如何从更改的数据框中创建查找 table?

How to create a look-up table from a dataframe of changes?

我想根据变化的数据框创建查找 table。原始数据框的每一行都表示给定地区编码的变化。 该数据集涵盖了从 2009 年到 2019 年的某个时间段。虽然一个地区在那个时间段内可能会经历一些变化,但我想要 2009 年和2019年各区编码。也就是第一个和最新的编码。

数据框涵盖数百个地区。一些地区可能只经历一次变化,而其他地区可能经历多次变化。一个区可以合并或拆分成多个区。

理想的查找 table 如下所示:

coding_2009 coding_2019
00QR S12000047
00QR S12000048
00RB S12000047
00RB S12000048

coding_2009 是截至 2009 年的地区编码,coding_2019 是截至 2019 年的最新编码。

原始数据框(子集),其中每一行显示一个变化,看起来像:

past new date
00QR S12000015 2009-01-01
S12000015 S12000047 2018-02-02
S12000015 S12000048 2018-02-02
00RB S12000015 2009-01-01
S12000024 S12000047 2018-02-02
S12000024 S12000048 2018-02-02

对于每一行,past 是自 date 起重新编码为 new 的代码。

比如区00QR变成了S12000015,后来分裂成S12000047S12000048

我已经处理这个问题好几个星期了,尝试了不同的临时版本,但 none 似乎始终如一。请注意,代码需要考虑到一些地区只经历了一次变化,而其他地区可能经历了两次或更多变化。区域也可以拆分或合并,如示例所示。

理想的答案是 tidyverse.

对于 reprex,我在下面选择了部分地区。

感谢您的帮助!将不胜感激。

代表数据: (您也可以超越并使用原始数据集,Changes.csv。请参阅下面的 link)

# Library tibble (a part of tidyverse) is needed to copy paste reprex data
#install.packages("tibble") # if you need to install it
library(tibble)

data <- tibble::tribble(
        ~past,        ~new,        ~date,
       "00RJ", "S12000013", "2009-01-01",
       "00QR", "S12000015", "2009-01-01",
       "00RB", "S12000024", "2009-01-01",
       "13UD", "E07000015", "2009-01-01",
       "15UH", "E07000025", "2009-01-01",
       "00HC", "E06000024", "2009-01-01",
       "00KG", "E06000034", "2009-01-01",
       "19UD", "E07000049", "2009-01-01",
       "19UE", "E07000050", "2009-01-01",
       "19UG", "E07000051", "2009-01-01",
       "19UH", "E07000052", "2009-01-01",
       "19UJ", "E07000053", "2009-01-01",
  "E07000017", "E06000049", "2009-04-01",
  "E07000025", "E06000053", "2009-04-01",
  "E07000014", "E06000049", "2009-04-01",
  "E07000015", "E06000049", "2009-04-01",
  "S12000013", "S12000013", "2015-06-16",
  "S12000013", "S12000013", "2015-11-01",
  "S12000015", "S12000047", "2018-02-02",
  "S12000024", "S12000047", "2018-02-02",
  "S12000015", "S12000048", "2018-02-02",
  "S12000024", "S12000048", "2018-02-02",
  "E07000049", "E06000059", "2019-04-01",
  "E07000050", "E06000059", "2019-04-01",
  "E07000053", "E06000059", "2019-04-01",
  "E07000051", "E06000059", "2019-04-01",
  "E07000052", "E06000059", "2019-04-01"
  )

# Convert date to Date (after being copy pasted as tibble)
data$date <- as.Date(data$date)

对于任何感兴趣的人,此数据来自英国的 Code History Database。您可以从下面的 link 下载 zip。这是名为 Changes.csv: https://geoportal.statistics.gov.uk/datasets/code-history-database-december-2019-for-the-united-kingdom 的文件。注意,在Changes.csv中,past被命名为geogcd_pnew被命名为geogcddate被命名为oper_date.

您实际上是在查看一个扁平的树结构。使用 igraph 包可以很容易地绘制它:


library(igraph)
g <- dat %>% select( past,new ) %>% t %>% c %>% graph
plot( g )

从现在开始,有很多方法可以解决这个问题,但归结为 深度优先宽度优先 方法问题。

假设我们有几个小图是合理的。一堆经过一些更改的不同代码,而不是经过多次更改的 select 几个代码。

这建议采用 宽度优先 方法,并且可以通过将数据与其自身连接来解决,希望不要太多次:


## work with data.table for that extra speed.
setDT(dat)

## remove duplicate entries of same code
dat <- dat[, .(date=max(date)), by=.(past,new) ]

## these are the roots, all `past` values never present in `new`
roots <- dat[ !past %in% new ]

## likewise, the leaves are those that never appear as `past` , unless they are self referencing.
leaves <- unique( dat[ !new %in% past | new == past, !"past" ], by="new" )


dd <- copy(roots)

## sucessively add next step from the source data till we have arrived at leaves only.
while( !all( dd$new %in% leaves$new ) ) {
    dd <- unique(
        merge( dd, dat, by.x="new", by.y="past", all.x=TRUE )[ , .(date.x, past, new=coalesce(new.y,new), date.y=coalesce(date.y,date.x) ) ]
    )
}

## final cleanup
dd[ order(past), .(coding_2009=past,coding_2019=new) ]

输出:


> dd[ order(past), .(coding_2009=past,coding_2019=new) ]
    coding_2009 coding_2019
 1:        00HC   E06000024
 2:        00KG   E06000034
 3:        00QR   S12000047
 4:        00QR   S12000048
 5:        00RB   S12000047
 6:        00RB   S12000048
 7:        00RJ   S12000013
 8:        13UD   E06000049
 9:        15UH   E06000053
10:        19UD   E06000059
11:        19UE   E06000059
12:        19UG   E06000059
13:        19UH   E06000059
14:        19UJ   E06000059
15:   E07000014   E06000049
16:   E07000017   E06000049

现在我只看了迷你数据集,所以我不知道代码在野外会如何形成,但你可以试一试。

看上图,我们看到每个图从根到叶最多有3步,这意味着上面的while循环只需要运行一次。

Sirius 使用 data.table 提供了一个惊人的答案。在这里,我将该答案翻译成 tidyverse:

# Remove duplicate entries of same code
data_sub <- data %>%
  group_by(past, new) %>%
  filter(date == max(date)) %>%
  ungroup()

# Create roots: All past values never present in new
roots <- data_sub %>%
  filter(!past %in% new)

# Create leaves: Those that never appear as past, unless they self reference
leaves <- data_sub %>%
  filter(!new %in% past | new == past) %>%
  select(-past) %>%
  distinct(new, .keep_all = TRUE)

# Copy before loop
dd <- roots

# Successively add next step from source data until we have arrived at leaves only
while(!all(dd$new %in% leaves$new)) {
  
  # Join
  dd_merge <- left_join(dd, data_sub, by = c("new" = "past"))
  
  # Coalesce
  dd_sub <- dd_merge %>%
    transmute(date.x,
              past,
              new = coalesce(new.y, new),
              date.y = coalesce(date.y, date.x))
  
  # Take unique
  dd <- unique(dd_sub)
  
}