数据帧子集内的计算[R]

3 回答

• 5

  此解决方案仅使用R的基数并保留输入的原始顺序：

  # Sort, calculate differences and unsort.
  # r is row indexes to use, order.by is ordering vector, col is vector to difference
  
  diffs <- function(r, order.by, col) {
      order.by <- order.by[r]
      col <- col[r]
      o <- order(order.by)
      replace(r, o, c(NA, diff(col[o])))
  }
  
  # fun specialized to arguments after first, i.e. subsequent arguments curried
  
  curry <- function (fun, ...) function(r) fun(r, ...)
  
  ix <- 1:nrow(DF)
  transform(DF, 
      days_since = ave(ix, customer, FUN = curry(diffs, date, date)),
      amt_diff = ave(ix, customer, FUN = curry(diffs, date, purchase_amt))
  )
  

  结果是：

  customer visit       date purchase_amt days_since amt_diff
  1     sarah     2 2013-08-09            5         19       -3
  2     sarah     3 2013-07-21            8         28       -1
  3     sarah     4 2013-06-23            9         21        8
  4     sarah     5 2013-06-02            1         NA       NA
  5     sarah     1 2013-08-20            8         11        3
  6     henry     1 2013-07-04            4         NA       NA
  7       che     1 2013-08-27            2         31        1
  8       che     2 2013-07-27            1         22       -7
  9       che     3 2013-07-05            8         21        5
  10      che     4 2013-06-14            3         NA       NA
  11       dt     3 2013-04-05            9         NA       NA
  12       dt     2 2013-06-07            1         63       -8
  13       dt     1 2013-07-11            6         34        5
  

  更新：对代码的微小改进 .

  回复于 2024-05-02T20:27:13+08:00
• 7

  像这样的东西？假设您的数据被称为 df ：

  library(plyr)
  
  # convert dates to class 'Date'
  df$date <- as.Date(df$date)
  
  # order by customer and date
  df <- df[order(df$customer, df$date), ]
  # or since plyr is loaded anyway:
  df <- arrange(df, customer, date) 
  
  # per customer, calculate differences in date and purchase, between consecutive visits
  # pad differences with a leading NA
  df2 <- ddply(.data = df, .variables = .(customer), mutate,
        days_since = c(NA, diff(date)),
        amt_diff = c(NA, diff(purchase_amt)))
  
  df2
  # customer visit       date purchase_amt days_since amt_diff
  # 1       che     4 2013-06-14            3         NA       NA
  # 2       che     3 2013-07-05            8         21        5
  # 3       che     2 2013-07-27            1         22       -7
  # 4       che     1 2013-08-27            2         31        1
  # 5        dt     3 2013-04-05            9         NA       NA
  # 6        dt     2 2013-06-07            1         63       -8
  # 7        dt     1 2013-07-11            6         34        5
  # 8     henry     1 2013-07-04            4         NA       NA
  # 9     sarah     5 2013-06-02            1         NA       NA
  # 10    sarah     4 2013-06-23            9         21        8
  # 11    sarah     3 2013-07-21            8         28       -1
  # 12    sarah     2 2013-08-09            5         19       -3
  # 13    sarah     1 2013-08-20            8         11        3
  

  回复于 2024-05-02T20:27:13+08:00
• 6

  这是与@Henrik一致的data.table解决方案：

  df<-structure(list(customer = structure(c(4L, 4L, 4L, 4L, 4L, 3L, 
  1L, 1L, 1L, 1L, 2L, 2L, 2L), .Label = c("che", "dt", "henry", 
  "sarah"), class = "factor"), visit = c(2L, 3L, 4L, 5L, 1L, 1L, 
  1L, 2L, 3L, 4L, 3L, 2L, 1L), date = structure(c(15926, 15907, 
  15879, 15858, 15937, 15890, 15944, 15913, 15891, 15870, 15800, 
  15863, 15897), class = "Date"), purchase_amt = c(5L, 8L, 9L, 
  1L, 8L, 4L, 2L, 1L, 8L, 3L, 9L, 1L, 6L)), .Names = c("customer", 
  "visit", "date", "purchase_amt"), row.names = c(NA, -13L), class =  
  "data.frame")
  
  library(data.table)
   df<-data.table(df)
  df[,list(visit=visit,date=date, purchase_amt=purchase_amt,days_since = c(NA, diff(date)),amt_diff = c(NA, diff(purchase_amt))),keyby="customer"]
      customer visit       date purchase_amt days_since amt_diff
   1:      che     1 2013-08-27            2         NA       NA
   2:      che     2 2013-07-27            1        -31       -1
   3:      che     3 2013-07-05            8        -22        7
   4:      che     4 2013-06-14            3        -21       -5
   5:       dt     3 2013-04-05            9         NA       NA
   6:       dt     2 2013-06-07            1         63       -8
   7:       dt     1 2013-07-11            6         34        5
   8:    henry     1 2013-07-04            4         NA       NA
   9:    sarah     2 2013-08-09            5         NA       NA
  10:    sarah     3 2013-07-21            8        -19        3
  11:    sarah     4 2013-06-23            9        -28        1
  12:    sarah     5 2013-06-02            1        -21       -8
  13:    sarah     1 2013-08-20            8         79        7
  

  回复于 2024-05-02T20:27:13+08:00