改变 r 中某些行的前导或滞后列

Mutate a lead or lag column for certain rows in r

我的预测与移动假期不太一致。我正在尝试寻找快速修复方法:

这是我的数据框的结构:

df1:
Date         City         Visitors      WKN    WKN_2015   Holiday
2016-11-06   New York     40000         45     46         No_Holiday
2016-11-13   New York     50000         46     47         No_Holiday
2016-11-20   New York     50000         47     48         Thanksgiving
2016-11-27   New York     100000        48     49         Cyber_Monday
2016-12-04   New York     100000        49     50         No_Holiday
2016-12-11   New York     70000         50     51         No_Holiday
.
.
.
2017-11-23   New York     120000        47     47         Thanksgiving

一般来说,感恩节和网络星期一会有更多游客到访这座城市。但我的预测并未反映这一点。现在我想用这样的东西快速修复:

df1:
Date         City         Visitors      WKN    WKN_2015   Holiday        New_Visitors
2016-11-06   New York     40000         45     46         No_Holiday     40000 
2016-11-13   New York     50000         46     47         No_Holiday     50000     
2016-11-20   New York     50000         47     48         Thanksgiving   100000
2016-11-27   New York     100000        48     49         Cyber_Monday   100000
2016-12-04   New York     100000        49     50         No_Holiday     70000
2016-12-11   New York     70000         50     51         No_Holiday     70000
.
.
.
2017-11-23   New York     120000        47     47         Thanksgiving   120000

如果你看到上面的数据 新卷只在感恩节、网络星期一和网络星期一后一周发生了变化。 有什么方法可以自动执行此操作,因为 2017 年的数据仍在继续,依此类推。

我一直在考虑快速解决方案,直到我制定出适合移动假期的预测。谁能指出我正确的方向?

我试过类似的方法,但这不起作用,因为我只需要 lag/lead 那 3 周:

df1 <- 
df1 %>%
mutate(New_Visitors = ifelse(Holiday == "Thanksgiving", lag(Visitors, (WKN - WKN_2015), Visitors)

逻辑:每年查找感恩节,看看 WKN 是否匹配。如果不这样做,则根据 WKN 之间的差异调整从感恩节开始的接下来 3 周的访客。如果 WKN-WKN_2015 == -1 则在接下来的 3 行中将访问者领先 1,如果 WKN-WKN_2015 == 1 那么在接下来的 3 行中将访问者落后 1

数据
df1 <- structure(list(Date = c("2016-11-06", "2016-11-13", "2016-11-20", 
"2016-11-27", "2016-12-04", "2016-12-11", "2017-11-23"), City = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L), .Label = "New York", class = "factor"), 
    Visitors = c(40000L, 50000L, 50000L, 100000L, 100000L, 70000L, 
    120000L), WKN = c(45L, 46L, 47L, 48L, 49L, 50L, 47L), WKN_2015 = c(46L, 
    47L, 48L, 49L, 50L, 51L, 47L), Holiday = structure(c(2L, 
    2L, 3L, 1L, 2L, 2L, 3L), .Label = c("Cyber_Monday", "No_Holiday", 
    "Thanksgiving"), class = "factor")), .Names = c("Date", "City", 
"Visitors", "WKN", "WKN_2015", "Holiday"), row.names = c(NA, 
7L), class = "data.frame")

您每年只对三周感兴趣,您可以在 "Thanksgiving" 行中计算滞后值。我认为不需要 dplyr

df1$New_Visitors <- df1$Visitors             # copy Visitors
ind <- which(df1$Holiday == "Thanksgiving")  # get number of "Thanksgiving" rows

invisible(sapply(ind, function(x) {
  lag <- df1[x, "WKN_2015"] - df1[x, "WKN"]       # calculate the lag
  df1[x:(x+2), "New_Visitors"] <<- df1[(x+lag):(x+lag+2), "Visitors"]  # rewrite
}))
> df1  # this method treats the three weeks as a unit, so made two NA rows in the example data)
        Date     City Visitors WKN WKN_2015      Holiday New_Visitors
1 2016-11-06 New York    40000  45       46   No_Holiday        40000
2 2016-11-13 New York    50000  46       47   No_Holiday        50000
3 2016-11-20 New York    50000  47       48 Thanksgiving       100000
4 2016-11-27 New York   100000  48       49 Cyber_Monday       100000
5 2016-12-04 New York   100000  49       50   No_Holiday        70000
6 2016-12-11 New York    70000  50       51   No_Holiday        70000
7 2017-11-23 New York   120000  47       47 Thanksgiving       120000
8       <NA>     <NA>       NA  NA       NA         <NA>           NA
9       <NA>     <NA>       NA  NA       NA         <NA>           NA