根据 R 中的组解析值

Question

我有一个非常大的数据集，其中的一个样本如下所示：

| Id | Name    | Start_Date | End_Date   |
|----|---------|------------|------------|
| 10 | Mark    | 4/2/1999   | 7/5/2018   |
| 10 |         | 1/1/2000   | 9/24/2018  |
| 25 |         | 5/3/1968   | 6/3/2000   |
| 25 |         | 6/6/2009   | 4/23/2010  |
| 25 | Anthony | 2/20/2010  | 7/21/2016  |
| 25 |         | 9/12/2014  | 11/26/2019 |

我需要根据 Id 解析 Name 列中的名称，以便输出 table 如下所示：

| Id | Name    | Start_Date | End_Date   |
|----|---------|------------|------------|
| 10 | Mark    | 4/2/1999   | 7/5/2018   |
| 10 | Mark    | 1/1/2000   | 9/24/2018  |
| 25 | Anthony | 5/3/1968   | 6/3/2000   |
| 25 | Antony  | 6/6/2009   | 4/23/2010  |
| 25 | Anthony | 2/20/2010  | 7/21/2016  |
| 25 | Anthony | 9/12/2014  | 11/26/2019 |

如何实现如上所示的输出？我经历了替代和解析功能，但无法理解它们如何应用于此问题。

我的数据集是：

df=data.frame(Id=c("10","10","25","25","25","25"),Name=c("Mark","","","","Anthony",""),
              Start_Date=c("4/2/1999", "1/1/2000","5/3/1968","6/6/2009","2/20/2010","9/12/2014"),
              End_Date=c("7/5/2018","9/24/2018","6/3/2000","4/23/2010","7/21/2016","11/26/2019"))

Answer 1

我们可以把空格("")改成NA，用fill把NA元素替换成之前的非NA元素

library(dplyr)
library(tidyr)
df1 %>%      
   mutate(Name = na_if(Name, "")) %>%
   group_by(Id) %>%
   fill(Name, .direction = "down") %>%
   fill(Name, .direction = "up)
# A tibble: 6 x 4
# Groups:   Id [2]
#  Id    Name    Start_Date End_Date  
#  <chr> <chr>   <chr>      <chr>     
#1 10    Mark    4/2/1999   7/5/2018  
#2 10    Mark    1/1/2000   9/24/2018 
#3 25    Anthony 5/3/1968   6/3/2000  
#4 25    Anthony 6/6/2009   4/23/2010 
#5 25    Anthony 2/20/2010  7/21/2016 
#6 25    Anthony 9/12/2014  11/26/2019

在 tidyr (‘0.8.3.9000’) 的 devel 版本中，这可以在单个 fill 语句中完成，因为 .direction = "downup" 也是一个选项

df1 %>%      
   mutate(Name = na_if(Name, "")) %>%
   group_by(Id) %>%
   fill(Name, .direction = "downup")

或者另一种选择是按 'Id' 分组，mutate 将 'Name' 作为 first 非空白元素

df1 %>%
    group_by(Id) %>%        
    mutate(Name = first(Name[Name!=""])) 
# A tibble: 6 x 4
# Groups:   Id [2]
#  Id    Name    Start_Date End_Date  
#  <chr> <chr>   <chr>      <chr>     
#1 10    Mark    4/2/1999   7/5/2018  
#2 10    Mark    1/1/2000   9/24/2018 
#3 25    Anthony 5/3/1968   6/3/2000  
#4 25    Anthony 6/6/2009   4/23/2010 
#5 25    Anthony 2/20/2010  7/21/2016 
#6 25    Anthony 9/12/2014  11/26/2019

数据

df1 <- structure(list(Id = c("10", "10", "25", "25", "25", "25"), Name = c("Mark", 
"", "", "", "Anthony", ""), Start_Date = c("4/2/1999", "1/1/2000", 
"5/3/1968", "6/6/2009", "2/20/2010", "9/12/2014"), End_Date = c("7/5/2018", 
"9/24/2018", "6/3/2000", "4/23/2010", "7/21/2016", "11/26/2019"
)), class = "data.frame", row.names = c(NA, -6L))

Answer 2

使用末尾注释中可重复定义的DF，将Name的每个零长度元素替换为NA，然后使用na.omit得到唯一的非NA用来填充。我们假设每个 Id 只有一个非 NA，问题就是这种情况。如果不是，我们可以将 na.omit 替换为 function(x) unique(na.omit(x))，假设非 NA 在 Id 中都是相同的。没有使用包。

transform(DF, Name = ave(replace(Name, !nzchar(Name), NA), Id, FUN = na.omit))

给予：

  Id    Name Start_Date   End_Date
1 10    Mark   4/2/1999   7/5/2018
2 10    Mark   1/1/2000  9/24/2018
3 25 Anthony   5/3/1968   6/3/2000
4 25 Anthony   6/6/2009  4/23/2010
5 25 Anthony  2/20/2010  7/21/2016
6 25 Anthony  9/12/2014 11/26/2019

na.strings

如果我们首先确保 Name 的零长度元素是 NA，我们可以稍微简化一下。我们将注释中的 read.table 行替换为下面的第一行。那么这只是使用 na.locf0.

的问题

DF <- read.table(text = Lines, header = TRUE, as.is = TRUE, sep = "|", 
  strip.white = TRUE, na.strings = "")

transform(DF, Name = ave(Name, Id, FUN = na.omit))

备注

可重现形式的输入：

Lines <- "
 Id | Name    | Start_Date | End_Date   
 10 | Mark    | 4/2/1999   | 7/5/2018   
 10 |         | 1/1/2000   | 9/24/2018  
 25 |         | 5/3/1968   | 6/3/2000   
 25 |         | 6/6/2009   | 4/23/2010  
 25 | Anthony | 2/20/2010  | 7/21/2016  
 25 |         | 9/12/2014  | 11/26/2019"
DF <- read.table(text = Lines, header = TRUE, as.is = TRUE, sep = "|", strip.white = TRUE)

根据 R 中的组解析值

parse values based on groups in R

grouping

r

substitution

gsub

数据

na.strings

备注