用 dplyr 中的数学条件进行总结

Question

基于这个问题：Summarize with conditions in dplyr 我想使用 dplyr 来根据数学条件汇总列（而不是链接 post 中的字符串匹配）。我需要在 measurement/time 的比率最高时找到最大值 measurement，同时创建一个新列 ratio。我还想遍历整行，我不确定如何使用 dplyr 的 summarize 函数。

示例数据框

print(df)

   sample     type time measurement
1       a bacteria   24     0.57561
2       a bacteria   44     1.67236
3       a bacteria   67     4.17100
4       a bacteria   88    11.51661
5       b bacteria   24     0.53269
6       b bacteria   44     1.24942
7       b bacteria   67     5.72147
8       b bacteria   88    11.04017
9       c bacteria    0     0.00000
10      c bacteria   24     0.47418
11      c bacteria   39     1.06286
12      c bacteria   64     3.59649
13      c bacteria   78     7.05190
14      c bacteria  108     7.27060

期望输出

  sample     type time measurement      ratio
1      a bacteria   88    11.51661 0.13087057
2      b bacteria   88    11.04017 0.12545648
3      c bacteria   78     7.05190 0.09040897

尝试失败

这只是returns由group_by和summarize函数定义的两列，希望将整行信息带入：

library(dplyr)
df %>% 
    group_by(sample) %>%
    summarize(ratio = max(measurement/time, na.rm = TRUE))

  sample  ratio
  <fct>   <dbl>
1 a      0.131 
2 b      0.125 
3 c      0.0904

可重现数据

structure(list(sample = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("a", "b", "c"), class = "factor"), 
    type = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L), .Label = "bacteria", class = "factor"), 
    time = c(24, 44, 67, 88, 24, 44, 67, 88, 0, 24, 39, 64, 78, 
    108), measurement = c(0.57561, 1.67236, 4.171, 11.51661, 
    0.53269, 1.24942, 5.72147, 11.04017, 0, 0.47418, 1.06286, 
    3.59649, 7.0519, 7.2706)), class = "data.frame", row.names = c(NA, 
-14L))

Answer 1

df %>%
  mutate(ratio = measurement/time) %>%
  group_by(sample) %>%
  filter(ratio == max(ratio, na.rm=TRUE))

Answer 2

这应该可以解决问题。

df %>%
   group_by(sample) %>%
   mutate(ratio = measurement/time) %>%
   filter(ratio == max(ratio))

Answer 3

一个选项是 filter 'measurement' 基于 measurement/time 的 max 位置，并用它来比较 (==) 与 'measurement' 按 'sample'

分组后的值

library(dplyr)
df %>%
   group_by(sample) %>% 
   mutate(ratio = measurement/time) %>%
   filter(measurement == measurement[which.max(ratio)])

用 dplyr 中的数学条件进行总结

Summarize with mathematical conditions in dplyr

r

conditional-statements

dplyr

summarize