我需要将一列日期转换为时间序列答案

【问题标题】：I need to convert a column of dates into a time series我需要将一列日期转换为时间序列
【发布时间】：2020-12-29 12:54:04
【问题描述】：

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

输出看起来像这样：

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地区的时间序列。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

或者类似的东西：

------------+------+-------+------+--+
|    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，我敢肯定没有 DPLYR 软件包可以做到这一点，但非常感谢任何帮助。如果需要任何其他详细信息，请询问。

【问题讨论】：

能否请您移动您的输入并期望以文本而不是图像的形式输出到问题中？如果需要，可以将其放入代码块中，使其看起来类似于表格。看不懂的看meta.stackoverflow.com/q/277716/12672179了解怎么办表
这样好吗？

标签： r datetime dplyr

【解决方案1】：

您可以使用tidyverse 函数来做到这一点：

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，获取长格式数据并为存在的日期分配 1/0 值。

【讨论】：

【解决方案2】：

使用pivot_longer，然后使用pivot_wider 来自tidyr。
您可以使用complete 获取每个area 的完整日期分布。

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

【讨论】：

【解决方案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

【讨论】：