根据另一个变量的老化情况生成一个新变量 [英] generate a new variable based on the aging of another variable

问题描述

我有一个像下面这样的数据集

I have a dataset likes like below

ID. Invoice. Date of Invoice.  paid or not.  

1    1         10/31/2019       yes
1    1         10/31/2019       yes
1    2         11/30/2019       no
1    3         12/31/2019       no

2    1         09/30/2019       no
2    2         10/30/2019       no
2    3         11/30/2019       yes

3    1         7/31/2019        no
3    2         9/30/2019        yes
3    3         12/31/2019       no

4    1         7/31/2019        yes
4    2         9/30/2019        no
4    3         12/31/2019       yes

我想知道客户是否愿意付款。只要客户支付了新发票而未支付的旧发票，我就会给他一个很好的分数。因此对于客户1和客户3，我给的评价是好，客户2的评价是差。

I would like to know whether the customers' willingness to pay. As long as a customer has paid a new invoice with an old invoice not paid, I will give him a good score. so for customer 1 and 3, I gave "good", customer 2 is a "bad" score.

，因此最终数据将再增加一列，其值包括好坏。

so the final data will have one more column, with values of good and bad.

ID。发票。发票日期。是否付款。不好或好

ID. Invoice. Date of Invoice. paid or not. Bad or good

1    1         10/31/2019       yes          bad
1    1         10/31/2019       yes          bad
1    2         11/30/2019       no           bad
1    3         12/31/2019       no           bad

2    1         09/30/2019       no           good
2    2         10/30/2019       no           good
2    3         11/30/2019       yes          good

3    1         7/31/2019        no           good
3    2         9/30/2019        yes          good
3    3         12/31/2019       no           good

4    1         7/31/2019        yes          good
4    2         9/30/2019        no           good
4    3         12/31/2019       yes          good

推荐答案

假设您的发票日期。已订购，然后这里是使用 ave

Assuming your Date of Invoice. is ordered already, then here is a base R solution using ave

df$`good or band.` <- ave(df$`paid or not.`,df$ID., FUN = function(v) ifelse(which(v=="yes")==1,"bad","good"))

> df
  ID. Invoice. Date of Invoice. paid or not. good or band.
1   1        1       09/30/2019           no          good
2   1        2       10/30/2019           no          good
3   1        3       11/30/2019          yes          good
4   2        1       10/31/2019          yes           bad
5   2        2       11/30/2019           no           bad
6   2        3       12/31/2019           no           bad
7   3        1        7/31/2019           no          good
8   3        2        9/30/2019          yes          good
9   3        3       12/31/2019           no          good

DATA

df <- structure(list(ID. = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L), Invoice. = c(1L, 
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), `Date of Invoice.` = c("09/30/2019", 
"10/30/2019", "11/30/2019", "10/31/2019", "11/30/2019", "12/31/2019", 
"7/31/2019", "9/30/2019", "12/31/2019"), `paid or not.` = c("no", 
"no", "yes", "yes", "no", "no", "no", "yes", "no")), class = "data.frame", row.names = c(NA, 
-9L))

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