当使用broom我用结合dplyr::group_by,并dplyr::do进行分组数据由于上行动@drob.例如,根据齿轮系统将线性模型拟合到汽车:
library("dplyr")
library("tidyr")
library("broom")
# using do()
mtcars %>%
group_by(am) %>%
do(tidy(lm(mpg ~ wt, data = .)))
# Source: local data frame [4 x 6]
# Groups: am [2]
# am term estimate std.error statistic p.value
# (dbl) (chr) (dbl) (dbl) (dbl) (dbl)
# 1 0 (Intercept) 31.416055 2.9467213 10.661360 6.007748e-09
# 2 0 wt -3.785908 0.7665567 -4.938848 1.245595e-04
# 3 1 (Intercept) 46.294478 3.1198212 14.838824 1.276849e-08
# 4 1 wt -9.084268 1.2565727 -7.229401 …Run Code Online (Sandbox Code Playgroud) 如何do.call使用变量参数和函数列表来使用summarise_dplyr中的标准评估版本?
## Some sample data, function, and variables to interpolate
set.seed(0)
dat <- data.frame(a=runif(10), b=runif(10))
fn <- function(x, y) IQR(x / y, na.rm = TRUE)
funs <- list(fn="fn")
targs <- list("a", "b")
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这是lazyeval::interp我正在努力工作
library(dplyr)
interp(~do.call(fn, xs), .values=list(fn=funs$fn, xs=targs))
# ~do.call("fn", list("a", "b"))
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但它不起作用,
dat %>%
summarise_(out = interp(~do.call(fn, xs), .values=list(fn=funs$fn, xs=targs)))
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预期结果
dat %>%
summarise(out = do.call(fn, list(a, b)))
# out
# 1 1.084402
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如果我添加一些打印语句,我知道问题是"a"和"b"没有被正确解释,但我无法弄清楚如何正确引用它们.
fn <- function(x, y) { print(x); print(y); IQR(x / …Run Code Online (Sandbox Code Playgroud)