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R自己函数中R中的动态变量引用的R准引用和tidyeval

如何解决《R自己函数中R中的动态变量引用的R准引用和tidyeval》经验,为你挑选了1个好方法。

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?



1> G. Grothendi..:

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%`
                                   
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 

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