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在编程中使用dplyr filter()

rlang tidyverse dplyr r

Kay · 技术社区 · 7 年前

我正在编写函数,希望使用dplyr的filter()函数来选择满足条件的数据帧行。这是我的代码:

library(tidyverse)

df <-data.frame(x = sample(1:100, 50), y = rnorm(50), z = sample(1:100,50), w = sample(1:100, 50),
            p = sample(1:100,50))

new <- function(ang,brad,drau){
  df%>%filter(!!drau %in% 1:50)%>%select(ang,brad) -> A
return(A)
}

brand <- c("z","w","p")
lapply(1:3, function(i) new(ang = "x", brad = "y", drau = brand[i]))%>%bind_rows()

每当我运行这个函数时,它看起来就像 filter

我该怎么做?

使现代化

new <- function(ang,brad,drau){
  df%>%filter(!!drau > 50)%>%select(ang,brad) -> A
return(A)
}

lapply(1:3, function(i) new(ang = "x", brad = "y", drau = brand[i]))%>%bind_rows()

然而,每个循环的结果都是相同的。为什么会这样?还有为什么我不能使用 %in% .

4 回复 | 直到 7 年前

dmi3kno 7 年前

我同意@hrbrmstr的标准评估解决方案。正如@hadley所建议的那样 today 以下是NSE解决方案:

library(tidyverse)

df <-data.frame(x = sample(1:100, 50), 
                y = rnorm(50), 
                z = sample(1:100,50), 
                w = sample(1:100, 50),
                p = sample(1:100,50))

new <- function(ang, brad, drau){
  ang  <- enquo(ang)
  brad <- enquo(brad)
  drau <- enquo(drau)

  df %>% filter(UQ(drau) %in% 1:50) %>%
    select(UQ(ang),UQ(brad)) 
}

brand <- c("z","w","p")
brand <- rlang::syms(brand)

map_df(brand, ~new(ang = x, brad = y, drau = UQ(.x)))

hrbrmstr 7 年前

这似乎是你想要的(但需要你确认):

library(tidyverse)
library(rlang)

set.seed(1492)

xdf <- data_frame(
  x = sample(1:100, 50),
  y = rnorm(50), 
  z = sample(1:100,50), 
  w = sample(1:100, 50),
  p = sample(1:100,50)
)

new_df <- function(ang, brad, drau) {
  drau <- sym(drau)
  filter(xdf, UQE(drau) %in% 1:50) %>% 
    select(ang, brad)
}

brand <- c("z", "w", "p")

map_df(brand, ~new_df(ang = "x", brad = "y", drau = .x))

df ,它是 stats pkg和我尽量避免再使用它。

既然你在使用tidyverse,不妨利用 map_df() 从…起 purrr

Kay 5 年前

具有 UQE !!sym()

library(rlang)
library(tidyverse)

df <-data.frame(x = sample(1:100, 50), y = rnorm(50), z = sample(1:100,50), w = sample(1:100, 50),
                p = sample(1:100,50))

new <- function(ang,brad,drau){
  df%>%filter(!!sym(drau) %in% 1:50)%>%select(!!sym(ang), !!sym(brad)) -> A
  return(A)
}

brand <- c("z","w","p")
lapply(1:3, function(i) new(ang = "x", brad = "y", drau = brand[i]))%>%bind_rows()

!!enquo()

new <- function(ang,brad,drau){
  df%>%filter(!!enquo(drau) %in% 1:50)%>%select(!!enquo(ang), !!enquo(brad)) -> A
  return(A)
}

> head(new(ang = x, brad = y, drau = z))
    x           y
1  44  0.47702540
2  84 -1.09670409
3  59 -0.20556334
4  81 -0.46306635
5  93  1.36845485
6   8  0.37392587

Dexterity 4 年前

我有一个类似的问题,对我有效的简单修复方法是在“filter”动词中使用“.data”代词,特别是在这种情况下:

filter(.data[[drau]] %in% 1:50)

更多信息请点击此处: https://tinyheero.github.io/2020/03/01/use-data-env-pronouns-tidyverse.html 莱昂内尔·亨利最近在Rstudio的youtube频道上的谈话也很有帮助。

library(tidyverse)

df <-data.frame(x = sample(1:100, 50), y = rnorm(50), z = sample(1:100,50), w = sample(1:100, 50),
                p = sample(1:100,50))

new <- function(ang,brad,drau){
  df%>%filter(.data[[drau]] %in% 1:50)%>%select(ang,brad) -> A
  return(A)
}

brand <- c("z","w","p")
lapply(1:3, function(i) new(ang = "x", brad = "y", drau = brand[i]))%>%bind_rows()