将“summarise”（dplyr）与需要公式的函数一起使用

Question

我正在尝试生成由基于 mblm 包的自定义函数生成的回归斜率表（此处示例中的函数是简化版本）。该函数需要一个公式作为参数，我想使用 dplyr summarise 将其应用于来自具有许多变量的大型数据帧的分组样本。 输出应该是我可以传递给热图函数的样本组和响应变量的回归斜率。

library (dplyr)

# Example data

test_data <- 
    rbind (
        data.frame(ID=paste0("someName", c(1:9)), Sample_Type="type1", 
           A=seq(1,17, length.out=9),
           I=0.1^seq(1,1.8,length.out=9), 
           J=1-0.1^seq(1,1.8,length.out=9)),
        data.frame(ID=paste0("someName", c(10:15)), Sample_Type="type2", 
           A=seq(1,7, length.out=6), 
           I=0.1^(1-seq(1,1.5,length.out=6)),
           J=1-0.1^(1-seq(1,1.5,length.out=6))))

# Define an independent and the responding variables - I would like to be able to easily test different independent variables
 
idpVar <- "A"
respVar <- test_data %>% .[!names(.) %in% c("ID", "Sample_Type", idpVar)] %>% names()

# Custom function generating numeric value of median slopes (simplified from mblm)

medianSlope <-
function (formula, dataframe) 
{
    if (missing(dataframe)) 
        dataframe <- environment(formula)
    term <- as.character(attr(terms(formula), "variables")[-1])
    x = dataframe[[term[2]]]
    y = dataframe[[term[1]]]
    if (length(term) > 2) {
        stop("Only linear models are accepted")
    }
    xx = sort(x)
    yy = y[order(x)]
    n = length(xx)
    slopes = c()
    smedians = c()
        for (i in 1:n) {
            slopes = c()
            for (j in 1:n) {
                if (xx[j] != xx[i]) {
                  slopes = c(slopes, (yy[j] - yy[i])/(xx[j] - 
                    xx[i]))
                 }
            }
            smedians = c(smedians, median(slopes))
        }
        slope = median(smedians)
    
    slope
}

# Custom function works with test dataframe and a single named dependent variable but "group_by" seems to be ignored:

test_data %>% group_by (Sample_Type) %>% medianSlope( formula(paste("J", "~", idpVar))  ,.)

暂时将分组问题放在一边，我尝试通过生成多个公式的列表来进行“总结”：

粘贴(respVar, "~", idpVar) [1]“B～A”“C～A”“D～A”“E～A”“F～A”“G～A”“H～A”“I～A”“J～A”“K 〜A”“L〜A”

然而

test_data %>% summarise_at (respVar, medianSlope(paste(respVar, "~", idpVar), .))

错误：$ 运算符对原子向量无效

test_data %>% summarise_at (respVar, medianSlope(paste(get(respVar), "~", get(idpVar)), .))

get(idpVar) 中的错误：找不到对象“A”

我对 R 比较陌生，有点迷茫。你能帮忙吗？

Answer 1

我不确定这是否能够使用summarise_at 函数来完成。但是，我们可以结合使用map_dbl、by 和其他一些数据清理函数来执行计算：

library(tidyverse)

# split the data using `by` (acts as a group_by)
# use `map_dbl` to iterate over the variables in respVar
# we use setNames so that the returned vector from map_dbl is named
# then, bind the rows together, convert to data frame
# finally convert row names (groups) to a column
by(test_data, test_data$Sample_Type,
       FUN = function(d) map_dbl(setNames(respVar, respVar), 
                             ~medianSlope(formula(paste(.x, "~", idpVar)), 
                                          data = d))) %>%
    do.call("rbind", .) %>%
    as.data.frame() %>%
    rownames_to_column(var = "Sample_Type")

  Sample_Type            I            J
1       type1 -0.004623987  0.004623987
2       type2  0.341974269 -0.341974269

Answer 2

Bouncyball，再次感谢您的帮助。似乎“总结”和“变异”确实不能调用使用公式作为输入的函数，尽管我没有看到其他地方对此进行了解释。解决方法很有启发性，但我遵循了您的替代建议并重写了被调用的函数。 仍然是一个学习者，我给自己设定了一个挑战，即替换 mblm 派生代码中的“for”循环，并消除看似冗余的计算（以对 RAM 的更高需求为代价——但它仍然运行得更快）我的 PC 上的数据，我计划在下一步开发代码中重新使用 dx 矩阵）。两种解决方案如下。 干杯，托马斯

mblm_2_short <-        # code adapted from mblm(y ~ x, repeated = T), for calculation of repeat median slope only 
function (x, y) 
{
xx = sort(x)
yy = y[order(x)]
n = length(xx)
slopes = c()
smedians = c()

    for (i in 1:n) {
        slopes = c()
        for (j in 1:n) {
            if (xx[j] != xx[i]) {
              slopes = c(slopes, (yy[j] - yy[i])/(xx[j] - 
                xx[i]))
             }
        }
        smedians = c(smedians, median(slopes))
    }
    slope = median(smedians)
}

.

med_slopesMed <-      # repeat median slope- like mblm(y ~ x, repeated = T), slope only                     
function (xx, yy) 
{
  x = sort(xx)
  y = yy[order(xx)]
  n = length(x)
  
  dx = matrix (rep (0,n^2), ncol=n)
  dy = c()
  z  = matrix (rep (0,n^2), ncol=n)
  
  for (i in 1:(n-1)) {                   ### x-axis distances (dx) and slopes (z) between points
    dxi             = x[-(1:i)]-x[i]
    dx [i, (i+1):n] = dxi                # for points 1:n, x-axis distances to all other points
    dyi             = y[-(1:i)]-y[i]
    zi              = dyi/dxi           
    z [i, (i+1):n]  = zi                 # for points 1:n, linear slopes connecting with all other points
  }
  
  z = replace(z, is.infinite(z), NA)           # removes +/-Inf and NaN generated by dx=0
  z  = t(z)[,-n] + z[,-1]

  median (apply(z, 1, median, na.rm=T)) 
}