根据一列中的条件在数据框中创建新变量，从另一列中提取？ (dplyr)

Question

我有以下数据框：

    df <- structure(list(country = c("Ghana", "Eritrea", "Ethiopia", "Ethiopia", 
"Congo - Kinshasa", "Ethiopia", "Ethiopia", "Ghana", "Botswana", 
"Nigeria"), CommodRank = c(1L, 2L, 3L, 1L, 3L, 1L, 1L, 1L, 1L, 
1L), topCommodInCountry = c(TRUE, FALSE, FALSE, TRUE, FALSE, 
TRUE, TRUE, TRUE, TRUE, TRUE), Main_Commod = c("Gold", "Copper", 
"Nickel", "Gold", "Gold", "Gold", "Gold", "Gold", "Diamonds", 
"Iron Ore")), row.names = c(NA, -10L), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"), vars = "country", drop = TRUE, indices = list(
    8L, 4L, 1L, c(2L, 3L, 5L, 6L), c(0L, 7L), 9L), group_sizes = c(1L, 
1L, 1L, 4L, 2L, 1L), biggest_group_size = 4L, labels = structure(list(
    country = c("Botswana", "Congo - Kinshasa", "Eritrea", "Ethiopia", 
    "Ghana", "Nigeria")), row.names = c(NA, -6L), class = "data.frame", vars = "country", drop = TRUE, .Names = "country"), .Names = c("country", 
"CommodRank", "topCommodInCountry", "Main_Commod"))

df

            country CommodRank topCommodInCountry Main_Commod
1             Ghana          1               TRUE        Gold
2           Eritrea          2              FALSE      Copper
3          Ethiopia          3              FALSE      Nickel
4          Ethiopia          1               TRUE        Gold
5  Congo - Kinshasa          3              FALSE        Gold
6          Ethiopia          1               TRUE        Gold
7          Ethiopia          1               TRUE        Gold
8             Ghana          1               TRUE        Gold
9          Botswana          1               TRUE    Diamonds
10          Nigeria          1               TRUE    Iron Ore  

我正在尝试添加另一列，显示此数据集中每个国家/地区的顶级商品（顶级 CommodRank），但我不确定如何。我可以用 CommodRank == 1 的“Main_Commod”标记“topcommod”，但我想将此相同的值复制到 CommodRank ！= 1 的情况。如下所示，第 3 行和第 4 行的埃塞俄比亚值都应为“黄金'。

df %>% mutate(topcommod = ifelse(CommodRank == 1, Main_Commod, 'unknown'))


            country CommodRank topCommodInCountry Main_Commod topcommod
1             Ghana          1               TRUE        Gold      Gold
2           Eritrea          2              FALSE      Copper   unknown
3          Ethiopia          3              FALSE      Nickel   unknown
4          Ethiopia          1               TRUE        Gold      Gold
5  Congo - Kinshasa          3              FALSE        Gold   unknown
6          Ethiopia          1               TRUE        Gold      Gold
7          Ethiopia          1               TRUE        Gold      Gold
8             Ghana          1               TRUE        Gold      Gold
9          Botswana          1               TRUE    Diamonds  Diamonds
10          Nigeria          1               TRUE    Iron Ore  Iron Ore

理想情况下，我正在寻找可以添加到现有长系列管道 %>% 函数调用的 dplyr 解决方案，但任何解决方案都会有所帮助。

Answer 1

IIUC，有多种方法可以做到这一点，例如：

df %>% mutate(topCom = if(!any(topCommodInCountry)) "unknown" 
                       else Main_Commod[which.max(topCommodInCountry)])

# A tibble: 10 x 5
# Groups:   country [6]
   country          CommodRank topCommodInCountry Main_Commod topCom  
   <chr>                 <int> <lgl>              <chr>       <chr>   
 1 Ghana                     1 TRUE               Gold        Gold    
 2 Eritrea                   2 FALSE              Copper      unknown 
 3 Ethiopia                  3 FALSE              Nickel      Gold    
 4 Ethiopia                  1 TRUE               Gold        Gold    
 5 Congo - Kinshasa          3 FALSE              Gold        unknown 
 6 Ethiopia                  1 TRUE               Gold        Gold    
 7 Ethiopia                  1 TRUE               Gold        Gold    
 8 Ghana                     1 TRUE               Gold        Gold    
 9 Botswana                  1 TRUE               Diamonds    Diamonds
10 Nigeria                   1 TRUE               Iron Ore    Iron Ore

关于 OP 在评论中的问题如何处理多个顶级商品的关系，您可以执行以下操作：

df %>% 
  mutate(topCom = if(!any(topCommodInCountry)) "unknown" 
              else paste(unique(Main_Commod[topCommodInCountry]), collapse = "/"))

如果一个国家有多个独特的顶级商品，它们将被粘贴到一个字符串中，以/分隔。

Answer 2

dplyr 的另一种模式...

df %>% arrange(CommodRank) %>%
    mutate(topCommod = Main_Commod[1])

Answer 3

这不是答案，而是从@docendo discimus 答案中学到了很多东西，我花了一秒钟来理解“如果是否定的”（!any(topCommodInCountry)），我想知道是只有我一个人还是我的电脑需要一秒钟也可以这样做:)

使用相同的数据集，我研究了使if else 为正的想法。首先，我在两个解决方案之间测试了identical：

identical(
  #Negative
  df %>% 
    mutate(topCom = if(!any(topCommodInCountry)) "unknown" 
           else Main_Commod[which.max(topCommodInCountry)]), 
  #Positive
  df %>% 
    mutate(topCom = if(any(topCommodInCountry)) Main_Commod[which.max(topCommodInCountry)] 
           else "unknown"))

[1] TRUE

接下来，我测试了两者的基准：

require(rbenchmark)

benchmark("Negative" = {
  df %>% 
    mutate(topCom = if(!any(topCommodInCountry)) "unknown" 
           else Main_Commod[which.max(topCommodInCountry)])
},
"Positive" = {
  df %>% 
    mutate(topCom = if(any(topCommodInCountry)) Main_Commod[which.max(topCommodInCountry)] 
           else  "unknown")
},
replications = 10000,
columns = c("test", "replications", "elapsed",
            "relative", "user.self", "sys.self"))

差异并不大，但我假设使用更大的数据集它会增加。

      test replications elapsed relative user.self sys.self
1 Negative        10000   12.59    1.015     12.44        0
2 Positive        10000   12.41    1.000     12.30        0