R自己函数中R中的动态变量引用的R准引用和tidyeval

Question

我试图在我自己的函数中使用R中的tidyverse的准引用。我在这里已经读过这一篇文章：将参数列表传递给带有准引号的函数，以及此处的全部内容：https : //tidyeval.tidyverse.org/

但是我仍然无法正常工作。

假设我有以下数据：

dat <- data.frame(time   = runif(20),
                  group1 = rep(1:2, times = 10),
                  group2 = rep(1:2, each = 10),
                  group3 = rep(3:4, each = 10))

我现在想做的是编写一个执行以下操作的函数：

• 取一个数据集
• 指定包含时间的变量（请注意，在另一个数据集中，这可能称为“小时”或“ qtime”或其他名称）
• 指定我要对哪些组进行操作/统计

因此，我希望用户使用的功能如下：

test_function(data = dat, time_var = "time", group_vars = c("group1", "group3")) 请注意，下次我可能选择其他分组变量，或者没有选择。

假设在我要执行的功能中：

• 计算有关时间变量的某些统计信息，例如分位数。注意：我想按我的分组变量进行拆分

---

这是我最近的尝试之一：

test_function <- function(data, time_var = NULL, group_vars = NULL)
{
# Note I initialize the variables with NULL, since e.g. the user might not specify a grouping 

and I want to check for that in my function at some point)
time_var <- enquo(time_var)
group_vars <- enquos(group_vars)

# Here I try to group by my grouping variables
temp_data <- data %>%
    group_by_at(group_vars) %>%
    mutate(!!sym(time_var) := !!sym(time_var) / 60)

# Here I'm calculating some stats  
time_stats <- temp_data %>%
    summarize_at(vars(!!time_var), list(p0.1_time   = ~quantile(., probs = 0.1, na.rm = T),
                                        p0.2_time   = ~quantile(., probs = 0.2, na.rm = T),
                                        p0.3_time   = ~quantile(., probs = 0.3, na.rm = T),
                                        p0.4_time   = ~quantile(., probs = 0.4, na.rm = T),
                                        p0.5_time   = ~quantile(., probs = 0.5, na.rm = T),
                                        p0.6_time   = ~quantile(., probs = 0.6, na.rm = T),
                                        p0.7_time   = ~quantile(., probs = 0.7, na.rm = T),
                                        p0.8_time   = ~quantile(., probs = 0.8, na.rm = T),
                                        p0.9_time   = ~quantile(., probs = 0.9, na.rm = T),
                                        p0.95_time  = ~quantile(., probs = 0.95, na.rm = T)))

}

---

我的代码有什么问题？即，我专门与!!，!!!，sym，enquo和enquos事物作斗争。为什么group_by_at东西不需要!! 东西，而我的摘要和变异确实需要它吗？

Answer 1

Make these changes:

• use sym and syms rather than enquo and enquos
• use !! and !!! respectively.
• createpo as a list and then use unnest_wider to expand into columns
• quantile is already vectorized so we don't need map
• the mutate can be incorporated right into the quantile call eliminating it
• consolidate the pipelines into a single pipeline
• use TRUE rather than T since the latter can be masked by a variable of that name whereas no variable may be called TRUE.
• we can use plain group_by and summarize
• there is no group3 in the sample data so we used group2 instead
• this does not make sense without time_var so remove the default of NULL

This gives the following code

test_function <- function(data, time_var, group_vars = NULL) {
  p <- c(1:9/10, 0.95)
  time_var <- sym(time_var)
  group_vars <- syms(group_vars)
  data %>%
    group_by(!!!group_vars) %>%
    summarize(po = list(quantile(!!time_var / 60, p, na.rm = TRUE))) %>%
    ungroup %>%
    unnest_wider(po)
}

test_function(data = dat, time_var = "time", group_vars = c("group1", "group2")) 

giving:

# A tibble: 4 x 12
  group1 group2   `10%`   `20%`   `30%`   `40%`   `50%`   `60%`   `70%`   `80%`   `90%`   `95%`
   <int>  <int>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>
1      1      1 0.00237 0.00432 0.00654 0.00903 0.0115  0.0120  0.0124  0.0133  0.0147  0.0154 
2      1      2 0.00244 0.00251 0.00281 0.00335 0.00388 0.00410 0.00432 0.00493 0.00591 0.00640
3      2      1 0.00371 0.00381 0.00468 0.00632 0.00796 0.0101  0.0122  0.0136  0.0143  0.0147 
4      2      2 0.00385 0.00538 0.00630 0.00660 0.00691 0.00725 0.00759 0.00907 0.0117  0.0130