新列REFERENCE_COST，其中REFERENCE_COST是相对于时间和DOC_NO排列的最高成本，这意味着如果成本随时间和DOC_NO增加，则较高的成本现在将被设置为REFERENCE_COST 因此，新的df将如下所示：

+-----+------+--------+--------------------+------+---------+----------+
| ID1 | ID2  | DOC_NO |        DATE        | COST | CLIENT  | REF_COST |
+-----+------+--------+--------------------+------+---------+----------+
| ABC | A123 |      1 | 2021-01-01 0:10:00 |   11 | ABC123  |       11 |
| DEF | B456 |      2 | 2021-01-01 0:10:00 |   12 | DEF256  |       12 |
| GHI | C789 |      3 | 2021-01-01 0:10:00 |   13 | GHI389  |       13 |
| JKL | D890 |      4 | 2021-01-01 0:10:00 |   14 | JKL490  |       14 |
| MNO | E012 |      5 | 2021-01-01 0:10:00 |   15 | MNO512  |       15 |
| ABC | A123 |      6 | 2021-01-01 0:15:00 |   11 | ABC623  |       11 |
| DEF | B456 |      7 | 2021-01-01 0:15:00 |   12 | DEF756  |       12 |
| GHI | C789 |      8 | 2021-01-01 0:15:00 |   13 | GHI889  |       13 |
| JKL | D890 |      9 | 2021-01-02 0:15:00 |   14 | JKL990  |       14 |
| MNO | E012 |     10 | 2021-01-03 0:15:00 |   15 | MNO1012 |       15 |
| ABC | A123 |     11 | 2021-01-03 0:20:00 |   10 | GHI890  |       11 |
| DEF | B456 |     12 | 2021-01-03 0:20:00 |   11 | JKL991  |       12 |
| GHI | C789 |     13 | 2021-01-03 0:20:00 |   12 | MNO1013 |       13 |
| JKL | D890 |     14 | 2021-01-03 0:20:00 |   13 | GHI891  |       14 |
| MNO | E012 |     15 | 2021-01-03 0:20:00 |   14 | JKL992  |       15 |
| ABC | A123 |     16 | 2021-01-03 0:20:00 |   12 | MNO1014 |       12 |
| DEF | B456 |     17 | 2021-01-03 0:20:00 |   13 | GHI892  |       13 |
| GHI | C789 |     18 | 2021-01-03 0:20:00 |   14 | JKL993  |       14 |
| JKL | D890 |     19 | 2021-01-03 0:20:00 |   15 | MNO1015 |       15 |
| MNO | E012 |     20 | 2021-01-03 0:20:00 |   16 | GHI893  |       16 |
| ABC | A123 |     21 | 2021-01-03 0:25:00 |   11 | ABC124  |       12 |
| DEF | B456 |     22 | 2021-01-03 0:25:00 |   12 | DEF257  |       13 |
| GHI | C789 |     23 | 2021-01-03 0:25:00 |   13 | GHI390  |       14 |
| JKL | D890 |     24 | 2021-01-03 0:25:00 |   14 | JKL491  |       15 |
| MNO | E012 |     25 | 2021-01-03 0:25:00 |   15 | MNO513  |       16 |
+-----+------+--------+--------------------+------+---------+----------+
否，我希望能够比较REFERENCE_COST和COST，并筛选成本小于REFERENCE_COST的所有行，并添加两个新列DATE_LAST_REF_COST_MET&；CLIENT_LAST_REF_COST_COST_MET，其中显示REFERENCE_COST的日期和该REFERENCE_COST中的客户编号 因此，结果df将如下所示：

+-----+------+--------+--------------------+------+---------+----------+------------------------+--------------------------+
| ID1 | ID2  | DOC_NO |        DATE        | COST | CLIENT  | REF_COST | DATE_LAST_REF_COST_MET | CLIENT_LAST_REF_COST_MET |
+-----+------+--------+--------------------+------+---------+----------+------------------------+--------------------------+
| ABC | A123 |     11 | 2021-01-03 0:20:00 |   10 | GHI890  |       11 | 2021-01-01 0:15:00     | ABC623                   |
| DEF | B456 |     12 | 2021-01-03 0:20:00 |   11 | JKL991  |       12 | 2021-01-01 0:15:00     | DEF756                   |
| GHI | C789 |     13 | 2021-01-03 0:20:00 |   12 | MNO1013 |       13 | 2021-01-01 0:15:00     | GHI889                   |
| JKL | D890 |     14 | 2021-01-03 0:20:00 |   13 | GHI891  |       14 | 2021-01-02 0:15:00     | JKL990                   |
| MNO | E012 |     15 | 2021-01-03 0:20:00 |   14 | JKL992  |       15 | 2021-01-03 0:15:00     | MNO1012                  |
| ABC | A123 |     21 | 2021-01-03 0:25:00 |   11 | ABC124  |       12 | 2021-01-03 0:20:00     | MNO1014                  |
| DEF | B456 |     22 | 2021-01-03 0:25:00 |   12 | DEF257  |       13 | 2021-01-03 0:20:00     | GHI892                   |
| GHI | C789 |     23 | 2021-01-03 0:25:00 |   13 | GHI390  |       14 | 2021-01-03 0:20:00     | JKL993                   |
| JKL | D890 |     24 | 2021-01-03 0:25:00 |   14 | JKL491  |       15 | 2021-01-03 0:20:00     | MNO1015                  |
| MNO | E012 |     25 | 2021-01-03 0:25:00 |   15 | MNO513  |       16 | 2021-01-03 0:20:00     | GHI893                   |
+-----+------+--------+--------------------+------+---------+----------+------------------------+--------------------------+

这是我能够做到的：

df %>%
  group_by(ID1, ID2) %>%
  arrange(DATE, DOC_NO, .by_group = TRUE) %>%
  mutate(diff = COST - lag(COST, default = first(COST)))%>%
  mutate(REF_COST = case_when(diff < 0~lag(COST), TRUE~diff)) %>%
  mutate(DATE_LAST_REF_COST_MET= case_when(diff < 0~lag(DATE), TRUE~DATE)) %>%
  mutate(CLIENT_LAST_REF_COST_MET= case_when(diff < 0~lag(CLIENT), TRUE~CLIENT))

这样做的限制是在进行计算时不会使用DATE和DOC_NO更改REFERENCE_COST

我不确定如何实现这一点

推荐答案

您可以使用cummax设置REF_COST和lag获取每组中的上一个值。使用filter仅保留引用成本高于成本的那些行。

library(dplyr)

df %>%
  group_by(ID1, ID2) %>%
  mutate(REF_COST = cummax(COST), 
         DATE_LAST_REF_COST_MET = lag(DATE), 
         CLIENT_LAST_REF_COST_MET = lag(CLIENT)) %>%
  ungroup() %>%
  filter(REF_COST > COST)

#    ID1   ID2   DOC_NO DATE                COST CLIENT  REF_COST DATE_LAST_REF_COST_MET CLIENT_LAST_REF_COST_MET
#   <chr> <chr>  <int> <chr>              <int> <chr>      <int> <chr>                  <chr>                   
# 1 ABC   A123      11 2021-01-03 0:20:00    10 GHI890        11 2021-01-01 0:15:00     ABC623                  
# 2 DEF   B456      12 2021-01-03 0:20:00    11 JKL991        12 2021-01-01 0:15:00     DEF756                  
# 3 GHI   C789      13 2021-01-03 0:20:00    12 MNO1013       13 2021-01-01 0:15:00     GHI889                  
# 4 JKL   D890      14 2021-01-03 0:20:00    13 GHI891        14 2021-01-02 0:15:00     JKL990                  
# 5 MNO   E012      15 2021-01-03 0:20:00    14 JKL992        15 2021-01-03 0:15:00     MNO1012                 
# 6 ABC   A123      21 2021-01-03 0:25:00    11 ABC124        12 2021-01-03 0:20:00     MNO1014                 
# 7 DEF   B456      22 2021-01-03 0:25:00    12 DEF257        13 2021-01-03 0:20:00     GHI892                  
# 8 GHI   C789      23 2021-01-03 0:25:00    13 GHI390        14 2021-01-03 0:20:00     JKL993                  
# 9 JKL   D890      24 2021-01-03 0:25:00    14 JKL491        15 2021-01-03 0:20:00     MNO1015                 
#10 MNO   E012      25 2021-01-03 0:25:00    15 MNO513        16 2021-01-03 0:20:00     GHI893

数据

如果以更易于复制的dput形式提供数据，则更容易获得帮助。

df <- structure(list(ID1 = c("ABC", "DEF", "GHI", "JKL", "MNO", "ABC", 
"DEF", "GHI", "JKL", "MNO", "ABC", "DEF", "GHI", "JKL", "MNO", 
"ABC", "DEF", "GHI", "JKL", "MNO", "ABC", "DEF", "GHI", "JKL", 
"MNO"), ID2 = c("A123", "B456", "C789", "D890", "E012", "A123", 
"B456", "C789", "D890", "E012", "A123", "B456", "C789", "D890", 
"E012", "A123", "B456", "C789", "D890", "E012", "A123", "B456", 
"C789", "D890", "E012"), DOC_NO = 1:25, DATE = c("2021-01-01 0:10:00", 
"2021-01-01 0:10:00", "2021-01-01 0:10:00", "2021-01-01 0:10:00", 
"2021-01-01 0:10:00", "2021-01-01 0:15:00", "2021-01-01 0:15:00", 
"2021-01-01 0:15:00", "2021-01-02 0:15:00", "2021-01-03 0:15:00", 
"2021-01-03 0:20:00", "2021-01-03 0:20:00", "2021-01-03 0:20:00", 
"2021-01-03 0:20:00", "2021-01-03 0:20:00", "2021-01-03 0:20:00", 
"2021-01-03 0:20:00", "2021-01-03 0:20:00", "2021-01-03 0:20:00", 
"2021-01-03 0:20:00", "2021-01-03 0:25:00", "2021-01-03 0:25:00", 
"2021-01-03 0:25:00", "2021-01-03 0:25:00", "2021-01-03 0:25:00"
), COST = c(11L, 12L, 13L, 14L, 15L, 11L, 12L, 13L, 14L, 15L, 
10L, 11L, 12L, 13L, 14L, 12L, 13L, 14L, 15L, 16L, 11L, 12L, 13L, 
14L, 15L), CLIENT = c("ABC123", "DEF256", "GHI389", "JKL490", 
"MNO512", "ABC623", "DEF756", "GHI889", "JKL990", "MNO1012", 
"GHI890", "JKL991", "MNO1013", "GHI891", "JKL992", "MNO1014", 
"GHI892", "JKL993", "MNO1015", "GHI893", "ABC124", "DEF257", 
"GHI390", "JKL491", "MNO513")), row.names = c(NA, -25L), class = "data.frame")

这篇关于根据日期按组比较值，并在值下降的位置创建新数据帧的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！

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