R如果函数用于时间序列中的每一天

Question

我的问题：

我有一个时间序列，看起来像这样：

v1 v2 t                   v3  day
1  46 33 2005-06-04 00:00:00 13 2005-06-04       
2  25 24 2005-06-04 01:00:00 15 2005-06-04       
3  18  9 2005-06-04 02:00:00 11 2005-06-04       
4  11 22 2005-06-04 03:00:00  1 2005-06-04       
5  11 31 2005-06-04 04:00:00  0 2005-06-04       
6  12 27 2005-06-04 05:00:00  3 2005-06-04       
7  46 33 2005-06-04 06:00:00 13 2005-06-04       
8  25 24 2005-06-04 07:00:00 15 2005-06-04       
9  18  9 2005-06-04 08:00:00 11 2005-06-04       
10 11 22 2005-06-04 09:00:00  1 2005-06-04       
11 11 31 2005-06-04 10:00:00 12 2005-06-04      
12 12 27 2005-06-04 11:00:00 13 2005-06-04    
13 46 33 2005-06-04 12:00:00 13 2005-06-04 
14 25 24 2005-06-04 13:00:00 15 2005-06-04 
15 18  9 2005-06-04 14:00:00 11 2005-06-04       
16 11 22 2005-06-04 15:00:00  1 2005-06-04       
17 11 31 2005-06-04 16:00:00  0 2005-06-04       
18 12 27 2005-06-04 17:00:00  3 2005-06-04                            

我想要"2005-06-04"的日期(格式(；v3"；10:00:00、11:00:00、12:00:00和13:00:00的值大于10。

我不知道如何实现这一点。

非常感谢。

Answer 1

使用dplyr和lubridate，请参阅代码中的注释以获得解释。

库(dplyr(图书馆

df1 %>% 
# create a counter
mutate(outcome = if_else(between(hour(t), 10, 13) & v3 > 10, 1, 0)) %>% 
group_by(day) %>% 
# rollup counter per day and filter the ones where the sum = 4
summarise(outcome = sum(outcome)) %>% 
filter(outcome == 4) %>% 
select(day)
# A tibble: 1 × 1
day       
<date>    
1 2005-06-04

数据：

df1 <- structure(list(v1 = c(46L, 25L, 18L, 11L, 11L, 12L, 46L, 25L, 
18L, 11L, 11L, 12L, 46L, 25L, 18L, 11L, 11L, 12L), v2 = c(33L, 
24L, 9L, 22L, 31L, 27L, 33L, 24L, 9L, 22L, 31L, 27L, 33L, 24L, 
9L, 22L, 31L, 27L), t = structure(c(1117843200, 1117846800, 1117850400, 
1117854000, 1117857600, 1117861200, 1117864800, 1117868400, 1117872000, 
1117875600, 1117879200, 1117882800, 1117886400, 1117890000, 1117893600, 
1117897200, 1117900800, 1117904400), class = c("POSIXct", "POSIXt"
), tzone = "UTC"), v3 = c(13L, 15L, 11L, 1L, 0L, 3L, 13L, 15L, 
11L, 1L, 12L, 13L, 13L, 15L, 11L, 1L, 0L, 3L), day = structure(c(12938, 
12938, 12938, 12938, 12938, 12938, 12938, 12938, 12938, 12938, 
12938, 12938, 12938, 12938, 12938, 12938, 12938, 12938), class = "Date")), row.names = c(NA, 
-18L), class = "data.frame")