如何根据 r 中的其他变量更新变量的信息？

Question

df_input 是我拥有的数据框，我想将其转换为df_output。

例如，因为 2001-2003 年是 assembly=1，我们在 2001 年有一个获胜者。这意味着只要程序集不变，我们就有一个获胜者。

    df_input <- data.frame(winner  = c(1,0,0,0,2,0,0,0,1,0,0,0,0),
                           party = c("A",0,0,0,"B",0,0,0,"C",0,0,0,0), 
                           assembly= c(1,1,1,2,2,2,3,3,3,3,4,4,4), 
                           year = c(2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012,2013))    
    
    df_output <- data.frame(winner  = c(1,1,1,0,2,2,0,0,1,1,0,0,0),
                            party = c("A","A","A",0,"B","B",0,0,"C","C",0,0,0),
                            assembly= c(1,1,1,2,2,2,3,3,3,3,4,4,4), 
                            year = c(2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012,2013))    
    

如何根据“大会”更新与获胜者列中相同的信息？

编辑：如果有一个额外的字符串变量“party”怎么办？看看编辑：

执行此代码后出现以下错误：

    df_output <- df_input %>%
      mutate(df_input$party = if_else(is.na(df_input$party)==FALSE, df_input$party, NA_real_)) %>%
      group_by(assembly) %>%
      fill(df_input$party) %>%
      ungroup() %>%
      replace_na(list(df_input$party = 0)) 

错误：

Error: unexpected '=' in:
"      ungroup() %>%
      replace_na(list(df_input$party ="

Answer 1

一种选择是像这样使用tidyr::fill：

library(dplyr)
library(tidyr)   

df_input %>%
  mutate(winner = if_else(winner > 0, winner, NA_real_)) %>% 
  group_by(assembly) %>% 
  fill(winner) %>% 
  ungroup() %>% 
  replace_na(list(winner = 0))
#> # A tibble: 13 × 3
#>    winner assembly  year
#>     <dbl>    <dbl> <dbl>
#>  1      1        1  2001
#>  2      1        1  2002
#>  3      1        1  2003
#>  4      0        2  2004
#>  5      2        2  2005
#>  6      2        2  2006
#>  7      0        3  2007
#>  8      0        3  2008
#>  9      1        3  2009
#> 10      1        3  2010
#> 11      0        4  2011
#> 12      0        4  2012
#> 13      0        4  2013

Answer 2

这是带有cumsum 和ave 的基本R 方式。
请注意 R 4.1.0 中引入的新 lambda 函数 \(x) 的使用。如果出现错误，请使用较旧的function(x)。

with(df_input, ave(winner, assembly, FUN = \(x){
  y <- cumsum(x != 0) != 0
  if(any(y)) x[y] <- x[min(which(y))]
  x
}))
# [1] 1 1 1 0 2 2 0 0 1 1 0 0 0

只需将结果分配回winner列。

df_output <- df_input
df_output$winner <- with(df_output, ave(winner, assembly, FUN = \(x){
  y <- cumsum(x != 0) != 0
  if(any(y)) x[y] <- x[min(which(y))]
  x
}))

---

编辑

在Henrik's comment 之后，这里是更简单的cummax 解决方案。

with(df_input, ave(winner, assembly, FUN = cummax))

Answer 3

更新：见 cmets：

library(dplyr)
df_input %>% 
  group_by(assembly) %>% 
  mutate(winner = case_when(first(winner) > 0 ~ first(winner),
                            lag(winner, default=0) > winner ~ lag(winner),
                            TRUE ~ winner))

   winner assembly  year
    <dbl>    <dbl> <dbl>
 1      1        1  2001
 2      1        1  2002
 3      1        1  2003
 4      0        2  2004
 5      2        2  2005
 6      2        2  2006
 7      0        3  2007
 8      0        3  2008
 9      1        3  2009
10      1        3  2010
11      0        4  2011
12      0        4  2012
13      0        4  2013

第一个答案（不考虑第 3 行） 通过assembly分组后就可以使用lag函数了

library(dplyr)
df_input %>% 
  group_by(assembly) %>% 
  mutate(winner = ifelse(lag(winner, default = 0) > winner, lag(winner), winner))

Groups:   assembly [4]
   winner assembly  year
    <dbl>    <dbl> <dbl>
 1      1        1  2001
 2      1        1  2002
 3      0        1  2003
 4      0        2  2004
 5      2        2  2005
 6      2        2  2006
 7      0        3  2007
 8      0        3  2008
 9      1        3  2009
10      1        3  2010
11      0        4  2011
12      0        4  2012
13      0        4  2013