根据给定的因子在 R 中按函数分组获取多列的百分比值

Posted 2023-02-16

【中文标题】根据给定的因子在 R 中按函数分组获取多列的百分比值

【英文标题】：Get percentage values across multiple columns based on factors given a group by function in R

【发布时间】：2021-12-03 06:40:00

【问题描述】：

我有一个只查看 1 个 ID 及其各自资产的 df：

ID  | Asset | CONF_1 |       CONF_2 |    CONF_3 | 
1       A     PERFECT        HIGH        LOW    
1       B     PERFECT        LOW         LOW
1       C     LOW            HIGH        VERY LOW
1       D     NA             MEDIUM      MEDIUM
1       E     MEDIUM         MEDIUM      PERFECT
1       F     MEDIUM         VERY LOW    NA
1       G     VERY LOW       VERY LOW    VERY LOW
1       H     NA             PERFECT     HIGH

目标是重新组织 df，这样我就可以在给定每个 ID 和 3 个 Conf 字段的情况下分解每个置信水平（PERFECT、HIGH、MEDIUM、ETC）的百分比。

期望的输出

ID | CONFIDENCE | CONF_1 % | CONF_2 % | CONF_3 %
1     PERFECT      25 %      12.5 %     12.5 %
1     HIGH         0         25 %       12.5 %
1     MEDIUM       25 %      25 %       12.5 %
1     LOW          12.5 %    12.5 %     25 %
1     VERY LOW     12.5 %    25 %       25 %
1     NA           25 %      0          12.5 %

【问题讨论】：

我相信@akrun - 分母是 8，我只是计算了每个 conf 级别出现的次数，共 8 次

请查看我发布的解决方案

【参考方案1】：

按“ID”分组，summariseacross“CONF”列，在factor 列上使用table 获取频率计数，并在顺序中指定levels，并找到proportions

library(dplyr)
df1 %>% 
   group_by(ID) %>% 
   summarise(lvls = c("PERFECT", "HIGH", "MEDIUM", "LOW", "VERY LOW", NA), 
    across(starts_with("CONF"), 
    ~ 100 * proportions(table(factor(., levels = na.omit(lvls)), 
        useNA = "always"))), .groups = 'drop') %>%
   rename(CONFIDENCE = lvls)

-输出

# A tibble: 6 × 5
     ID CONFIDENCE CONF_1  CONF_2  CONF_3 
  <int> <chr>      <table> <table> <table>
1     1 PERFECT    25.0    12.5    12.5   
2     1 HIGH        0.0    25.0    12.5   
3     1 MEDIUM     25.0    25.0    12.5   
4     1 LOW        12.5    12.5    25.0   
5     1 VERY LOW   12.5    25.0    25.0   
6     1 <NA>       25.0     0.0    12.5   

--

或者另一种选择是使用pivot_longer 重塑为“长”格式，执行count 并使用pivot_wider 重塑为“宽”格式

library(tidyr)
df1 %>% 
   select(-Asset) %>% 
   pivot_longer(cols = starts_with("CONF"), values_to = 'CONFIDENCE') %>% 
   count(ID, name, CONFIDENCE) %>%
   group_by(ID, name) %>%
   mutate(n = 100 *n/sum(n) ) %>%
   ungroup %>%
   pivot_wider(names_from = name, values_from = n, values_fill = 0)

-输出

# A tibble: 6 × 5
     ID CONFIDENCE CONF_1 CONF_2 CONF_3
  <int> <chr>       <dbl>  <dbl>  <dbl>
1     1 LOW          12.5   12.5   25  
2     1 MEDIUM       25     25     12.5
3     1 PERFECT      25     12.5   12.5
4     1 VERY LOW     12.5   25     25  
5     1 <NA>         25      0     12.5
6     1 HIGH          0     25     12.5

数据

df1 <- structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), Asset = c("A", 
"B", "C", "D", "E", "F", "G", "H"), CONF_1 = c("PERFECT", "PERFECT", 
"LOW", NA, "MEDIUM", "MEDIUM", "VERY LOW", NA), CONF_2 = c("HIGH", 
"LOW", "HIGH", "MEDIUM", "MEDIUM", "VERY LOW", "VERY LOW", "PERFECT"
), CONF_3 = c("LOW", "LOW", "VERY LOW", "MEDIUM", "PERFECT", 
NA, "VERY LOW", "HIGH")), class = "data.frame", row.names = c(NA, 
-8L))

【讨论】：

很棒 - 我开始使用 pivot_wider 但这太棒了。谢谢。

有没有办法按 ID 排序，然后是 CONFIDENCE 级别？因此，每个 ID 具有相同的 CONFIDENCE 级别顺序，并且每个 ID 都会重置。

@Dinho 你可以指定arrange(ID, factor(CONFIDENCE, levels = c("PERFECT", "HIGH", "MEDIUM", "LOW", "VERY LOW"))) in betwen

【参考方案2】：

基于reshape2的解决方案：

library(dplyr)
library(reshape2)

df %>% 
  melt(id.vars="ID", measure.vars=paste0("CONF_",1:3), variable.name="X") %>% 
  dcast(ID + X ~ value, fun.aggregate = length) %>% 
  melt(id.vars=c("ID","X"), measure.vars=3:ncol(.)  ) %>% 
  dcast(ID+variable ~ X) %>% 
  group_by(ID) %>% 
  mutate(across(starts_with("CONF_"), ~ .x*100 /sum(.x))) %>% 
  rename(CONFIDENCE=variable) %>% 
  arrange(ID,CONFIDENCE)