的R - 跨月普通用戶

Question

我有以下的列的事務表：

TransactionId UserId YearMonth Group 

我試圖做到的是在不同的幾個月得到的唯一用戶。 如：

YearMonth Group UsersCountMonth1 UsersCountMonth2 UsersCountMonth3 
201301 A  1000    900    800 
201301 B  1200    940    700 
201302 B  1300    1140    900 
201303 A  12e0    970    706 

基本上MONTH1和MONTH2是根據備案YearMonth值增量個月。

我正在使用此結果執行保留分析。

Answer 1

我記得你昨天正在尋找分析訂閱羣組的可能性。所以我猜你可以做

library(tidyverse) 
set.seed(1) 
n <- 100 
df <- data.frame(
    user = sample(1:20, n, T), 
    transDate = sample(seq(as.Date("2016-01-01"), as.Date("2016-12-31"), "1 month"), n, T), 
    group = sample(LETTERS[1:2], n, T) 
) 
diffmonth <- function(d1, d2) { 
# http://stackoverflow.com/questions/1995933/number-of-months-between-two-dates 
    monnb <- function(d) { 
    lt <- as.POSIXlt(as.Date(d, origin="1900-01-01")) 
    lt$year*12 + lt$mon 
    } 
    monnb(d2) - monnb(d1) + 1L 
} 
df %>% 
    group_by(user, group) %>% 
    mutate(cohort = min(transDate), month = diffmonth(cohort, transDate)) %>% 
    unite(cohort, cohort, group, remove = T) %>% 
    group_by(month, cohort) %>% 
    summarise(n=n()) %>% 
    spread(month, n, fill = 0, drop = F) 
# # A tibble: 16 × 12 
#   cohort `1` `2` `3` `4` `5` `6` `7` `8` `9` `10` `11` 
# *   <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> 
# 1 2016-01-01_A  5  1  0  1  1  1  1  0  2  0  0 
# 2 2016-02-01_A  1  0  0  0  0  0  0  0  1  0  1 
# 3 2016-02-01_B  4  1  2  1  0  1  2  0  1  1  0 
# 4 2016-03-01_A  5  0  3  1  2  2  2  0  1  2  0 
# 5 2016-03-01_B  4  0  0  0  2  0  1  0  0  0  0 
# 6 2016-04-01_A  4  0  2  1  0  1  0  2  1  0  0 
# 7 2016-04-01_B  1  0  0  0  0  0  0  0  0  0  0 
# 8 2016-05-01_A  2  0  2  2  0  0  2  0  0  0  0 
# 9 2016-05-01_B  1  0  0  1  0  0  2  0  0  0  0 
# 10 2016-06-01_A  1  0  2  0  0  1  0  0  0  0  0 
# 11 2016-06-01_B  4  0  0  0  0  1  1  0  0  0  0 
# 12 2016-07-01_A  1  0  1  0  0  0  0  0  0  0  0 
# 13 2016-08-01_B  4  1  1  0  0  0  0  0  0  0  0 
# 14 2016-09-01_A  1  0  0  0  0  0  0  0  0  0  0 
# 15 2016-10-01_B  1  0  0  0  0  0  0  0  0  0  0 
# 16 2016-12-01_A  3  0  0  0  0  0  0  0  0  0  0