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r - R中具有多个分组因子的多个变量的均值和标准差

问题描述

我一直在寻找答案，但仍然没有找到解决方案，我对 R 还是新手。我的数据框显示了在不同条件下对约 70 种植物物种的一种生态特征（相对土壤覆盖率）的测量：不同年份，不同的化学处理和温室的存在/不存在。

我需要将这些数据汇总到一个新的数据框中，该数据框显示每个物种和每种因素组合（条件）的性状平均值和标准差。我知道aggregate或lapply可以提供帮助，但我很难将 3 个不同的因素和多个物种的分组结合起来，这意味着需要“自动化”代码。

如果我错过了回答我问题的帖子，我很抱歉

感谢您的耐心和帮助

编辑：这是一个可重现的例子，希望我能正确地做到这一点：

mydata<-structure(list(Year = c(2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 
2010L, 2010L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 
2011L), Replicate = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 
1L, 1L, 2L, 2L, 2L, 2L), Treatment = structure(c(1L, 2L, 1L, 
2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L), .Label = c("A", 
"B"), class = "factor"), Greenhouse = structure(c(2L, 2L, 1L, 
1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L), .Label = c("No", 
"Yes"), class = "factor"), Sp_1 = c(4L, 0L, 2L, 5L, 4L, 0L, 2L, 
5L, 0L, 0L, 4L, 6L, 4L, 0L, 2L, 5L), Sp_2 = c(7L, 0L, 1L, 1L, 
7L, 0L, 1L, 1L, 7L, 0L, 1L, 1L, 6L, 0L, 1L, 1L), Sp_3 = c(8L, 
2L, 2L, 1L, 8L, 2L, 2L, 1L, 10L, 2L, 1L, 1L, 4L, 2L, 2L, 1L)), class = "data.frame", row.names = c(NA, 
-16L))

我在那个例子中只放了 3 个物种，但正如我所说，我有 70 多个物种，所以我需要一些可以选择所有物种列的mydata[,5:75]东西（？沿着这些线的东西）超过c("sp_1","sp_2",..., "sp_70").

我希望输出看起来像这样：

Year   Treatment   Greenhouse   Sp_1_mean   Sp_1_sd   Sp_2_mean   Sp_2_sd 
2010   A           Yes          x           x         x           x
2010   A           No           x           x         x           x
2010   B           Yes          x           x         x           x
2010   B           No           x           x         x           x
2011   A           Yes          x           x         x           x
2011   A           No           x           x         x           x
2011   B           Yes          x           x         x           x
2011   B           No           x           x         x           x

这是dput()显示所需的输出应该是什么样子的

    desired_output<-structure(list(Year = c(2010L, 2010L, 2010L, 2010L, 2011L, 2011L, 
2011L, 2011L), Treatment = structure(c(1L, 1L, 2L, 2L, 1L, 1L, 
2L, 2L), .Label = c("A", "B"), class = "factor"), Greenhouse = structure(c(2L, 
1L, 2L, 1L, 2L, 1L, 2L, 1L), .Label = c("No", "Yes"), class = "factor"), 
    Sp_1_mean = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "x", class = "factor"), 
    Sp_1_sd = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "x", class = "factor"), 
    Sp_2_mean = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "x", class = "factor"), 
    Sp_2_sd = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "x", class = "factor"), 
    Sp_3_mean = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "x", class = "factor"), 
    Sp_3_sd = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "x", class = "factor")), class = "data.frame", row.names = c(NA, 
-8L))

我希望这更清楚！谢谢

标签： rgroup-byaggregatelapplymean

解决方案

使用data.table，您可以以这种方式做一些事情：

library(data.table)
setDT(df)
df[, lapply(.SD, function(x) return(c(mean(x), sd(x))),
                 by = c("col1","col2"),
                .SDcols = c("x1","x2")]

（没有可重现的例子，很难给你更精确的语法）

这意味着：按组（和）对每个数据子集（此处为列x1和）应用均值和标准差x2col1col2

例子

library(data.table)
df <- as.data.table(mtcars)
output <- df[, lapply(.SD, function(x) return(c(mean(x, na.rm = TRUE), sd(x, na.rm = TRUE)))),
            .SDcols = c("disp","drat"),
            by = c("cyl","gear")]
output[, 'stat' := c("mean","sd"), by = c("cyl","gear")]

output
    cyl gear       disp       drat stat
 1:   6    4 163.800000 3.91000000 mean
 2:   6    4   4.387862 0.01154701   sd
 3:   4    4 102.625000 4.11000000 mean
 4:   4    4  30.742699 0.37156042   sd
 5:   6    3 241.500000 2.92000000 mean
 6:   6    3  23.334524 0.22627417   sd
 7:   8    3 357.616667 3.12083333 mean
 8:   8    3  71.823494 0.23027487   sd
 9:   4    3 120.100000 3.70000000 mean
10:   4    3         NA         NA   sd
11:   4    5 107.700000 4.10000000 mean
12:   4    5  17.819091 0.46669048   sd
13:   8    5 326.000000 3.88000000 mean
14:   8    5  35.355339 0.48083261   sd
15:   6    5 145.000000 3.62000000 mean
16:   6    5         NA         NA   sd

在这里，我有一列可以了解每行的统计信息

使用可重现的示例进行编辑

setDT(mydata)
output <- mydata[, lapply(.SD, function(x) return(c(mean(x, na.rm = TRUE), sd(x, na.rm = TRUE)))),
       .SDcols = c("Sp_1", "Sp_2", "Sp_3"),
       by = c("Year", "Treatment", "Greenhouse")
       ]
output[, 'stat' := c('mean','sd') ,
       by = c("Year", "Treatment", "Greenhouse")]

由于您对宽格式感兴趣，您可以使用它dcast来重塑您的数据。

output <- dcast(output, Year + Treatment + Greenhouse ~  ...,
      value.var = c("Sp_1", "Sp_2", "Sp_3"))

output

Year Treatment Greenhouse Sp_1_mean   Sp_1_sd Sp_2_mean   Sp_2_sd Sp_3_mean   Sp_3_sd
1: 2010         A         No       2.0 0.0000000       1.0 0.0000000       2.0 0.0000000
2: 2010         A        Yes       4.0 0.0000000       7.0 0.0000000       8.0 0.0000000
3: 2010         B         No       5.0 0.0000000       1.0 0.0000000       1.0 0.0000000
4: 2010         B        Yes       0.0 0.0000000       0.0 0.0000000       2.0 0.0000000
5: 2011         A         No       3.0 1.4142136       1.0 0.0000000       1.5 0.7071068
6: 2011         A        Yes       2.0 2.8284271       6.5 0.7071068       7.0 4.2426407
7: 2011         B         No       5.5 0.7071068       1.0 0.0000000       1.0 0.0000000
8: 2011         B        Yes       0.0 0.0000000       0.0 0.0000000       2.0 0.0000000

这种从长到宽的转换可以通过稍微修改聚合来避免。

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