R:计算特定事件之间的时差 [英] R: calculate time difference between specific events
问题描述
我有以下数据集:
df = data.frame(cbind(user_id = c(rep(1, 4), rep(2,4)),
complete_order = c(rep(c(1,0,0,1), 2)),
order_date = c('2015-01-28', '2015-01-31', '2015-02-08', '2015-02-23', '2015-01-25', '2015-01-28', '2015-02-06', '2015-02-21')))
library(lubridate)
df$order_date = as_date(df$order_date)
user_id complete_order order_date
1 1 2015-01-28
1 0 2015-01-31
1 0 2015-02-08
1 1 2015-02-23
2 1 2015-01-25
2 0 2015-01-28
2 0 2015-02-06
2 1 2015-02-21
我正在尝试计算每个用户仅完成的订单之间的天数差异.理想的结果如下所示:
I'm trying to calculate the difference in days between only completed orders for each user. The desirable outcome would look like this:
user_id complete_order order_date complete_order_time_diff
<fctr> <fctr> <date> <time>
1 1 2015-01-28 NA days
1 0 2015-01-31 3 days
1 0 2015-02-08 11 days
1 1 2015-02-23 26 days
2 1 2015-01-25 NA days
2 0 2015-01-28 3 days
2 0 2015-02-06 12 days
2 1 2015-02-21 27 days
当我尝试此解决方案时:
when I try this solution:
library(dplyr)
df %>%
group_by(user_id) %>%
mutate(complete_order_time_diff = order_date[complete_order==1]-lag(order_date[complete_order==1))
它返回错误:
Error: incompatible size (3), expecting 4 (the group size) or 1
任何对此的帮助都将非常有用，谢谢！
Any help with this will be great, thank you!
推荐答案
似乎您正在寻找每个订单与最后一个已完成订单之间的距离.具有二元向量x和c(NA, cummax(x * seq_along(x))[-length(x)])给出在每个元素之前看到的最后一个"1"的索引.然后，从该相应索引处的"order_date"中减去"order_date"的每个元素，即可得到所需的输出.例如
It seems that you're looking for the distance of each order from the last completed one. Having a binary vector, x, c(NA, cummax(x * seq_along(x))[-length(x)]) gives the indices of the last "1" seen before each element. Then, subtracting each element of "order_date" from the "order_date" at that respective index gives the desired output. E.g.
set.seed(1453); x = sample(0:1, 10, TRUE)
set.seed(1821); y = sample(5, 10, TRUE)
cbind(x, y,
last_x = c(NA, cummax(x * seq_along(x))[-length(x)]),
y_diff = y - y[c(NA, cummax(x * seq_along(x))[-length(x)])])
# x y last_x y_diff
# [1,] 1 3 NA NA
# [2,] 0 3 1 0
# [3,] 1 5 1 2
# [4,] 0 1 3 -4
# [5,] 0 3 3 -2
# [6,] 1 5 3 0
# [7,] 1 1 6 -4
# [8,] 0 3 7 2
# [9,] 0 4 7 3
#[10,] 1 5 7 4
为方便起见，为方便起见，请先格式化df
On your data, first format df for convenience:
df$order_date = as.Date(df$order_date)
df$complete_order = df$complete_order == "1" # lose the 'factor'
然后，在group_by之后应用上述方法:
And, then, either apply the above approach after a group_by:
library(dplyr)
df %>% group_by(user_id) %>%
mutate(time_diff = order_date -
order_date[c(NA, cummax(complete_order * seq_along(complete_order))[-length(complete_order)])])
，或者尝试在考虑"user_id"更改的索引后避免分组(假设排序为"user_id")的操作:
, or, perhaps give a try on operations that avoid grouping (assuming ordered "user_id") after accounting for the indices where "user_id" changes:
# save variables to vectors and keep a "logical" of when "id" changes
id = df$user_id
id_change = c(TRUE, id[-1] != id[-length(id)])
compl = df$complete_order
dord = df$order_date
# accounting for changes in "id", locate last completed order
i = c(NA, cummax((compl | id_change) * seq_along(compl))[-length(compl)])
is.na(i) = id_change
dord - dord[i]
#Time differences in days
#[1] NA 3 11 26 NA 3 12 27
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