Question

我的预测与移动假期并不完全一致。我想找到一个快速修复：

这是我数据框的结构：

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

一般情况下，感恩节和网络星期一会有更多的游客来到这个城市。但我的预测并未反映出这一点。现在我想快速修复一下这样的事情：

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

如果您看到上述数据新的音量仅在感恩节，网络星期一和网络星期一后一周更改。有没有办法实现自动化，因为2017年的数据仍在继续，依此类推。

我正在考虑快速修复，直到我制定出适合移动假期的预测。有人能指出我正确的方向吗？

我尝试过这样的事情，但这不起作用，因为我只需要3个星期就需要滞后/领先：

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

逻辑：每年都要寻找感恩节，看看WKN是否匹配。如果不是根据WKN之间的差异从感恩节开始调整未来3周的访客。如果WKN-WKN_2015 == -1然后将访问者引导为接下来的3行，如果WKN-WKN_2015 == 1，那么接下来的3行将访问者滞后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")

Answer 1

您每年只有三个星期的兴趣，您可以计算“感恩节”行中的滞后值。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

为r中的某些行改变超前或滞后列

1 个答案: