我需要将一列日期转换为时间序列

Question

所以我有以下类型的数据集。 我有三个区域，每个区域恰好在一个日期下雨和冰雹，

area<-c("A","B","C")
rain<-c("1994/08/01","1994/08/01","1994/08/03")
hail<-c("1994/08/03","1994/08/04","1994/08/05")

data1<-as.data.frame(cbind(area,rain,hail))

data1

输出如下所示：

我拥有的数据类型

看起来像：

+-------+------------+------------+--+--+
|       |            |            |  |  |
+-------+------------+------------+--+--+
| area  | rain       | hail       |  |  |
| A     | 1994/08/01 | 1994/08/03 |  |  |
| B     | 1994/08/01 | 1994/08/04 |  |  |
| C     | 1994/08/03 | 1994/08/05 |  |  |
+-------+------------+------------+--+--+

我想将其转换为每个区域的时间序列。 有点像长数据：

date<-as.Date(c("1994/08/01","1994/08/02","1994/08/03","1994/08/04","1994/08/05"))
date<-c(date,date,date)
area<-c("A","A","A","A","A","B","B","B","B","B","C","C","C","C","C")

rain<-c(1,0,0,0,0,1,0,0,0,0,0,0,1,0,0)
hail<-c(0,0,1,0,0,0,0,0,1,0,0,0,0,0,1)
data2<-as.data.frame(date)
data2<-cbind(data2,area,rain,hail)
data2

我想要的数据类型

或类似的东西：

------------+------+-------+------+--+
|    date    | area | rain  | hail |  |
+------------+------+-------+------+--+
| 1994-08-01 | A    |     1 |    0 |  |
| 1994-08-02 | A    |     0 |    0 |  |
| 1994-08-03 | A    |     0 |    1 |  |
| 1994-08-04 | A    |     0 |    0 |  |
| 1994-08-05 | A    |     0 |    0 |  |
| 1994-08-01 | B    |     1 |    0 |  |
| 1994-08-02 | B    |     0 |    0 |  |
| 1994-08-03 | B    |     0 |    0 |  |
| 1994-08-04 | B    |     0 |    1 |  |
| 1994-08-05 | B    |     0 |    0 |  |
+------------+------+-------+------+--+

这是非常非常规的，我确信没有 DPLYR 包可以做到这一点，但非常感谢任何帮助。 如果需要任何其他详细信息，请务必询问。

Answer 1

您可以使用tidyverse函数执行此tidyverse ：

library(dplyr)
data1 %>%
  mutate(across(c(rain, hail), lubridate::ymd),
         date = list(seq(min(rain, hail), max(rain, hail), 'day'))) %>%
  tidyr::unnest(date) %>%
  mutate(across(c(rain, hail), ~+(. == date)))

# A tibble: 15 x 4
#   area   rain  hail date      
#   <chr> <int> <int> <date>    
# 1 A         1     0 1994-08-01
# 2 A         0     0 1994-08-02
# 3 A         0     1 1994-08-03
# 4 A         0     0 1994-08-04
# 5 A         0     0 1994-08-05
# 6 B         1     0 1994-08-01
# 7 B         0     0 1994-08-02
# 8 B         0     0 1994-08-03
# 9 B         0     1 1994-08-04
#10 B         0     0 1994-08-05
#11 C         0     0 1994-08-01
#12 C         0     0 1994-08-02
#13 C         1     0 1994-08-03
#14 C         0     0 1994-08-04
#15 C         0     1 1994-08-05

制作rain和hail日期列，在最小和最大日期之间创建一个序列，以长格式获取数据并为存在的日期分配 1/0 值。

Answer 2

使用pivot_longer然后pivot_wider从tidyr 。
您可以使用complete获取每个area complete日期分布。

data1 %>%
  pivot_longer(c(rain, hail), names_to = "weather", values_to = "date",
               values_transform = list(date = as.Date)) %>%
  mutate(min_date = min(date),
         max_date = max(date)) %>%
  group_by(area) %>%
  complete(date = seq.Date(first(min_date), last(max_date), by="day")) %>%
  pivot_wider(names_from = weather, 
              values_from = weather, 
              values_fn = length,
              values_fill = 0) %>%
  select(date, area, rain, hail)

# A tibble: 15 x 4
# Groups:   area [3]
   date       area   rain  hail
   <date>     <fct> <int> <int>
 1 1994-08-01 A         1     0
 2 1994-08-02 A         0     0
 3 1994-08-03 A         0     1
 4 1994-08-04 A         0     0
 5 1994-08-05 A         0     0
 6 1994-08-01 B         1     0
 7 1994-08-02 B         0     0
 8 1994-08-03 B         0     0
 9 1994-08-04 B         0     1
10 1994-08-05 B         0     0
11 1994-08-01 C         0     0
12 1994-08-02 C         0     0
13 1994-08-03 C         1     0
14 1994-08-04 C         0     0
15 1994-08-05 C         0     1

Answer 3

如果您更喜欢基础 R，并且想从基础函数中了解更多 R，您可以尝试：

data1[-1] <- lapply(data1[-1], as.Date) # Change the rain and hail columns to date
data2 <- merge(data1, do.call(seq,c(as.list(do.call(range,data1[-1])), by = 1))) #create data2
names(data2)[4] <- "date" # change the name of column y
data2[2:3] <- +sapply(data2[2:3],`==`,data2[,4]) # find the 1, 0
data.frame(data2[order(data2$area),], row.names = NULL) # Just arrange the data


 area rain hail       date
1     A    1    0 1994-08-01
2     A    0    0 1994-08-02
3     A    0    1 1994-08-03
4     A    0    0 1994-08-04
5     A    0    0 1994-08-05
6     B    1    0 1994-08-01
7     B    0    0 1994-08-02
8     B    0    0 1994-08-03
9     B    0    1 1994-08-04
10    B    0    0 1994-08-05
11    C    0    0 1994-08-01
12    C    0    0 1994-08-02
13    C    1    0 1994-08-03
14    C    0    0 1994-08-04
15    C    0    1 1994-08-05