使用 dplyr 汇总自定义函数以进行许多独特的测量

Question

我想计算不同个体在不同级别的热和冷温度处理之间的面积变化百分比（时间 T1 和 T9 之间）。

一些样本数据：

library(dplyr)

Individual<-c("a1.2", "a1.2","bd3.d","bd3.d", "k20.d","k20.d", "dfd.2","dfd.2", "d3.d","d3.d", "df3.1","df3.1")
Treat <- c('hot','hot','hot','hot','hot','hot','cold',"cold",'cold',"cold",'cold',"cold")
Time <- c("T1", "T9", "T1", "T9","T1", "T9","T1", "T9","T1", "T9","T1", "T9")
Area<- c("0.1", "0.5", "0.1", "0.645","0.1", "0.54","0.1", "0.587","0.1", "0.78","0.23", "0.78")
df.Area <- data.frame(Individual, Treat,Time,Area)
head(df.Area, n=20)

head(df.Area, n=20)
   Individual Treat Time  Area
1        a1.2   hot   T1   0.1
2        a1.2   hot   T9   0.5
3       bd3.d   hot   T1   0.1
4       bd3.d   hot   T9 0.645
5       k20.d   hot   T1   0.1
6       k20.d   hot   T9  0.54
7       dfd.2  cold   T1   0.1
8       dfd.2  cold   T9 0.587
9        d3.d  cold   T1   0.1
10       d3.d  cold   T9  0.78
11      df3.1  cold   T1  0.23
12      df3.1  cold   T9  0.78

例如：(T9-T1/T9)*100

首先找到相同的个体，例如第1行和第2行的a1.2，在T9和T1之间做计算：(0.5-0.1/0.1)*100=400%增加。

输出将是：

       Individual Treat Ch.Area  
    1        a1.2   hot    400    
    2        bd3.d  hot     num.etc 
    3       k20.d   hot     num.etc 
    4       dfd.2   cold    num.etc
    5       d3.d    cold    num.etc ....

df1 <- df.Area %>% group_by(Treat, Time, Individual)

这是对结构的疯狂猜测：

df2 <- df1 %>%  summarise(Ch.Area = T9[!Individual == "??"] - T1[!Individual == "??"])/T9([!Individual == "??"])*100)

我希望 dplyr 将具有相同名称的每个人分组在一起以计算百分比，同时仍保留 Treat 的组变量。这可能吗？如果更好的话，我也很乐意使用其他包/方法。

任何帮助都会很棒！

Answer 1

使用dplyr：

解决方案 1：假设 T9 在 Area 变量上的值始终高于 T1

Individual<-c("a1.2", "a1.2","bd3.d","bd3.d", "k20.d","k20.d", "dfd.2","dfd.2", "d3.d","d3.d", "df3.1","df3.1")
Treat <- c('hot','hot','hot','hot','hot','hot','cold',"cold",'cold',"cold",'cold',"cold")
Time <- c("T1", "T9", "T1", "T9","T1", "T9","T1", "T9","T1", "T9","T1", "T9")
Area<- c("0.1", "0.5", "0.1", "0.645","0.1", "0.54","0.1", "0.587","0.1", "0.78","0.23", "0.78")

df <- data.frame(Individual, Treat,Time, Area)

df %>%
  group_by(Individual) %>%
  mutate(Ch.Area = ((last(as.numeric(as.character(Area)))-first(as.numeric(as.character(Area))))/first(as.numeric(as.character(Area))))*100) %>% #Setting them as.numeric because in your data.frame they are stored as factors 
  summarise(Treat = last(Treat),
            Ch.Area = last(Ch.Area))

# A tibble: 6 x 3
  Individual Treat Ch.Area
  <fct>      <fct>   <dbl>
1 a1.2       hot      400.
2 bd3.d      hot      545.
3 d3.d       cold     680.
4 df3.1      cold     239.
5 dfd.2      cold     487.
6 k20.d      hot      440.

解决方案 2：不假设 T9 的面积变量值总是高于 T1

df %>%
  group_by(Individual) %>%
  mutate(Ch.Area = ((as.numeric(as.character(Area[Time=="T9"]))-as.numeric(as.character(Area[Time=="T1"])))/as.numeric(as.character(Area[Time=="T1"])))*100) %>% 
  summarise(Treat = last(Treat),
            Ch.Area = last(Ch.Area))

# A tibble: 6 x 3
  Individual Treat Ch.Area
  <fct>      <fct>   <dbl>
1 a1.2       hot      400.
2 bd3.d      hot      545.
3 d3.d       cold     680.
4 df3.1      cold     239.
5 dfd.2      cold     487.
6 k20.d      hot      440.

Answer 2

library(dplyr)
df.Area %>%  mutate_at('Area', as.numeric) %>% 
             group_by(Individual,Treat) %>% 
             summarise(Ch.Area = (Area[Time=='T9']/Area[Time=='T1']-1)*100)


# A tibble: 6 x 3
# Groups:   Individual [?]
Individual Treat Ch.Area
<chr>      <chr>   <dbl>
1 a1.2       hot      400 
2 bd3.d      hot      545 
3 d3.d       cold     680 
4 df3.1      cold     239.
5 dfd.2      cold     487.
6 k20.d      hot      440.

Answer 3

这是一个tidyverse 选项，它使用了一些整形：

library(tidyverse)

Individual<-c("a1.2", "a1.2","bd3.d","bd3.d", "k20.d","k20.d", "dfd.2","dfd.2", "d3.d","d3.d", "df3.1","df3.1")
Treat <- c('hot','hot','hot','hot','hot','hot','cold',"cold",'cold',"cold",'cold',"cold")
Time <- c("T1", "T9", "T1", "T9","T1", "T9","T1", "T9","T1", "T9","T1", "T9")
Area<- c("0.1", "0.5", "0.1", "0.645","0.1", "0.54","0.1", "0.587","0.1", "0.78","0.23", "0.78")
df.Area <- data.frame(Individual, Treat,Time,Area)

df.Area %>%
  spread(Time, Area, convert = T) %>%
  mutate(Ch.Area = 100*(T9/T1-1))

#   Individual Treat   T1    T9  Ch.Area
# 1       a1.2   hot 0.10 0.500 400.0000
# 2      bd3.d   hot 0.10 0.645 545.0000
# 3       d3.d  cold 0.10 0.780 680.0000
# 4      df3.1  cold 0.23 0.780 239.1304
# 5      dfd.2  cold 0.10 0.587 487.0000
# 6      k20.d   hot 0.10 0.540 440.0000