如何用R量化图的下载斜率

Question

我有以下数据框架：

df <- structure(list(
  peptide = structure(c(
    1L, 1L, 1L, 1L, 2L, 2L,
    2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L
  ), levels = c(
    "P1",
    "P2", "P3", "P4", "P5"
  ), class = "factor"), reaction_time = c(
    0,
    3, 5, 10, 0, 3, 5, 10, 0, 3, 5, 10, 0, 3, 5, 10, 0, 3, 5, 10
  ),
  mean_residual_quantity = c(
    100, 110, 114, 110.5, 100, 91,
    84.5, 69.5, 100, 75, 70, 59, 100, 63.5, 58, 43, 100, 44,
    28, 12
  )
), class = c("grouped_df", "tbl_df", "tbl", "data.frame"), row.names = c(NA, -20L), groups = structure(list(peptide = structure(1:5, levels = c(
  "P1",
  "P2", "P3", "P4", "P5"
), class = "factor"), .rows = structure(list(
  1:4, 5:8, 9:12, 13:16, 17:20
), ptype = integer(0), class = c(
  "vctrs_list_of",
  "vctrs_vctr", "list"
))), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -5L), .drop = TRUE))

使用此代码：

ggpubr::ggline(df,
       x = "reaction_time",
       y = "mean_residual_quantity",
       color = "peptide", 
       xlab = "Reaction Time",
       palette = "jco", 
       size = 1,
       ylab = "Residual Quantity (%)"
) +
  scale_y_continuous(breaks = get_breaks(n = 10)) + 
  grids() +
  rremove("legend.title") 

我可以创造出这样的情节：

​

​

从视觉上看，我们可以看到，从P1到P5，图形的斜率一直在下降。对于每一个P1到P5，是否有一个单一的度量来量化这种下降趋势？

最后，如果我们对这个值进行排序，那么顺序应该是P1，P2，P3，P4，P5。

我如何用R来实现它呢？

Answer 1

假设斜率是第一个值和最后一个值之间的斜率。

如果每个mean_residual_quantity的第一个peptide值与数据相同，

library(dplyr)
df %>%
  arrange(reaction_time) %>%
  group_by(peptide) %>%
  filter(row_number() == n()) %>%
  arrange(desc(mean_residual_quantity)) %>% pull(peptide)

如果不是，

library(dplyr)
library(data.table)
df %>%
  arrange(reaction_time) %>%
  group_by(peptide) %>%
  filter(row_number() == 1 | row_number() == n()) %>%
  summarize(slope = (last(mean_residual_quantity) - first(mean_residual_quantity))/
              (last(reaction_time) - first(reaction_time))) %>%
  arrange(desc(slope)) %>% pull(peptide)

会给出你想要的结果

[1] P1 P2 P3 P4 P5
Levels: P1 P2 P3 P4 P5

这里还有另一个选项，就是在每两个连续的点之间按斜率排序。

df %>%
  arrange(peptide, mean_residual_quantity) %>%
  group_by(peptide) %>%
  mutate(slope = (mean_residual_quantity - lag(mean_residual_quantity))/
           (reaction_time - lag(reaction_time))) %>%
  summarize(slope_mean = mean(slope, na.rm = T)) %>%
  arrange(desc(slope_mean)) %>% pull(peptide)

Answer 2

另一种可能的、完全任意的方法是，对每个时间点对每个P进行排序，然后对每个P的秩进行求和，得到一种总秩。从图表中转录你的数据

df=data.frame(
  "RT"=c(0,3,5,10),
  "P1"=c(100,110,115,110),
  "P2"=c(100,90,85,70),
  "P3"=c(100,75,70,60),
  "P4"=c(100,65,60,40),
  "P5"=c(100,45,30,10)
)
tmp=t(apply(df[,-1],1,rank))
colSums(tmp)

P1 P2 P3 P4 P5 
18 15 12  9  6