Question

我是一个看起来像这样的数据库：

userId              Screen         Platform       Version
01                  first          IOS            1.0.1
01                  main           IOS            1.0.1
02                  first          Android        1.0.2
03                  first          IOS            1.0.2
03                  main           IOS            1.0.2
03                  detail         IOS            1.0.2

基本上我想知道有多少人＆＃34;掉线＆＃34;在第一个屏幕之后，我的想法是创建一个新列，它告诉用户通过userId访问的屏幕数量，理想的数据库看起来像这样：

userId              DifferentScreen        Platform      Version
01                  2                     IOS            1.0.1
02                  1                     Android        1.0.2
03                  3                     IOS            1.0.2

我试过了：

setDT(database)[order(userId) ,. (DifferentScreen = uniqueN(Screen), Version = Version[1L], Platform = Platform[1L], by = userId)]

但它不起作用，我发现的问题是：它没有按用户列分组，因为列数保持不变，我使用命令uniqueN因为我还没找到命令只做.N（）。

Answer 1

你基本上就在那里。关于缺少括号只有一个小问题。尝试：

dt[order(userId) ,. (DifferentScreen = uniqueN(Screen), Version = Version[1L], Platform = Platform[1L]), by = userId]

   userId DifferentScreen Version Platform
1:      1               2   1.0.1      IOS
2:      2               1   1.0.2  Android
3:      3               3   1.0.2      IOS

您需要在by = userID之前关闭括号。这种方式data.table将by=...读取为分组而不是名为by的新变量。目前，您的输出数据集没有按任何分组，它认为您要创建一个名为by的变量。

您可以在旧代码的结果中看到这一点：

dt[order(userId) ,. (DifferentScreen = uniqueN(Screen), Version = Version[1L], Platform = Platform[1L], by = userId)]

#See how this creates a variable "by"?
   DifferentScreen Version Platform by
1:               3   1.0.1      IOS  1
2:               3   1.0.1      IOS  1
3:               3   1.0.1      IOS  2
4:               3   1.0.1      IOS  3
5:               3   1.0.1      IOS  3
6:               3   1.0.1      IOS  3

数据：

dt <- structure(list(userId = c(1L, 1L, 2L, 3L, 3L, 3L), Screen = structure(c(2L, 3L, 2L, 2L, 3L, 1L), .Label = c("detail", "first", "main"), class = "factor"), Platform = structure(c(2L, 2L, 1L, 2L, 2L, 2L), .Label = c("Android", "IOS"), class = "factor"), Version = structure(c(1L, 1L, 2L, 2L, 2L, 2L), .Label = c("1.0.1", "1.0.2"), class = "factor")), .Names = c("userId", "Screen", "Platform", "Version"), class = c("data.table", "data.frame" ), row.names = c(NA, -6L), .internal.selfref = <pointer: 0x0000000000250788>)

确定发生后用户丢弃的次数

1 个答案: