denis's questions

当然已经问过了，但我正在努力寻找适合我的情况的明确解决方案。

如何在分组的箱线图之间移动刻度（而不是以它们为中心）以定义小时间隔？

初始数据：

期望输出：

数据：

dat0 <- 
structure(list(group = c("A", "B", "A", "B", "A", "B", "A", "B", 
"A", "B", "A", "B", "A", "B", "A", "B", "A", "B", "A", "B", "A", 
"B", "A", "B", "A", "B", "A", "B", "A", "B", "A", "B", "A", "B", 
"A", "B", "A", "B", "A", "B", "A", "B", "A", "B", "A", "B", "A", 
"B"), hour = structure(c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 5L, 
5L, 6L, 6L, 7L, 7L, 8L, 8L, 9L, 9L, 10L, 10L, 11L, 11L, 12L, 
12L, 13L, 13L, 14L, 14L, 15L, 15L, 16L, 16L, 17L, 17L, 18L, 18L, 
19L, 19L, 20L, 20L, 21L, 21L, 22L, 22L, 23L, 23L, 24L, 24L), levels = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
"14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24"
), class = "factor"), q05 = c(38, 42, 38, 42, 34, 42, 35, 40, 
39, 40, 42, 49, 50, 56, 53, 61, 55, 59, 50, 58, 50, 58, 47, 55, 
46, 54, 48, 54, 48, 50, 46, 50, 46, 49, 44, 46, 44, 48, 41, 45, 
42, 44, 39, 45, 38, 42, 38, 41), q25 = c(48, 55, 47, 53, 44, 
53, 43, 51, 48, 53, 54, 64, 60, 74, 80, 96, 72, 81, 64, 77, 63, 
78, 61, 77, 60, 75, 62, 73, 60, 65, 58, 63, 59, 63, 55, 61, 53, 
61, 51, 55, 51, 59, 47, 58, 46, 54, 46, 53), q50 = c(56, 67, 
55, 66, 50, 65, 51, 63, 58, 65, 68, 84, 71, 94, 122, 135, 93, 
102, 81, 97, 80, 99, 80, 98, 75, 96, 82, 95, 74, 84, 72, 82, 
70, 79, 69, 74, 64, 76, 62, 70, 62, 76, 56, 71, 56, 69, 55, 66
), q75 = c(74, 88, 70, 88, 60, 83, 65, 84, 84, 88, 101, 129, 
93, 125, 167, 184, 131, 139, 117, 126, 111, 134, 112, 132, 124, 
139, 115, 128, 99, 112, 98, 113, 90, 105, 92, 100, 84, 102, 95, 
120, 86, 100, 73, 87, 76, 90, 70, 84), q95 = c(147, 135, 172, 
150, 124, 144, 136, 165, 320, 189, 265, 256, 239, 249, 268, 307, 
215, 262, 201, 215, 200, 216, 201, 233, 225, 237, 176, 233, 193, 
205, 165, 186, 170, 194, 160, 170, 222, 223, 203, 210, 166, 162, 
178, 152, 142, 167, 145, 129)), row.names = c(NA, -48L), class = "data.frame")    

# plot
ggplot(dat0, aes(x = hour)) +
  theme(
    legend.position="top",
    legend.title = element_blank(),
    panel.grid.major = element_blank()
  ) +
  geom_boxplot(
    aes(y = q50,
        fill = factor(group, levels = c('A', 'B')),
        ymin = q05, lower = q25, middle = q50, upper = q75, ymax = q95),
    stat = "identity",
    width = 0.8,
    position = position_dodge2(width = 1)
  )

感谢帮助

前 3 列中有重复项，我想在pivot_wider转置后保留这些重复项，但不要在列表格式中保留。
该怎么做？

有重复的初始数据：

dat0 <-
structure(list(id = c("P1", "P1", "P1", "P1", "P1", "P1", "P1", 
"P1", "P1", "P1", "P2", "P2", "P2", "P2", "P2", "P2"), analyte = c("A", 
"A", "B", "B", "B", "B", "C", "C", "D", "D", "B", "B", "B", "B", 
"D", "D"), analyzer = c("X", "Y", "X", "Y", "X", "Y", "X", "Y", 
"X", "Y", "X", "Y", "X", "Y", "X", "Y"), result = c(0.7, 0.9, 
1.26, 1.23, 1.24, 1.22, 5.7, 5.3, 4.1, 4.2, 1.22, 1.23, 1.21, 
1.22, 4.4, 4.5)), row.names = c(NA, -16L), class = c("tbl_df", 
"tbl", "data.frame")) 

pivot_wider运行后生成以下消息：

dat1 <- dat0 %>% 
  pivot_wider(names_from = analyzer, values_from = result)  

Values from `result` are not uniquely identified; output will contain list-cols.
• Use `values_fn = list` to suppress this warning.
• Use `values_fn = {summary_fun}` to summarise duplicates.
• Use the following dplyr code to identify duplicates.
  {data} |>
  dplyr::summarise(n = dplyr::n(), .by = c(id, analyte, analyzer)) |>
  dplyr::filter(n > 1L) 

期望输出重复项：

感谢帮助

根据以下按id和分组的数据visit（其中包括数字和文本/字符的混合）values，如何创建这 3 个新列：

• count_wotxt：按id和进行计数visit，但不考虑文本/字符值
• diff_value_first：计算每个数值与第一个数值visit之间的差异id，忽略文本/字符values
• diff_value_previous：计算每个数值与前一个数值visit之间的差异id，忽略文本/字符values

数据：

dat <-  
structure(list(id = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("1", "2"), class = "factor"), 
    visit = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 
    1L, 1L, 2L, 2L, 2L), .Label = c("1", "2"), class = "factor"), 
    value = c("5", "7", "10", "20", "15", "text0", "25", "text1", 
    "100", "text2", "text3", "120", "text4", "50", "45"), count = c(1L, 
    2L, 3L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 1L, 2L, 3L)), class = "data.frame", row.names = c(NA, 
-15L)) 

期望的输出：

> dat2
   id visit value count count_wotxt diff_value_first diff_value_previous
1   1     1     5     1           1                0                   0
2   1     1     7     2           2                2                   2
3   1     1    10     3           3                5                   3
4   1     2    20     1           1                0                   0
5   1     2    15     2           2               -5                  -5
6   1     2 text0     3          NA               NA                  NA
7   1     2    25     4           3                5                  10
8   1     2 text1     5          NA               NA                  NA
9   2     1   100     1           1                0                   0
10  2     1 text2     2          NA               NA                  NA
11  2     1 text3     3          NA               NA                  NA
12  2     1   120     4           2               20                  20
13  2     2 text4     1          NA               NA                  NA
14  2     2    50     2           1               NA                   0
15  2     2    45     3           2               NA                  -5 

感谢帮助