输入:
目标:
使用以下规则创建名为“dayDifference”的新列:对于每对“item-city”对,计算相关对的日差。
期望的输出:
信息:当然可以有超过2行的对[例如我可以让'pizza-berlin'对100行:如果是这样的话总是取最大值(日期)并减去最小(日期)比萨 - 柏林对。
约束:
需要在R [例如没有与数据库的外部连接]
源代码:
df <- structure(list(id = c(4848L, 4887L, 4899L, 4811L, 4834L, 4892L
), item = structure(c(2L, 2L, 2L, 1L, 1L, 1L), .Label = c("Pasta",
"Pizza"), class = "factor"), city = structure(c(1L, 1L, 2L, 2L,
2L, 1L), .Label = c("Berlin", "Hamburg"), class = "factor"),
date = structure(c(17199, 17201, -643892, 17449, 17459, 17515
), class = "Date")), .Names = c("id", "item", "city", "date"
), row.names = c(NA, -6L), class = "data.frame")
Reduce
是一个很棒的功能
library(dplyr)
df %>%
group_by(item, city) %>%
mutate(dayDifference=abs(Reduce(`-`, as.numeric(range(date)))))
# A tibble: 6 x 5
# Groups: item, city [4]
id item city date dayDifference
<int> <fctr> <fctr> <date> <dbl>
1 4848 Pizza Berlin 2017-02-02 2
2 4887 Pizza Berlin 2017-02-04 2
3 4899 Pizza Hamburg 0207-02-01 0
4 4811 Pasta Hamburg 2017-10-10 10
5 4834 Pasta Hamburg 2017-10-20 10
6 4892 Pasta Berlin 2017-12-15 0
我会用data.table
做到这一点:
library(data.table)
setDT(df)
df[, min_date := min(date), by = c("item", "city")]
df[, max_date := max(date), by = c("item", "city")]
df[, dayDifference := difftime(max_date, min_date, units = "days")]
df[, c("min_date", "max_date") := NULL]
它会给你想要的输出:
id item city date dayDifference
1: 4848 Pizza Berlin 2017-02-02 2 days
2: 4887 Pizza Berlin 2017-02-04 2 days
3: 4899 Pizza Hamburg 0207-02-01 0 days
4: 4811 Pasta Hamburg 2017-10-10 10 days
5: 4834 Pasta Hamburg 2017-10-20 10 days
6: 4892 Pasta Berlin 2017-12-15 0 days
您也可以使用df[, dayDifference := max_date - min_date]
而不是df[, dayDifference := difftime(max_date, min_date, units = "days")]
。
不漂亮,但......
i<-unique(lapply(1:nrow(df),function(x) which(paste(df[,2],df[,3]) %in% paste(df[x,2],df[x,3]))))
for(j in 1:length(i)) df[i[[j]],"days"]<-abs(difftime(df[i[[j]],][1,"date"],df[i[[j]],][2,"date"]))
> df
id item city date days
1 4848 Pizza Berlin 2017-02-02 2
2 4887 Pizza Berlin 2017-02-04 2
3 4899 Pizza Hamburg 0207-02-01 NA
4 4811 Pasta Hamburg 2017-10-10 10
5 4834 Pasta Hamburg 2017-10-20 10
6 4892 Pasta Berlin 2017-12-15 NA