Question

plyr包具有多种_ply函数，其中前两个字母表示输入和输出，以便ddply获取数据帧输入并生成数据帧输出，以及dlply获取数据框输入并生成列表输出。出于各种原因，我通常更喜欢使用dplyr包，而plyr和dplyr在单个环境中无法很好地协同工作。有没有办法在dlply的管道语法中从plyr复制dplyr函数的“数据框，列出”功能？

我想要复制的功能的一个简单示例：

data = data.frame(x = rep(seq(from = 1, to = 100, by = 1), times = 3), 
              y = rnorm(n = 300), 
              group_var = c(rep("A", 100), rep("B", 100), rep("C", 100)))

spline.fun = function(x, xvar, yvar, ...) {
  smooth.spline(x = x[,xvar], y = x[,yvar], ...)
}

spline_list = dlply(data, "group_var", spline.fun, xvar = "x", yvar = "y")

我想写的代码就是这样的：

spline_list = data %>%
    group_by(group_var) %>%
    list_mutate(list_element = spline.fun, xvar = x, yvar = y)

但据我所知，没有dplyr函数以mutate创建新列的方式创建列表元素

Answer 1

我们可以按group_var拆分数据框，使用purrr包中的map来应用您的功能。

library(tidyverse)

data2 <- data %>%
  split(f = data$group_var) %>%
  map(~spline.fun(.x, xvar = "x", yvar = "y"))
# $`A`
# Call:
#   smooth.spline(x = x[, xvar], y = x[, yvar])
# 
# Smoothing Parameter  spar= 1.315545  lambda= 14.95228 (20 iterations)
# Equivalent Degrees of Freedom (Df): 2.016214
# Penalized Criterion (RSS): 74.08271
# GCV: 0.7716288
# 
# $B
# Call:
#   smooth.spline(x = x[, xvar], y = x[, yvar])
# 
# Smoothing Parameter  spar= 1.499963  lambda= 321.1298 (29 iterations)
# Equivalent Degrees of Freedom (Df): 2.000764
# Penalized Criterion (RSS): 77.98068
# GCV: 0.8119731
# 
# $C
# Call:
#   smooth.spline(x = x[, xvar], y = x[, yvar])
# 
# Smoothing Parameter  spar= 1.499953  lambda= 321.0788 (27 iterations)
# Equivalent Degrees of Freedom (Df): 2.000764
# Penalized Criterion (RSS): 104.8997
# GCV: 1.092268

在dplyr中复制dlply（）函数

1 个答案: