按列分组的数据框上R中的行之间的差异

Question

我正在寻找由app_name按版本进行计数的差异。 我的数据集如下所示：app_name，version_id，count，[difference]

这是数据集

    data = structure(list(app_name = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 
2L, 3L, 3L), .Label = c("a", "b", "c"), class = "factor"), version_id = c(1, 
1.1, 2.3, 2, 3.1, 3.3, 4, 1.1, 2.4), count = c(600L, 620L, 620L, 
200L, 200L, 250L, 250L, 15L, 36L)), .Names = c("app_name", "version_id", 
"count"), class = "data.frame", row.names = c(NA, -9L))

给定此data.frame，我如何才能获得app_name和version_id计数的滞后差异？ 每个应用程序的初始（第一）版本差异为零，因为两者之间没有差异。

这是最终的“ diff”列的最终结果的示例

structure(list(app_name = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 
2L, 3L, 3L), .Label = c("a", "b", "c"), class = "factor"), version_id = c(1, 
1.1, 2.3, 2, 3.1, 3.3, 4, 1.1, 2.4), count = c(600L, 620L, 620L, 
200L, 200L, 250L, 250L, 15L, 36L), diff = c(0, 20, 0, 0, 0, 1.25, 
0, 0, 2.4)), .Names = c("app_name", "version_id", "count", "diff"
), class = "data.frame", row.names = c(NA, -9L))

Answer 1

尝试使用dplyr和lag ：

library(dplyr)
data %>% group_by(app_name) %>%
         mutate(diffvers = version_id - dplyr::lag(version_id, default = version_id[1]),
                diffcount = count - dplyr::lag(count, default = count[1]))

Source: local data frame [9 x 5]
Groups: app_name [3]

  app_name version_id count diffvers diffcount
    (fctr)      (dbl) (int)    (dbl)     (int)
1        a        1.0   600      0.0         0
2        a        1.1   620      0.1        20
3        a        2.3   620      1.2         0
4        b        2.0   200      0.0         0
5        b        3.1   200      1.1         0
6        b        3.3   250      0.2        50
7        b        4.0   250      0.7         0
8        c        1.1    15      0.0         0
9        c        2.4    36      1.3        21

Answer 2

我们可以使用data.table 。 我们的“data.frame”转换为“data.table”（ setDT(data) ），通过“APP_NAME”，循环（分组lapply(.. ）在指定的列.SDcols ，获取当前元素之间的差异，它的lag （默认为shift为type='lag' ）并分配（ := ）输出以创建新列。

library(data.table)#v1.9.6
setDT(data)[, c('diffvers', 'diffcount') := lapply(.SD, 
              function(x) x-shift(x, fill=x[1L])), by = app_name, .SDcols=2:3]

data
#   app_name version_id count diffvers diffcount
#1:        a        1.0   600      0.0         0
#2:        a        1.1   620      0.1        20
#3:        a        2.3   620      1.2         0
#4:        b        2.0   200      0.0         0
#5:        b        3.1   200      1.1         0
#6:        b        3.3   250      0.2        50
#7:        b        4.0   250      0.7         0
#8:        c        1.1    15      0.0         0
#9:        c        2.4    36      1.3        21