R 中数据集中具有最小值和最大值的列名

Question

我有这个数据集：

   Year  January February March April   May  June  July August 
   <chr>   <dbl>    <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl>         
 1 2018     45        51    63    61    79    85    88     85         
 2 2017     51        60    65    69    75    82    86     84          
 3 2016     47        55    61    68    72    84    87     85        
... with 20 more rows

我想得到每一行的最小值和最大值对应的月份，以及最大值和最小值的差值。这是我的最小值和最大值代码，

x <- colnames(data)[apply(data[,c(2:9)],1,which.max)]
y <- colnames(data)[apply(data[,c(2:9)],1,which.min)]
data$MaxMonth <- x
data$MinMonth <- y

但是，它给我 Year 作为某些 which.min 函数的输出。

   Year    January February March April May  June  July  August   MaxMonth  MinMonth    Diff
   <chr>   <dbl>    <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl>          
 1 2018     45        51    63    61    79    85    88     85      July      January    43
 2 2017     51        60    65    69    75    82    86     84      July      Year       35
 3 2016     47        55    61    68    72    84    87     85      July      Year       40
... with 20 more rows

Answer 1

无需执行 3 个应用功能。你可以这样做：

nms <- names(df)[-1]
n <- seq(nrow(df))
maxMonth = max.col(df[-1])
minMonth = max.col(-df[-1]) 
diff <-  df[-1][cbind(n, maxMonth)] - df[-1][cbind(n, minMonth)]
cbind(df, maxMonth = nms[maxMonth], minMonth = nms[minMonth], diff)

  Year January February March April May June July August maxMonth minMonth diff
1 2018      45       51    63    61  79   85   88     85     July  January   43
2 2017      51       60    65    69  75   82   86     84     July  January   35
3 2016      47       55    61    68  72   84   87     85     July  January   40

Answer 2

我认为对您 post 的评论强调了问题所在

你应该写

x <- colnames(data)[2:9][apply(data[,c(2:9)],1,which.max)]
y <- colnames(data)[2:9][apply(data[,c(2:9)],1,which.min)]
data$MaxMonth <- x
data$MinMonth <- y

这样是不是效果更好？

Answer 3

我们可以用pivot_longer整形为长格式，按'Year'分组，得到'value'的max/min对应的列名（用which.max/which.min) 然后加入原始数据

library(dplyr)
library(tidyr)
df %>% 
    pivot_longer(cols = -1) %>%
    group_by(Year) %>%
    summarise(maxMonth = name[which.max(value)],
           minMonth = name[which.min(value)]) %>%
    left_join(df, .)

Answer 4

library(tidyverse)
df %>% 
  mutate(max_month = pmap(across(January:August), ~ names(c(...)[which.max(c(...))])),
         min_month = pmap(across(January:August), ~ names(c(...)[which.min(c(...))]))
         ) %>% 
    unnest(cols = c(max_month, min_month)) %>%
  rowwise() %>% 
  mutate(Diff = max(c_across(January:August)) - min(c_across(January:August)))

输出：

   Year January February March April   May  June  July August max_month min_month  Diff
  <dbl>   <dbl>    <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl> <chr>     <chr>     <dbl>
1  2018      45       51    63    61    79    85    88     85 July      January      43
2  2017      51       60    65    69    75    82    86     84 July      January      35
3  2016      47       55    61    68    72    84    87     85 July      January      40

R 中数据集中具有最小值和最大值的列名

Column name with the min and max values in a dataset in R

r

dataframe

max

apply

min