分组变量差异的计算平均值

Question

我有一个数据集，对响应变量有5个级别的处理。

假设我在土壤含水量的40%、30%、20%和10%的5个水平上测定了土壤氮含量。每级我都有5个副本。

现在，我想计算每个复制的非标准化(最优- 40%，最优- 30%，最优- 20%，最优- 10%)和标准化(最- 40% /最，最- 30% /最，等等)。

有没有办法在R和Tidyverse一起做这件事？我在做“循环”功能方面有问题。每个治疗水平重复5个。

df<- data.frame(Soilwater = c("optimal", "optimal", "optimal", "optimal", "optimal", 
       "40", "40", "40", "40", "40", 
       "30","30","30","30","30", 
       "20", "20","20","20","20",
       "10","10","10","10","10", 
       "optimal", "optimal", "optimal", "optimal", "optimal", 
       "40", "40", "40", "40", "40", 
       "30","30","30","30","30", 
       "20", "20","20","20","20",
       "10","10","10","10","10"), 
Diversity = c("High","High","High","High","High","High","High","High","High","High",   "High","High","High","High","High","High","High","High","High","High",
       "High","High","High","High","High", 
       "Low", "Low", "Low","Low","Low","Low","Low","Low","Low","Low",
       "Low","Low","Low","Low","Low","Low","Low","Low","Low","Low",
       "Low","Low","Low","Low","Low"),
Soil_N = c(50,45, 49, 48, 49, 69, 68, 69, 70, 67, 79, 78, 79, 78, 77, 89, 89, 87, 88, 89, 99, 98, 97, 98, 98, 120,    
   121,    121,    120,    122,    134,    131,    132,    134,    131,    145,    148,    149,    147,    
   148,    159,    159,    157,    156,    157,    169,    167,    167,    168,    164))

我使用了@JonSpring建议的下面的代码，这是非常有用的。

df %>%
    # First, we can add a `Replicate` number based on position within 
    # each Soilwater/Diversity cohort.
    group_by(Soilwater, Diversity) %>%
    mutate(Replicate = row_number()) %>%

    # Calc diff vs. experiment with same Diversity & Replicate, optimal Soilwater 
    group_by(Diversity, Replicate) %>%
    mutate(Difference = Soil_N - Soil_N[Soilwater == "optimal"]) %>%

    # Summarize avg diffs
    group_by(Soilwater, Diversity) %>%
    summarize(Mean_Diff = mean(Difference), .groups = "drop")

但是，我意识到，首先我需要为optimal土壤水位做一个平均值，然后计算这个平均值与其他土壤水位之间的差异，我尝试了下面的代码(用mean函数计算差值之前的optimal土壤水位平均值)。但这是行不通的。

df%>%       group_by(Soilwater, Diversity)%>%       mutate(Replicate = row_number())%>%        
group_by(Diversity, Replicate)%>%       mutate(Difference = mean(Soil_N[Soilwater=="optimal"])- Soil_N)

Answer 1

很难理解你的问题。所以我从对你起作用的东西开始。

作为R和tidyverse的新手，请注意%>% (管道)将您的操作链在(启动)对象上。

您可以将操作的任何状态/阶段分配给新对象(也称为变量)。

我还建议您在处理问题时创建几个“临时”对象，以存储问题/算法的步骤。这会让你更好地感受到你所拥有的。随着时间的推移，您将获得足够的经验来连锁操作，并避免某些临时阶段/对象。

为此，我介绍了一个“临时”结果/对象，正如您的描述所建议的那样--直到那时为止，我为您指定了interim_df <- ...。

library(dplyr)

interim_df <- df %>%
    group_by(Soilwater, Diversity)  %>%
    mutate(Replicate = row_number()) %>%
    group_by(Diversity, Replicate)

这将产生一个对象interim_df。让我们来看看

interim_df
# A tibble: 50 x 4
# Groups:   Diversity, Replicate [10]
   Soilwater Diversity Soil_N Replicate
   <chr>     <chr>      <dbl>     <int>
 1 optimal   High          50         1
 2 optimal   High          45         2
 3 optimal   High          49         3
 4 optimal   High          48         4
 5 optimal   High          49         5
 6 40        High          69         1
 7 40        High          68         2
 8 40        High          69         3
 9 40        High          70         4
10 40        High          67         5

好的。我们有一个有4个变量的第50行..。这似乎是你满意的数据结构。您还拥有一个“分组数据”。当您想要对整个数据(或数据文件的其他部分)进行操作时，一定要使用ungroup()。

interim_df <- interim_df %>% ungroup()

您可以“提取”您的“最优”度量，并计算这个“新”df/tibble的平均值。

mean_optimal <- interim_df %>%
    filter(Soilwater == "optimal") %>%
    summarise(MeanOptimal = mean(SoilN)   # we calculate/summarise the mean over the part we want

这给了你

# A tibble: 1 x 1
  MeanOptimal
        <dbl>
1        84.5

要明确的是，我们现在有了另一个带有一个变量/列的tibble。这可以在您的interim_df中使用。但是，请确保了解如何从tibble中“提取”列(也可以将其作为可重用的向量)。基-R表示法$允许您直接访问列(向量)；tidyverse提供pull()函数。

final <- interim_df %>% mutate(Difference = mean_optimal$MeanOptimal - Soil_N)
final
# A tibble: 50 x 5
   Soilwater Diversity Soil_N Replicate Difference
   <chr>     <chr>      <dbl>     <int>      <dbl>
 1 optimal   High          50         1       34.5
 2 optimal   High          45         2       39.5
 3 optimal   High          49         3       35.5
 4 optimal   High          48         4       36.5
 5 optimal   High          49         5       35.5
 6 40        High          69         1       15.5
 7 40        High          68         2       16.5
 8 40        High          69         3       15.5
 9 40        High          70         4       14.5
10 40        High          67         5       17.5

您还可以将mean_optimal$MeanOptimal作为interim_df %>% mutate(MeanOptimal = mean_optimal$MeanOptimal)的新列“添加”到interim_df中，然后执行不同的操作。