我编写了 R 代码来生成定期报告,该报告需要对周数进行重新排序,以便我可以按最近 10 周进行过滤和排序。为了防止错误并最大程度地减少硬编码值,我更喜欢在脚本顶部声明本周顺序,以获取所使用的其他几个脚本。因此,我想定义一个有序因子列表,然后使用它来对周数列进行排序。 RepEx 如下,但通常我会重新排序所有 52 周,以便最近的 10 周周期是最后/最大的,例如
new_levels <- factor(1:52, levels = c(29:52, 1:28), ordered=TRUE)
。
旁注:欢迎就如何更好地处理最近(不一定是最长)的 10 周期间提出任何建议。我过去的挣扎是由于临近年底的延期(51、52、1、2、3...)。
示例:
new_levels <- factor(1:10, levels = c(8:10, 1:7), ordered=TRUE)
data <- tibble(Week = 1:10, ID = c("A","A","B","B","C","A","D","B","D","A"))
data <- data %>% mutate(Week2 = factor(Week, levels = new_levels, ordered = TRUE)) %>% arrange(Week2)
有序因子(new_levels)似乎是正确的,但排列()和str()的行为表明我想要的排序没有发生:
> new_levels
[1] 1 2 3 4 5 6 7 8 9 10
Levels: 8 < 9 < 10 < 1 < 2 < 3 < 4 < 5 < 6 < 7
> data
# A tibble: 10 × 3
Week ID Week2
<int> <chr> <ord>
1 1 A 1
2 2 A 2
3 3 B 3
4 4 B 4
5 5 C 5
6 6 A 6
7 7 D 7
8 8 B 8
9 9 D 9
10 10 A 10
> str(data)
tibble [10 × 3] (S3: tbl_df/tbl/data.frame)
$ Week : int [1:10] 1 2 3 4 5 6 7 8 9 10
$ ID : chr [1:10] "A" "A" "B" "B" ...
$ Week2: Ord.factor w/ 10 levels "1"<"2"<"3"<"4"<..: 1 2 3 4 5 6 7 8 9 10
谢谢!
levels 参数的顺序必须正确。另请注意,它通过
factor
函数转换为字符向量。您可以按如下方式简化代码:
new_levels <- c(8:10, 1:7)
data <- tibble(Week = 1:10, ID = c("A","A","B","B","C","A","D","B","D","A"))
data <- data %>%
mutate(Week2 = factor(Week, levels = new_levels, ordered = TRUE)) %>%
arrange(Week2)