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r - 为订阅之间的间隙添加空行

转载 作者：行者123 更新时间：2023-12-01 04:25:06 27 4
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我已经为此苦苦挣扎了一段时间，但在任何地方都找不到类似的问题，因此我在这里提出了第一个问题!

我对 R 相当陌生，所以请原谅我犯的任何明显错误。

我有一个数据集，对于用户拥有或曾经拥有的每个订阅都有一行。一些用户有多行，而另一些用户只有一行。仅存在事件或以前事件的订阅。

我有两个变量分别说明订阅何时开始和何时结束调用，分别是 Begindate 和 Enddate。我已经创建了关系长度变量，这些变量说明了每种订阅类型的这两个变量之间的天数。这意味着relationlength 变量只给出订阅处于事件状态的天数。

我想要做的是在没有订阅处于事件状态的时间段的不同订阅行之间创建空行，从特定用户已知的最早开始日期开始，到所有订阅结束的给定日期(20- 04-2022)。

我试图比较用户已知的第一个开始日期和最终日期的日期差异，并减去其他订阅类型已知的关系长度。但是，我无法完成这项工作。

df 当前外观的示例:

(rl 代表关系长度)

ID Begindate Enddate Subscrtype active rl_fixed rl_promotional Productgroup

1 2019-08-26 2022-04-20 fixed   1      968      0              1
1 2018-08-24 2019-08-23 fixed   0      364      0              1
1 2015-08-24 2016-08-23 promo   0      0        364            2
2 2019-08-26 2019-09-12 fixed   0      17       0              1
2 2018-08-24 2019-08-23 fixed   0      364      0              1


我希望它看起来像什么: 
ID Begindate Enddate Subscrtype active rl_fixed rl_promo rl_none Productgroup

1 2019-08-26 2022-04-20 fixed   1      968      0        0       1
1 2019-08-24 2019-08-25 none    0      0        0        2       NA
1 2018-08-24 2019-08-23 fixed   0      364      0        0       1
1 2016-08-24 2018-08-23 none    0      0        0        729     NA
1 2015-08-24 2016-08-23 promo   0      0        364      0       2
2 2019-09-13 2022-04-20 none    0      0        0        950     NA
2 2019-08-26 2019-09-12 fixed   0      17       0        0       1
2 2019-08-24 2019-08-25 none    0      0        0        2       NA
2 2018-08-24 2019-08-23 fixed   0      364      0        0       1


最终目标是汇总并清楚地了解用户可能存在的不同类型关系的特定关系长度。

先感谢您!

实际 df 中一个特定用户的 dput: 
structure(list(ï..CRM.relatienummer = structure(c(1L, 1L, 1L, 
1L, 1L, 1L), .Label = "1", class = "factor"), Begindatum = c("2019-08-26", 
"2018-08-24", "2017-08-24", "2016-08-24", "2015-08-20", "2016-06-01"
), Einddatum = c("2022-04-20", "2019-08-23", "2018-08-23", "2017-08-23", 
"2016-05-31", "2016-08-19"), Type.abonnement = structure(c(1L, 
1L, 1L, 1L, 1L, 1L), .Label = "Actie", class = "factor"), Status_dummy = c(1, 
0, 0, 0, 0, 0), relationlength_fixed = c(0, 0, 0, 0, 0, 0), relationlength_promo = c(968, 
364, 364, 364, 285, 79), relationlength_trial = c(0, 0, 0, 0, 
0, 0), fixed_dummy = c(0, 0, 0, 0, 0, 0), trial_dummy = c(0, 
0, 0, 0, 0, 0), promotional_dummy = c(1, 1, 1, 1, 1, 1)), row.names = c("1:20610", 
"2:38646", "2:39231", "2:39232", "2:39248", "2:39837"), class = "data.frame")

编辑:

我试图运行此代码: 
dfs <- split(testdata,testdata$ï..CRM.relatienummer)

r <- lapply(seq(length(dfs)), function(k){
  v <- dfs[[k]]
  vt <- data.frame(unique(v$ï..CRM.relatienummer), 
                   as.character((as.Date(v$Einddatum)+1)[-1]), 
                   as.character((as.Date(v$Begindatum)-1)[-nrow(v)]), 
                   0,
                   0,
                   0,
                   0,
                   (as.Date(v$Begindatum)-1)[-nrow(v)] - (as.Date(v$Einddatum)+1)[-1],
                   NA,
                   0,
                   0,
                   0,
                   0,
                   0)
  colnames(vt) <- c(colnames(v)[-ncol(v)],"rl_none",colnames(v)[ncol(v)])
  (testdata <- rbind(data.frame(v[-ncol(v)],rl_none = 0,v[ncol(v)]),vt))[order(as.Date(testdata$Begindatum),decreasing = T),]
})

res <- data.frame(Reduce(rbind,r),row.names = NULL)

在这个数据帧上，不幸的是没有运气: 
structure(list(ï..CRM.relatienummer = structure(c("d45248b8974dc4f8ff948779e0fd07e20f304e929ada4e14c0420aebed81e9b5", 
"2ab04e80b3e64601147df977d6054c04ffa80014b3691b25dd1cc8ef85cea06a", 
"2ab04e80b3e64601147df977d6054c04ffa80014b3691b25dd1cc8ef85cea06a", 
"bcf2c99e6dc974380f967204b9623dce2c8a3fad694dc0b4430fcbf77f8f39f3", 
"bcf2c99e6dc974380f967204b9623dce2c8a3fad694dc0b4430fcbf77f8f39f3", 
"f8610cd0237858ac9384d6ba209759ae306860ffabb3f8e6c3d6fc68dbaddc51", 
"e5b8b3f46165e48aec8bbe65ed1cb29d18a0492fbcac44803372f672348459db", 
"c737815b2365b01a8a85c380364a0f721685a131de98cd7790b4d40bb8c4e05b", 
"b9c0272caa8d5d3497d28cce3bda5d3d17c22f18c5f65c5e82c572b410a8ea71", 
"b9c0272caa8d5d3497d28cce3bda5d3d17c22f18c5f65c5e82c572b410a8ea71", 
"539c6c3e604245008daefbe500ff29357bee91f82a7896126bd0f69848524cb7", 
"d361338bed51cb9c8aa73fd8914cbf392f4e05e7b073f637f7b150cf02b89c8c", 
"505d3df3f1298e07aa96073490b72acd2391da06ad4cfbd5a9fbde3a3de79684", 
"826443481cbb5b4e061040d443a0ce8d94322615d8ffae1e68b2ff7d896afcf7", 
"2b59a1ec028c261c0f22cd6a49220dc7cec9a9fb0fabe2296b4ba77a60cfdaae"
), class = c("hash", "sha256")), Begindatum = c("2019-06-14", 
"2019-03-01", "2019-09-02", "2019-03-03", "2019-04-01", "2019-09-21", 
"2019-02-02", "2019-06-11", "2019-02-05", "2019-02-09", "2019-07-24", 
"2019-05-08", "2019-09-27", "2019-08-03", "2019-04-03"), Einddatum = c("2022-04-20", 
"2019-09-01", "2022-04-20", "2019-03-31", "2022-04-20", "2022-04-20", 
"2019-02-14", "2019-07-08", "2019-02-11", "2020-02-08", "2019-09-03", 
"2019-06-18", "2019-11-07", "2019-08-16", "2022-04-20"), Status_dummy = c(1, 
0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1), relationlength_fixed = c(0, 
184, 961, 28, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0), relationlength_promo = c(1041, 
0, 0, 0, 1115, 942, 12, 0, 0, 364, 0, 0, 0, 0, 1113), relationlength_trial = c(0, 
0, 0, 0, 0, 0, 0, 27, 0, 0, 41, 41, 41, 13, 0), rl_none = c(NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), fixed_dummy = c(0, 
1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0), trial_dummy = c(0, 
0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0), promotional_dummy = c(1, 
0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1), active_subscr_dummy = c(3, 
0, 5, 0, 3, 3, 0, 0, 0, 3, 0, 0, 1, 0, 3), hashedEmail = c(NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c("1:1", 
"1:2", "1:3", "1:4", "1:5", "1:6", "1:7", "1:8", "1:9", "1:10", 
"1:11", "1:12", "1:13", "1:14", "1:15"), class = "data.frame")

最佳答案

希望这是你所期待的

dfs <- split(df,df$ID)

r <- lapply(seq(length(dfs)), function(k){
  v <- dfs[[k]]
  vt <- data.frame(unique(v$ID), 
                   as.character((as.Date(v$Enddate)+1)[-1]), 
                   as.character((as.Date(v$Begindate)-1)[-nrow(v)]), 
                   "none",
                   0,
                   0,
                   0,
                   (as.Date(v$Begindate)-1)[-nrow(v)] - (as.Date(v$Enddate)+1)[-1],
                   NA)
  colnames(vt) <- c(colnames(v)[-ncol(v)],"rl_none",colnames(v)[ncol(v)])
  (df <- rbind(data.frame(v[-ncol(v)],rl_none = 0,v[ncol(v)]),vt))[order(as.Date(df$Begindate),decreasing = T),]
})

res <- data.frame(Reduce(rbind,r),row.names = NULL)

这使 
> res
  ID  Begindate    Enddate Subscrtype active rl_fixed rl_promo rl_none Productgroup
1  1 2019-08-26 2022-04-20      fixed      1      968        0       0            1
2  1 2019-08-24 2019-08-25       none      0        0        0       1           NA
3  1 2018-08-24 2019-08-23      fixed      0      364        0       0            1
4  1 2016-08-24 2018-08-23       none      0        0        0     729           NA
5  1 2015-08-24 2016-08-23      promo      0        0      364       0            2
6  2 2019-08-26 2019-09-12      fixed      0       17        0       0            1
7  2 2019-08-24 2019-08-25       none      0        0        0       1           NA
8  2 2018-08-24 2019-08-23      fixed      0      364        0       0            1

数据 
structure(list(ID = c(1L, 1L, 1L, 2L, 2L), Begindate = structure(c(3L, 
2L, 1L, 3L, 2L), .Label = c("2015-08-24", "2018-08-24", "2019-08-26"
), class = "factor"), Enddate = structure(c(4L, 2L, 1L, 3L, 2L
), .Label = c("2016-08-23", "2019-08-23", "2019-09-12", "2022-04-20"
), class = "factor"), Subscrtype = structure(c(1L, 1L, 2L, 1L, 
1L), .Label = c("fixed", "promo"), class = "factor"), active = c(1L, 
0L, 0L, 0L, 0L), rl_fixed = c(968L, 364L, 0L, 17L, 364L), rl_promo = c(0L, 
0L, 364L, 0L, 0L), Productgroup = c(1L, 1L, 2L, 1L, 1L)), class = "data.frame", row.names = c(NA, 
-5L))

关于r - 为订阅之间的间隙添加空行，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/59106396/

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