如何计算数据框中许多列的排名/百分位数，同时过滤每列的标准

问题描述 投票：0回答：2

我们必须计算数据框中 100 列的百分位数。在下面的示例中，需要百分位数的列名称是 

pctile_columns
。接收百分位数的标准是 (1) 该列不是 
NA
，并且 (1) 
min_pg
列是 
>= 12
。我们正在努力获取正确的百分位数集：

数据+尝试

temp_df = structure(list(group_var = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), 
                         min_pg = c(11, 15, 19, 7, 5, 34, 32, 27, 24, 18, 13, 10), 
                         stat1 = c(0.35, 0.32, 0.27, NA, NA, 0.42, 0.45, 0.47, 0.33, NA, 0.24, 0.39)), 
                    row.names = c(NA, -12L), class = "data.frame")

pctile_columns <- c('stat1') 
library(dplyr)
temp_output <- temp_df %>%
  group_by(group_var) %>%
  mutate(across(.cols = all_of(pctile_columns),
                .fns = ~ if_else(is.na(.) | min_pg < 12, as.numeric(NA),
                                 rank(., ties.method = "max")), 
                .names = "{.col}__rank")) %>%
  mutate(across(.cols = all_of(pctile_columns),
                .fns = ~ if_else(is.na(.) | min_pg < 12, as.numeric(NA),
                                 round((rank(., ties.method = "max") - 1) / (n() - 1) * 100, 0)),
                .names = "{.col}__pctile"))

输出

# Groups:   group_var [1]
   group_var min_pg stat1 stat1__rank stat1__pctile
       <dbl>  <dbl> <dbl>       <dbl>         <dbl>
 1         1     11  0.35          NA            NA
 2         1     15  0.32           3            18
 3         1     19  0.27           2             9
 4         1      7 NA             NA            NA
 5         1      5 NA             NA            NA
 6         1     34  0.42           7            55
 7         1     32  0.45           8            64
 8         1     27  0.47           9            73
 9         1     24  0.33           4            27
10         1     18 NA             NA            NA
11         1     13  0.24           1             0
12         1     10  0.39          NA            NA

此输出的问题在于，排名从 1 到 9，而它们应该从 1 到 7。即使具有 

stat1
的 
min_pg < 12
值已正确分配为 
NA
值，但在计算所有其他行的排名时，这些 
stat1
值仍会被考虑到 
rank
方程中。在这种情况下，正确的排名集应该是 1-7，因为有 7 个指标满足 
stat1
接收排名/百分位数的标准。

我们如何修改代码以根据我们的标准正确计算排名/百分位数？

r dataframe ranking
2个回答
1
投票

您可以编写 

statfun
并在 
by
中使用它。

> statfun <- \(x, stat) {
+   rk <- \(x, m, z=12) rank(replace(x, m < z, NA), 'keep', 'max')  ## rank fun
+   pctl <- \(x) round((x - 1L)/length(na.omit(x) - 1L)*100)  ## perc fun
+   o <- lapply(stat, \(s) {
+     r <- with(x, rk(get(s), x$min_pg))
+     p <- pctl(r)
+     data.frame(r, p) |> setNames(paste(s, c('rank', 'percentile'), sep='_'))
+   })
+   cbind(x, o)
+ }
> by(temp_df, ~group_var, statfun, stat=c('stat1', 'stat2')) |> do.call(what='rbind')
     group_var min_pg stat1 stat2 stat1_rank stat1_percentile stat2_rank stat2_percentile
1.1          1     11  0.35    NA         NA               NA         NA               NA
1.2          1     15  0.32  0.45          3               29          3               29
1.3          1     19  0.27  0.89          2               14          5               14
1.4          1      7    NA    NA         NA               NA         NA               NA
1.5          1      5    NA  0.27         NA               NA         NA               NA
1.6          1     34  0.42  0.63          5               57          4               57
1.7          1     32  0.45    NA          6               71         NA               71
1.8          1     27  0.47  0.24          7               86          1               86
1.9          1     24  0.33    NA          4               43         NA               43
1.10         1     18    NA  0.27         NA               NA          2               NA
1.11         1     13  0.24    NA          1                0         NA                0
1.12         1     10  0.39  0.43         NA               NA         NA               NA
2.13         2     11  0.35  0.42         NA               NA         NA               NA
2.14         2     12  0.31    NA          2               29         NA               29
2.15         2     13  0.27  0.47          1               14          5               14
2.16         2      6    NA  0.45         NA               NA         NA               NA
2.17         2      5    NA  0.39         NA               NA         NA               NA
2.18         2     31  0.43  0.45          3               57          4               57
2.19         2     22  0.45  0.35          5               71          3               71
2.20         2     29  0.45  0.27          5               86          1               86
2.21         2     24  0.63  0.31          6               43          2               43
2.22         2     11    NA  0.35         NA               NA         NA               NA
2.23         2     11  0.27  0.32         NA                0         NA                0
2.24         2      9  0.89  0.33         NA               NA         NA               NA

数据：

> dput(temp_df)
structure(list(group_var = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
), min_pg = c(11L, 15L, 19L, 7L, 5L, 34L, 32L, 27L, 24L, 18L, 
13L, 10L, 11L, 12L, 13L, 6L, 5L, 31L, 22L, 29L, 24L, 11L, 11L, 
9L), stat1 = c(0.35, 0.32, 0.27, NA, NA, 0.42, 0.45, 0.47, 0.33, 
NA, 0.24, 0.39, 0.35, 0.31, 0.27, NA, NA, 0.43, 0.45, 0.45, 0.63, 
NA, 0.27, 0.89), stat2 = c(NA, 0.45, 0.89, NA, 0.27, 0.63, NA, 
0.24, NA, 0.27, NA, 0.43, 0.42, NA, 0.47, 0.45, 0.39, 0.45, 0.35, 
0.27, 0.31, 0.35, 0.32, 0.33)), class = "data.frame", row.names = c(NA, 
-24L))
0
投票

怎么样：

## helper function:
rank_special <- \(xs, reject = FALSE){
  xs[reject] <- NA
  list(rank = rank(xs, ties.method = 'max', na.last = 'keep'),
       pctile = round(findInterval(xs, quantile(xs, 1:100 * .01, na.rm = TRUE)))
       )
}

library(dplyr)

pctile_columns <- c('stat1') 
LB <- 12 ## lower bound (set values below to NA)

temp_df %>%
  mutate(across(.cols = all_of(pctile_columns),
                .fns = list(rank = ~ rank_special(.x, reject = min_pg < LB )$rank,
                            ptile = ~ rank_special(.x, reject = min_pg < LB )$ptile
                           ),
                .names = '{.col}_{.fn}'
                ),
         .by = group_var
         )

##    group_var min_pg stat1 stat1_rank stat1_pctile
## 1          1     11  0.35         NA           NA
## 2          1     15  0.32          3           33
## 3          1     19  0.27          2           16
## 4          1      7    NA         NA           NA
## 5          1      5    NA         NA           NA
## 6          1     34  0.42          5           66
## 7          1     32  0.45          6           83
## 8          1     27  0.47          7          100
## 9          1     24  0.33          4           50
## 10         1     18    NA         NA           NA
## 11         1     13  0.24          1            0
## 12         1     10  0.39         NA           NA

（注意调整百分位数计算以适应是否考虑 NA）

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