计算购买之间的时间间隔（带 NA 的日期时间变量）

Question

df 是一个数据框，显示不同客户的访问和购买时间。

id<-c(1,1,1,1,1,1,2,2,2,2,2)
visit<-rep("yes",11)
purchase<-c("2015-04-27 13:57:06","2015-04-27 13:59:19","2015-04-27  14:03:35","NA","NA","2015-04-27 16:59:42","2015-05-18 17:01:09","2015-05-18 17:03:40","2015-05-18 17:04:00","NA","NA") 
df<-data.frame(id,visit,purchase)

当 purchase 变量为 NA 时，表示客户已访问该网站但未进行购买。

现在我需要创建一个名为 time.gap 的新变量来计算每个客户购买之间的时间间隔，如下所示：

   id visit            purchase time.gap
1   1   yes 2015-04-27 13:57:06       NA
2   1   yes 2015-04-27 13:59:19      133
3   1   yes 2015-04-27 14:03:35      256
4   1   yes                  NA       NA
5   1   yes                  NA       NA
6   1   yes 2015-04-27 16:59:42    10567
7   2   yes 2015-05-18 17:01:09       NA
8   2   yes 2015-05-18 17:03:40      151
9   2   yes 2015-05-18 17:04:00       20
10  2   yes                  NA       NA
11  2   yes                  NA       NA

感谢您的帮助

Answer 1

我会说每个用户 ID 的时间间隔需要一个额外的步骤，在 ID 级别分组。

-EDITED 错过了日期转换。

使用 dplyr lubridate 和 zoo

的非常简洁的方法

# libraries
library(dplyr)
library(zoo)
library(lubridate)

# the data
id<-c(1,1,1,1,1,1,2,2,2,2,2)
visit<-rep("yes",11)
purchase<-c("2015-04-27 13:57:06","2015-04-27 13:59:19","2015-04-27 14:03:35","NA","NA","2015-04-27 16:59:42","2015-05-18 17:01:09","2015-05-18 17:03:40","2015-05-18 17:04:00","NA","NA") 
df<-data.frame(id,visit,purchase) 
df$purchase <- lubridate::ymd_hms(df$purchase)
# helper column
df$purch <- zoo::na.locf(df$purchase)
df

#> df
#   id visit            purchase               purch
#1   1   yes 2015-04-27 13:57:06 2015-04-27 13:57:06
#2   1   yes 2015-04-27 13:59:19 2015-04-27 13:59:19
#3   1   yes 2015-04-27 14:03:35 2015-04-27 14:03:35
#4   1   yes                <NA> 2015-04-27 14:03:35
#5   1   yes                <NA> 2015-04-27 14:03:35
#6   1   yes 2015-04-27 16:59:42 2015-04-27 16:59:42
#7   2   yes 2015-05-18 17:01:09 2015-05-18 17:01:09
#8   2   yes 2015-05-18 17:03:40 2015-05-18 17:03:40
#9   2   yes 2015-05-18 17:04:00 2015-05-18 17:04:00
#10  2   yes                <NA> 2015-05-18 17:04:00
#11  2   yes                <NA> 2015-05-18 17:04:00

# run it
df%>%
  group_by(id)%>%
  mutate(dif=c(NA, diff(purch)))%>%
  select(-purch)

#Source: local data frame [11 x 4]
#Groups: id
#
#   id visit            purchase   dif
#1   1   yes 2015-04-27 13:57:06    NA
#2   1   yes 2015-04-27 13:59:19   133
#3   1   yes 2015-04-27 14:03:35   256
#4   1   yes                <NA>     0
#5   1   yes                <NA>     0
#6   1   yes 2015-04-27 16:59:42 10567
#7   2   yes 2015-05-18 17:01:09    NA
#8   2   yes 2015-05-18 17:03:40   151
#9   2   yes 2015-05-18 17:04:00    20
#10  2   yes                <NA>     0
#11  2   yes                <NA>     0

或一发

df%>%
  mutate(purch=zoo::na.locf(lubridate::ymd_hms(df$purchase))) %>%
  group_by(id) %>%
  mutate(dif=c(NA, diff(purch))) %>%
  select(-purch)

# packages [1] lubridate_1.3.3 zoo_1.7-12      dplyr_0.4.2

Answer 2

df$purchase <- strptime(df$purchase, "%Y-%m-%d %H:%M:%S")
df1 <- df

library(dplyr)
df %>% 
  filter(!is.na(purchase)) %>% 
  group_by(id) %>%
  mutate(time.gap = c(NA, difftime(purchase[-1], 
                                   purchase[-length(purchase)], units="secs"))) %>%
  left_join(df1, .)
#    id visit            purchase time.gap
# 1   1   yes 2015-04-27 13:57:06       NA
# 2   1   yes 2015-04-27 13:59:19      133
# 3   1   yes 2015-04-27 14:03:35      256
# 4   1   yes                <NA>       NA
# 5   1   yes                <NA>       NA
# 6   1   yes 2015-04-27 16:59:42    10567
# 7   2   yes 2015-05-18 17:01:09       NA
# 8   2   yes 2015-05-18 17:03:40      151
# 9   2   yes 2015-05-18 17:04:00       20
# 10  2   yes                <NA>       NA
# 11  2   yes                <NA>       NA

这是一种添加了合并的 dplyr 方法。使用 difftime 而不是 diff 因为它允许单位参数。

数据

id<-c(1,1,1,1,1,1,2,2,2,2,2)
visit<-rep("yes",11)
purchase<-c("2015-04-27 13:57:06","2015-04-27 13:59:19","2015-04-27  14:03:35","NA","NA","2015-04-27 16:59:42","2015-05-18 17:01:09","2015-05-18 17:03:40","2015-05-18 17:04:00","NA","NA") 
df<-data.frame(id,visit,purchase) 
is.na(df$purchase) <- df$purchase == "NA"
df$purchase <- as.POSIXct(df$purchase)

疑难解答

如果您运行发现您的真实数据有问题，我们可以通过测试找出问题的来源：

s <- split(df, df$id)
test <- list()
for(i in 1:length(s)) {
  s1 <- s[[i]]
test[[i]] <- s[[i]] %>% 
  filter(!is.na(purchase)) %>% 
  group_by(id) %>%
  mutate(time.gap = c(NA, difftime(purchase[-1], 
                                   purchase[-length(purchase)], units="secs"))) %>%
  left_join(s1, .)

}

现在 test 是一个包含代码所有迭代的列表。如果抛出错误，我们将知道它发生在哪里，因为所有其他成功的运行s 都将被保存。因此，如果我得到一个错误并查看 test 并且它具有直到 id 3 的所有 id，我知道 id 4 导致了错误。

计算购买之间的时间间隔（带 NA 的日期时间变量）

calculating time gap between purchases (datetime variable with NAs)

r

datetime-format