是否有更快的方法根据另一列中的水平来分解和重新排列列？

Question

我下面的代码有效，但我觉得我缺少一种更快地完成此操作的方法（即我不知道更好的功能）。当我根据另一专栏搜索 relevel 论坛时，我得到的只是

factor(metric, levels = c(...)

的答案——我已经通过一系列电话完成了——但我觉得有一个更优雅的解决方案。

# Order of the week I would like to display in ggplot
week_order <- c('Wed', 'Thu', 'Fri', 'Sat', 'Sun', 'Mon', 'Tue' )

# Data
df <- tibble(metric = c(0, 8, 12, 18, 6, 12, 20),
                  label_day = c('Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun'),
                  legend_text = c('Off', 'ReEntry', 'Strength', 'Match 1', 'Recovery', 'Activation', 'Match 2'))

# Change `label_day` into a factor and order based on `week_order` 
df <- df |>
  mutate(label_day = factor(label_day, levels = week_order)) |>
  arrange(label_day)

# Is there a more elegant way to do this ? -------------------------------------------------------
# Now that the df is in the correct order, pull out the order that `legend_text` is in to use in ggplot
legend_text_order <- df |>
  distinct(legend_text) |>
  pull()

# Change `legend_text` into a factor and order based on `legend_text_order` 
df <- df |>
  mutate(legend_text = factor(legend_text, levels = legend_text_order))
#-------------------------------------------------------------------------------------------------

# Plot
ggplot(df, aes(x = legend_text, y = metric)) +
  geom_col()

Answer 1

您可以在一个管道中完成所有操作：

df <- df |>
  mutate(label_day = factor(label_day, levels = week_order)) |>
  arrange(label_day) |>
  mutate(legend_text = factor(legend_text, levels = unique(legend_text)))

并且因为您实际上不需要保留

label_day

作为一个因素，只需使用它的顺序来排列数据集，您可以进一步缩小它：

df <- df |>
  arrange(factor(label_day, levels = week_order)) |>
  mutate(legend_text = factor(legend_text, levels = unique(legend_text)))

输出：