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的已完成订单之间的天数差异。理想的结果如下：
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
当我尝试这个解决方案时：
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
对此的任何帮助都会很棒，谢谢！
答案 0 :(得分：2)
试试这个
library(dplyr)
df %>% group_by(user_id, complete_order) %>%
mutate(c1 = order_date - lag(order_date)) %>%
group_by(user_id) %>% mutate(c2 = order_date - lag(order_date)) %>% ungroup %>%
mutate(complete_order_time_diff = ifelse(complete_order==0, c2, c1)) %>%
select(-c(c1, c2))
获取多个已取消的订单
df %>% mutate(c3=cumsum( complete_order != "0")) %>% group_by(user_id, complete_order) %>%
mutate(c1 = order_date - lag(order_date)) %>%
group_by(user_id) %>% mutate(c2 = order_date - lag(order_date)) %>%
mutate(c2=as.numeric(c2)) %>% group_by(user_id, c3) %>%
mutate(c2=cumsum(ifelse(complete_order==1, 0, c2))) %>% ungroup %>%
mutate(complete_order_time_diff = ifelse(complete_order==0, c2, c1)) %>%
select(-c(c1, c2, c3))
c3）增加1时， id为complete_order not 0。
c1计算日差bu user_id（但对于非完整订单，结果错误）
c2修复了c1与非完整订单的不一致。
我建议你使用group_by()和mutate(cumsum())的组合来更好地理解拥有多个分组变量的结果。
答案 1 :(得分：2)
您似乎正在寻找每个订单与上一个订单的距离。具有二元向量x，c(NA, cummax(x * seq_along(x))[-length(x)])给出在每个元素之前看到的最后“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，得到所需的输出。 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
在您的数据上，为方便起见，首先格式化df：
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：
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，在考虑“user_id”改变的索引之后，或者尝试避免分组（假设有序的“user_id”）的操作：
# 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$order_date
# accounting for changes in "id", locate last completed order
i = c(NA, cummax((compl | id_change) * seq_along(compl))[-length(compl)])
is.na(i) = id_change
dord - dord[i]
#Time differences in days
#[1] NA 3 11 26 NA 3 12 27
答案 2 :(得分：0)
我认为您可以添加filter函数来代替order_date[complete_order == 1]的子集，并通过添加{{1}确保order_date（和其他变量）是正确的数据类型转到stringsAsFactors = F）：
data.frame()这将返回下一个完整订单的时间（如果没有，则返回df = data.frame(cbind(user_id = c(rep(1, 4), rep(2,4)),
complete_order = c(rep(c(1,1,0,1), 2)),
order_date = c('2015-01-28', '2015-01-31', '2015-02-08', '2015-02-23', '2015-01-25', '2015-01-28', '2015-02-06', '2015-02-21')),
stringsAsFactors = F)
df$order_date <- lubridate::ymd(df$order_date)
df %>%
group_by(user_id) %>%
filter(complete_order == 1) %>%
mutate(complete_order_time_diff = order_date - lag(order_date))
）：
NA