在函数内使用dplyr，进行非标准评估 [英] Using dplyr within a function, non-standard evaluation

问题描述

试图围绕非标准评估由dplyr使用，但没有成功。我想要一个简短的函数返回一组特定变量的汇总统计信息（N，mean，sd，median，IQR，min，max）。

Trying to get my head around Non-Standard Evaluation as used by dplyr but without success. I'd like a short function that returns summary statistics (N, mean, sd, median, IQR, min, max) for a specified set of variables.

简化版的功能...

my_summarise <- function(df = temp,
                         to.sum = 'eg1',
                         ...){
    ## Summarise
    results <- summarise_(df,
                          n = ~n(),
                          mean = mean(~to.sum, na.rm = TRUE))
    return(results)
}

并运行一些虚拟数据...

And running it with some dummy data...

set.seed(43290)
temp <- cbind(rnorm(n = 100, mean = 2, sd = 4),
              rnorm(n = 100, mean = 3, sd = 6)) %>% as.data.frame()
names(temp) <- c('eg1', 'eg2')
mean(temp$eg1)
  [1] 1.881721
mean(temp$eg2)
  [1] 3.575819
my_summarise(df = temp, to.sum = 'eg1')
    n mean
1 100   NA

N是计算的，但平均值不是，不能弄清楚为什么。

N is calculated, but the mean is not, can't figure out why.

最终我希望我的功能更加一般，沿着...

Ultimately I'd like my function to be more general, along the lines of...

my_summarise <- function(df = temp,
                         group.by = 'group'
                         to.sum = c('eg1', 'eg2'),
                         ...){
    results <- list()
    ## Select columns
    df <- dplyr::select_(df, .dots = c(group.by, to.sum))
    ## Summarise overall
    results$all <- summarise_each(df,
                                  funs(n = ~n(),
                                       mean = mean(~to.sum, na.rm = TRUE)))
    ## Summarise by specified group
    results$by.group <- group_by_(df, ~to.group) %>%
                        summarise_each(df,
                                       funs(n = ~n(),
                                       mean = mean(~to.sum, na.rm = TRUE)))        
    return(results)
}

...但在我移动到这个更复杂的版本之前（我正在使用这个例子的指导）我需要让评估工作在简单的版本首先是绊脚板，调用 dplyr :: select（）工作正常。

...but before I move onto this more complex version (which I was using this example for guidance) I need to get the evaluation working in the simple version first as thats the stumbling block, the call to dplyr::select() works ok.

欣赏任何关于我错误的地方的建议。

Appreciate any advice as to where I'm going wrong.

提前

推荐答案

基本思想是您必须自己实际构建适当的调用，最容易使用 lazyeval 包。

The basic idea is that you have to actually build the appropriate call yourself, most easily done with the lazyeval package.

在这种情况下，您想以编程方式创建一个看起来像〜mean（eg1 ，na.rm = TRUE）。这是如何：

In this case you want to programmatically create a call that looks like ~mean(eg1, na.rm = TRUE). This is how:

my_summarise <- function(df = temp,
                         to.sum = 'eg1',
                         ...){
  ## Summarise
  results <- summarise_(df,
                        n = ~n(),
                        mean = lazyeval::interp(~mean(x, na.rm = TRUE),
                                                x = as.name(to.sum)))
  return(results)
}

这是我努力让事情工作时所做的：

Here is what I do when I struggle to get things working:

记住，就像您已经拥有的〜n（）一样，通话必须以〜。

1. 使用实际变量编写正确的电话，看看它是否有效（〜mean（eg1，na.rm = TRUE））。


2. 使用 lazyeval :: interp 重新创建该调用，并通过运行 interp 可以直观地看到它在做什么。

1. Remember that, just like the ~n() you already have, the call will have to start with a ~.
2. Write the correct call with the actual variable and see if it works (~mean(eg1, na.rm = TRUE)).
3. Use lazyeval::interp to recreate that call, and check this by running only the interp to visually see what it is doing.

在这个c我可能会经常写 interp（〜mean（x，na.rm = TRUE），x = to.sum）。但是运行它会给我们〜mean（eg1，na.rm = TRUE）这是将 eg1 一个字符而不是一个变量名。所以我们使用 as.name ，正如我们在 vignette（nse）中所教导的那样。

In this case I would probably often write interp(~mean(x, na.rm = TRUE), x = to.sum). But running that will give us ~mean("eg1", na.rm = TRUE) which is treating eg1 as a character instead of a variable name. So we use as.name, as is taught to us in vignette("nse").

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