需要按一天中的每小时进行更广泛的调整，而无需创建多个日条目

Question

我有一个数据集，其中包含一年中每一天的每小时空气质量数据。我正在尝试扩大范围，但当 NULLCODE 列中存在空代码时遇到问题。这些特定日期会更广泛地分为同一天的多个条目（例如，下面的一小部分数据分为 2016 年 2 月 2 日的 3 个条目），但我每天只需要一个条目。

我想更广泛地旋转，以便每个小时都有一个包含 PM25 值的列和一个包含 NULLCODE 值的列，以防止每天分成多行，但我不确定如何做到这一点。

复制我的数据：

library(tidyr)
data <- data.frame(DATE = c("2019-02-09"),
                   TIME = c("00:00", "01:00", "02:00","03:00", "04:00","05:00","06:00","07:00","08:00","09:00","10:00","11:00","12:00","13:00","14:00","15:00","16:00","17:00","18:00","19:00","20:00","21:00","22:00","23:00"),
                   PM25 = c("2","1","0","-1","-4","NA","NA","NA","-4","1","6","28","23","3","3","5","NA","9","24","24","17","16","9","15"),
                   NULLCODE = c("","","","","","DA","DA","DA","","","","","","","","","AM","","","","","","",""))

我的原始数据框是什么样的

> print(data)
         DATE  TIME PM25 NULLCODE
1  2019-02-09 00:00    2         
2  2019-02-09 01:00    1         
3  2019-02-09 02:00    0         
4  2019-02-09 03:00   -1         
5  2019-02-09 04:00   -4         
6  2019-02-09 05:00   NA       DA
7  2019-02-09 06:00   NA       DA
8  2019-02-09 07:00   NA       DA
9  2019-02-09 08:00   -4         
10 2019-02-09 09:00    1         
11 2019-02-09 10:00    6         
12 2019-02-09 11:00   28         
13 2019-02-09 12:00   23         
14 2019-02-09 13:00    3         
15 2019-02-09 14:00    3         
16 2019-02-09 15:00    5         
17 2019-02-09 16:00   NA       AM
18 2019-02-09 17:00    9         
19 2019-02-09 18:00   24         
20 2019-02-09 19:00   24         
21 2019-02-09 20:00   17         
22 2019-02-09 21:00   16         
23 2019-02-09 22:00    9         
24 2019-02-09 23:00   1

这是我用来旋转更宽的代码，为 2016-02-09 创建三行：

data2 <- data %>% pivot_wider( 
  names_from = TIME,
  names_prefix = "Time_",
  values_from = PM25
)

旋转后我的数据框是什么样子

> print(data2)
        DATE NULLCODE Time_00:00 Time_01:00 Time_02:00 Time_03:00 Time_04:00
1 2019-02-09                   2          1          0         -1         -4
2 2019-02-09       DA       <NA>       <NA>       <NA>       <NA>       <NA>
3 2019-02-09       AM       <NA>       <NA>       <NA>       <NA>       <NA>
  Time_05:00 Time_06:00 Time_07:00 Time_08:00 Time_09:00 Time_10:00 Time_11:00
1       <NA>       <NA>       <NA>         -4          1          6         28
2         NA         NA         NA       <NA>       <NA>       <NA>       <NA>
3       <NA>       <NA>       <NA>       <NA>       <NA>       <NA>       <NA>
  Time_12:00 Time_13:00 Time_14:00 Time_15:00 Time_16:00 Time_17:00 Time_18:00
1         23          3          3          5       <NA>          9         24
2       <NA>       <NA>       <NA>       <NA>       <NA>       <NA>       <NA>
3       <NA>       <NA>       <NA>       <NA>         NA       <NA>       <NA>
  Time_19:00 Time_20:00 Time_21:00 Time_22:00 Time_23:00
1         24         17         16          9         15
2       <NA>       <NA>       <NA>       <NA>       <NA>
3       <NA>       <NA>       <NA>       <NA>       <NA>

Answer 1

能够通过以下方式获得正确的旋转：

data2 <- data %>% pivot_wider( 
  names_from = TIME,
  names_prefix = "Time_",
  values_from = c(PM25, NULLCODE)
)