我有如下数据：
month shop product sales sales_per_shop
1 1 1 1 10 90
2 1 1 2 20 90
3 1 2 1 40 120
4 1 3 2 50 150
5 2 1 1 10 90
6 2 1 2 20 90
7 2 2 1 40 120
8 2 3 2 50 150
9 3 1 1 10 90
10 3 1 2 20 90
11 3 2 1 40 120
12 3 3 2 50 150
我的目标是为sales和sales_per_shop列创建一个月的滞后时间。
对于销售而言，这没问题，因为每一行都是不同的。
z %>%
group_by(shop, product) %>%
mutate(lag_sales_per_shop = lag(sales, 1)) %>%
head(5)
# A tibble: 5 x 6
# Groups: shop, product [4]
month shop product sales sales_per_shop lag_sales
<int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 10 90 NA
2 1 1 2 20 90 NA
3 1 2 1 40 120 NA
4 1 3 2 50 150 NA
5 2 1 1 10 90 10
但是，对于sales_per_shop，我无法这样做：
z %>%
group_by(shop) %>%
mutate(lag_sales_per_shop = lag(sales_per_shop, 1))
# A tibble: 5 x 6
# Groups: shop [3]
month shop product sales sales_per_shop lag_sales_per_shop
<int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 10 90 NA
2 1 1 2 20 90 90
3 1 2 1 40 120 NA
4 1 3 2 50 150 NA
5 2 1 1 10 90 90
如您所见，第一个月仍然有一个值。由于我落后了一个月，因此不应该有任何价值。有可能滞后于另一个值吗？
结果应如下所示：
# A tibble: 12 x 7
# Groups: shop, product [4]
month shop product sales sales_per_shop lag_sales lag_sales_per_shop
<int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 10 90 NA NA
2 1 1 2 20 90 NA NA
3 1 2 1 40 120 NA NA
4 1 3 2 50 150 NA NA
5 2 1 1 10 90 10 90
6 2 1 2 20 90 20 90
7 2 2 1 40 120 40 120
8 2 3 2 50 150 50 150
9 3 1 1 10 90 10 90
10 3 1 2 20 90 20 90
11 3 2 1 40 120 40 120
12 3 3 2 50 150 50 150
structure(list(month = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L), shop = c(1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3), product = c(1,
2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2), sales = c(10, 20, 40, 50, 10,
20, 40, 50, 10, 20, 40, 50), sales_per_shop = c(90, 90, 120,
150, 90, 90, 120, 150, 90, 90, 120, 150)), row.names = c(NA,
-12L), class = "data.frame")
答案 0 :(得分：1)
您可能需要left_join-
df %>%
left_join(
df %>%
mutate(month = month + 1) %>%
distinct(shop, month, sales_per_shop) %>%
rename(lag_sales_per_shop = sales_per_shop),
by = c("shop", "month")
)
month shop product sales sales_per_shop lag_sales_per_shop
1 1 1 1 10 90 NA
2 1 1 2 20 90 NA
3 1 2 1 40 120 NA
4 1 3 2 50 150 NA
5 2 1 1 10 90 90
6 2 1 2 20 90 90
7 2 2 1 40 120 120
8 2 3 2 50 150 150
9 3 1 1 10 90 90
10 3 1 2 20 90 90
11 3 2 1 40 120 120
12 3 3 2 50 150 150
答案 1 :(得分：1)
这是带有filter和bind_rows的另一个版本
library(dplyr)
z %>%
filter(month == first(month)) %>%
bind_rows(z %>%
filter(month != first(month)) %>%
mutate(lag_sales = sales, lag_sales_per_shop = sales_per_shop))
# month shop product sales sales_per_shop lag_sales lag_sales_per_shop
#1 1 1 1 10 90 NA NA
#2 1 1 2 20 90 NA NA
#3 1 2 1 40 120 NA NA
#4 1 3 2 50 150 NA NA
#5 2 1 1 10 90 10 90
#6 2 1 2 20 90 20 90
#7 2 2 1 40 120 40 120
#8 2 3 2 50 150 50 150
#9 3 1 1 10 90 10 90
#10 3 1 2 20 90 20 90
#11 3 2 1 40 120 40 120
#12 3 3 2 50 150 50 150