如果它们之间的所有值都是NA，则连接R中数据框的两列？

Question

我有一个如下所示的数据框：

> sample
# A tibble: 6 x 10
  Level_1 Level_2 Level_3 Level_4 Level_5 Level_6 Level_7 Level_8 Level_9 Supplier 
    <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl> <lgl>   <chr>    
1       1       2       3       4       8      NA      NA      NA NA      orioles  
2       1       2       3       4       9      13      NA      NA NA      nationals
3       1       2       3       5      10      14      16      18 NA      dodgers  
4       1       2       3       5      10      14      17      19 NA      cardinals
5       1       2       3       6      11      NA      NA      NA NA      giants   
6       1       2       3       7      12      15      NA      NA NA      padres  

如果它们之间的所有值都是NA，我想做的是将供应商列与任何级别列连接起来。我想到的另一种方法是，如果 Level 列右侧的列是 NA，则将该列与供应商列连接。

我在考虑一个 for 循环，但我还没有弄清楚如何实现逻辑。我想的逻辑是这样的：

for (level in levels) {
   if is.na(level n + 1) {
     paste0(level, Supplier)
     }
   else {
    level}
   }

我也可以像这样打一堆mutate，但它似乎超级重复且不必要：

sample %>%
  mutate(
    Level_5 = ifelse(
      is.na(Level_6),
      paste0(Supplier, "<br>", Level_5),
      Level_5)
  )

这是数据的输出：

structure(list(Level_1 = c(1, 1, 1, 1, 1, 1), Level_2 = c(2, 
2, 2, 2, 2, 2), Level_3 = c(3, 3, 3, 3, 3, 3), Level_4 = c(4, 
4, 5, 5, 6, 7), Level_5 = c(8, 9, 10, 10, 11, 12), Level_6 = c(NA, 
13, 14, 14, NA, 15), Level_7 = c(NA, NA, 16, 17, NA, NA), Level_8 = c(NA, 
NA, 18, 19, NA, NA), Level_9 = c(NA, NA, NA, NA, NA, NA), Supplier = c("orioles", 
"nationals", "dodgers", "cardinals", "giants", "padres")), row.names = c(NA, 
-6L), class = c("tbl_df", "tbl", "data.frame"))

Answer 1

说实话，我不能 100% 确定您想要的输出。使用dplyr 和tidyr：

library(tidyr)
library(dplyr)

sample %>%
  pivot_longer(cols=starts_with("Level_"), names_prefix="Level_", names_to="level") %>%
  drop_na() %>%
  group_by(Supplier) %>%
  mutate(new_val=ifelse(level==max(level), paste0(Supplier, "<br>", value), value)) %>%
  select(-value) %>%
  pivot_wider(names_from=level, names_prefix="Level_", values_from=new_val)

返回

# A tibble: 6 x 9
# Groups:   Supplier [6]
  Supplier  Level_1 Level_2 Level_3 Level_4 Level_5      Level_6         Level_7 Level_8        
  <chr>     <chr>   <chr>   <chr>   <chr>   <chr>        <chr>           <chr>   <chr>          
1 orioles   1       2       3       4       orioles<br>8 NA              NA      NA             
2 nationals 1       2       3       4       9            nationals<br>13 NA      NA             
3 dodgers   1       2       3       5       10           14              16      dodgers<br>18  
4 cardinals 1       2       3       5       10           14              17      cardinals<br>19
5 giants    1       2       3       6       giants<br>11 NA              NA      NA             
6 padres    1       2       3       7       12           padres<br>15    NA      NA  

我丢失了 Level_9 列，因为它只包含 NA。您可以轻松地再次添加它。

Answer 2

另一种方法：

library(tidyr)
library(dplyr)

df %>% mutate(across(contains('Level'), ~ as.character(.))) %>% 
   mutate(across(contains('Level'), ~ coalesce(., Supplier))) %>% select(-Supplier) %>% 
   mutate(ID = row_number()) %>% 
     pivot_longer(cols = -ID) %>% group_by(ID) %>% 
       mutate(value = case_when(duplicated(value) ~ NA_character_, TRUE ~ value)) %>% pivot_wider(names_from = name, values_from = value) %>% 
   ungroup() %>% select(-ID)
# A tibble: 6 x 9
  Level_1 Level_2 Level_3 Level_4 Level_5 Level_6 Level_7   Level_8 Level_9  
  <chr>   <chr>   <chr>   <chr>   <chr>   <chr>   <chr>     <chr>   <chr>    
1 1       2       3       4       8       orioles NA        NA      NA       
2 1       2       3       4       9       13      nationals NA      NA       
3 1       2       3       5       10      14      16        18      dodgers  
4 1       2       3       5       10      14      17        19      cardinals
5 1       2       3       6       11      giants  NA        NA      NA       
6 1       2       3       7       12      15      padres    NA      NA       

Answer 3

非常有趣的问题。这是我没有pivot_longer的方法：

library(dplyr)
# check is na
ind <- !is.na(df1)

# save vector who fullfill assumption value before first NA
values <- as.vector(tapply(df1[ind], row(df1)[ind], tail, 1))

# bind to dataframe
df2 <- cbind(df, values)

# accomplish the task
df2 %>% 
  mutate(across(Level_1:Level_9, ~ case_when(. == values ~ str_c(Supplier ,.),
                                             . != values ~ as.character(.)))) %>% 
  select(-values)

输出：

  Level_1 Level_2 Level_3 Level_4  Level_5     Level_6 Level_7     Level_8 Level_9  Supplier
1       1       2       3       4 orioles8        <NA>    <NA>        <NA>    <NA>   orioles
2       1       2       3       4        9 nationals13    <NA>        <NA>    <NA> nationals
3       1       2       3       5       10          14      16   dodgers18    <NA>   dodgers
4       1       2       3       5       10          14      17 cardinals19    <NA> cardinals
5       1       2       3       6 giants11        <NA>    <NA>        <NA>    <NA>    giants
6       1       2       3       7       12    padres15    <NA>        <NA>    <NA>    padres

Answer 4

最终更新 我意识到我试图在每一行中找到最大值并将其替换为所需的连接字符串时的错误。所以我想出了另一个解决方案，它只替换最后一个非NA 值（它也可以不是该行的最大值），因为所有值都不是数字。所以这是我的最终解决方案：

library(dplyr)
library(stringr)
library(purrr)

df %>%
  pmap_dfr(., ~ {x <- c(...)[-10][!is.na(c(...)[-10])];
  ind <- which(c(...) == x[length(x)]);
  replace(c(...), ind[length(ind)], str_c(..10, x[length(x)], sep = "_"))}
  )

# A tibble: 6 x 10
  Level_1 Level_2 Level_3 Level_4 Level_5   Level_6      Level_7 Level_8      Level_9 Supplier 
  <chr>   <chr>   <chr>   <chr>   <chr>     <chr>        <chr>   <chr>        <chr>   <chr>    
1 1       2       3       4       orioles_8 NA           NA      NA           NA      orioles  
2 1       2       3       4       9         nationals_13 NA      NA           NA      nationals
3 1       2       3       5       10        14           16      dodgers_18   NA      dodgers  
4 1       2       3       5       10        14           17      cardinals_19 NA      cardinals
5 1       2       3       6       giants_11 NA           NA      NA           NA      giants   
6 1       2       3       7       12        padres_15    NA      NA           NA      padres 

Answer 5

结合c_across和across的方法

library(tidyverse)
df %>% rowwise() %>%
  mutate(dummy = max(which(!is.na(c_across(starts_with('Level')))))) %>% ungroup() %>%
  mutate(across(starts_with('Level_'), 
                ~ifelse(as.numeric(str_remove(cur_column(), 'Level_')) == dummy, paste(Supplier, ., sep = '_'), .)))

# A tibble: 6 x 11
  Level_1 Level_2 Level_3 Level_4 Level_5   Level_6      Level_7 Level_8      Level_9 Supplier  dummy
    <dbl>   <dbl>   <dbl>   <dbl> <chr>     <chr>          <dbl> <chr>        <lgl>   <chr>     <int>
1       1       2       3       4 orioles_8 NA                NA NA           NA      orioles       5
2       1       2       3       4 9         nationals_13      NA NA           NA      nationals     6
3       1       2       3       5 10        14                16 dodgers_18   NA      dodgers       8
4       1       2       3       5 10        14                17 cardinals_19 NA      cardinals     8
5       1       2       3       6 giants_11 NA                NA NA           NA      giants        5
6       1       2       3       7 12        padres_15         NA NA           NA      padres        6

结合上面我朋友Anoushiravan的answer使用的which策略可以简化为：

• 在purrr::pmap_dfr 里面做这些
  • 收集/存储列数到临时变量n
  • 收集/存储所需的索引到临时变量i
  • 创建一个长度为n 的temp 变量，并在i 和F 其他地方有一个T
  • 使用replace 将ith 变量（使用temp）替换为所需值

df %>%
  pmap_dfr(., ~ {n <- ncol(df); i <- max(which(!is.na(c(...)[-n])));
  tmp <- rep(F, n); tmp[i] <- T;
  replace(c(...), tmp, paste(c(...)[n], c(...)[i], sep = '_'))})

# A tibble: 6 x 10
  Level_1 Level_2 Level_3 Level_4 Level_5   Level_6      Level_7 Level_8      Level_9 Supplier 
  <chr>   <chr>   <chr>   <chr>   <chr>     <chr>        <chr>   <chr>        <chr>   <chr>    
1 1       2       3       4       orioles_8 NA           NA      NA           NA      orioles  
2 1       2       3       4       9         nationals_13 NA      NA           NA      nationals
3 1       2       3       5       10        14           16      dodgers_18   NA      dodgers  
4 1       2       3       5       10        14           17      cardinals_19 NA      cardinals
5 1       2       3       6       giants_11 NA           NA      NA           NA      giants   
6 1       2       3       7       12        padres_15    NA      NA           NA      padres