【问题标题】：Check if condition is met in the days before current (R)检查当前 (R) 前几天是否满足条件
【发布时间】：2021-04-25 23:02:47
【问题描述】：

我有一个按日期排列的数据库表格，每小时测量一次。如果此度量大于 5，则符合标准 1（我已经在 dplyr 中使用了 mutate）像这样：

date	hour	Measure	Meets_criteria_1
01/01/2020	00:00:00	7	1
01/01/2020	01:00:00	12	1
01/01/2020	02:00:00	3	0
............
10/01/2020	21:00:00	2	0
10/01/2020	22:00:00	15	1
10/01/2020	23:00:00	20	1

现在，我想知道一天中有多少次满足条件。为此，我按天对“Meets_criteria_1”分组中的所有“1”求和：

date	hour	Meets_criteria_1	Sum_criteria_1
01/01/2020	00:00:00	1	3
01/01/2020	01:00:00	1	3
01/01/2020	02:00:00	0	3
............
10/01/2020	21:00:00	0	11
10/01/2020	22:00:00	1	11
10/01/2020	23:00:00	1	11

但是，我需要第三个条件来满足条件 2，即“在前两天，“Sum_criteria_1”至少为 6”

我的问题是：我如何告诉 R 检查前两天到当前？因此，例如，如果我正在查看日期 08/01/2020，则仅当 sum_criteria_1 在 2020 年 6 月 1 日和 2020 年 7 月 1 日至少为 6 时才满足条件。

编辑：我尝试使用滞后，但我只是得到 NA：

mutate(yesterday = lag(sum_criteria_1, n=1), 
       day_before = lag(sum_criteria_1, n = 2))

我们非常欢迎任何帮助！谢谢

【问题讨论】：

能不能把dput(dat[1:48,])的输出贴出来（我用48保证两天。
使用dplyr::lag()

标签： r date conditional-statements

【解决方案1】：

使用类似的东西

df %>% mutate(meets_criteria_1 = ifelse(Measure > 5, 1, 0)) %>%
  group_by(Date) %>%
  summarise(sum_criteria = sum(meets_criteria_1)) %>%
  mutate(criteria2 = ifelse(lag(sum_criteria) >= 6 & lag(sum_criteria,2), 1, 0)) %>%
  right_join(df, by = "Date")

样本数据

df <- structure(list(Date = c("01-01-2020", "01-01-2020", "01-01-2020", 
"01-01-2020", "01-01-2020", "01-01-2020", "01-01-2020", "01-01-2020", 
"01-01-2020", "01-01-2020", "01-01-2020", "01-01-2020", "01-01-2020", 
"01-01-2020", "01-01-2020", "01-01-2020", "01-01-2020", "01-01-2020", 
"01-01-2020", "01-01-2020", "01-01-2020", "01-01-2020", "01-01-2020", 
"01-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", 
"02-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", 
"02-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", 
"02-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", 
"02-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", 
"03-01-2020", "03-01-2020", "03-01-2020", "03-01-2020", "03-01-2020", 
"03-01-2020", "03-01-2020", "03-01-2020", "03-01-2020", "03-01-2020", 
"03-01-2020", "03-01-2020", "03-01-2020", "03-01-2020", "03-01-2020", 
"03-01-2020", "03-01-2020", "03-01-2020", "03-01-2020", "03-01-2020", 
"03-01-2020", "03-01-2020", "03-01-2020", "03-01-2020", "04-01-2020", 
"04-01-2020", "04-01-2020", "04-01-2020", "04-01-2020", "04-01-2020", 
"04-01-2020", "04-01-2020", "04-01-2020", "04-01-2020", "04-01-2020", 
"04-01-2020", "04-01-2020", "04-01-2020", "04-01-2020", "04-01-2020", 
"04-01-2020", "04-01-2020", "04-01-2020", "04-01-2020", "04-01-2020", 
"04-01-2020", "04-01-2020", "04-01-2020", "05-01-2020", "05-01-2020", 
"05-01-2020", "05-01-2020", "05-01-2020", "05-01-2020", "05-01-2020", 
"05-01-2020", "05-01-2020", "05-01-2020", "05-01-2020", "05-01-2020", 
"05-01-2020", "05-01-2020", "05-01-2020", "05-01-2020", "05-01-2020", 
"05-01-2020", "05-01-2020", "05-01-2020", "05-01-2020", "05-01-2020", 
"05-01-2020", "05-01-2020"), Hour = c("00:00:00", "01:00:00", 
"02:00:00", "03:00:00", "04:00:00", "05:00:00", "06:00:00", "07:00:00", 
"08:00:00", "09:00:00", "10:00:00", "11:00:00", "12:00:00", "13:00:00", 
"14:00:00", "15:00:00", "16:00:00", "17:00:00", "18:00:00", "19:00:00", 
"20:00:00", "21:00:00", "22:00:00", "23:00:00", "00:00:00", "01:00:00", 
"02:00:00", "03:00:00", "04:00:00", "05:00:00", "06:00:00", "07:00:00", 
"08:00:00", "09:00:00", "10:00:00", "11:00:00", "12:00:00", "13:00:00", 
"14:00:00", "15:00:00", "16:00:00", "17:00:00", "18:00:00", "19:00:00", 
"20:00:00", "21:00:00", "22:00:00", "23:00:00", "00:00:00", "01:00:00", 
"02:00:00", "03:00:00", "04:00:00", "05:00:00", "06:00:00", "07:00:00", 
"08:00:00", "09:00:00", "10:00:00", "11:00:00", "12:00:00", "13:00:00", 
"14:00:00", "15:00:00", "16:00:00", "17:00:00", "18:00:00", "19:00:00", 
"20:00:00", "21:00:00", "22:00:00", "23:00:00", "00:00:00", "01:00:00", 
"02:00:00", "03:00:00", "04:00:00", "05:00:00", "06:00:00", "07:00:00", 
"08:00:00", "09:00:00", "10:00:00", "11:00:00", "12:00:00", "13:00:00", 
"14:00:00", "15:00:00", "16:00:00", "17:00:00", "18:00:00", "19:00:00", 
"20:00:00", "21:00:00", "22:00:00", "23:00:00", "00:00:00", "01:00:00", 
"02:00:00", "03:00:00", "04:00:00", "05:00:00", "06:00:00", "07:00:00", 
"08:00:00", "09:00:00", "10:00:00", "11:00:00", "12:00:00", "13:00:00", 
"14:00:00", "15:00:00", "16:00:00", "17:00:00", "18:00:00", "19:00:00", 
"20:00:00", "21:00:00", "22:00:00", "23:00:00"), Measure = c(2L, 
14L, 4L, 19L, 0L, 15L, 13L, 17L, 3L, 19L, 0L, 17L, 15L, 14L, 
8L, 7L, 13L, 14L, 4L, 18L, 18L, 14L, 8L, 1L, 3L, 12L, 18L, 7L, 
13L, 15L, 12L, 17L, 2L, 8L, 1L, 18L, 19L, 14L, 2L, 7L, 12L, 17L, 
14L, 20L, 1L, 15L, 18L, 1L, 12L, 5L, 0L, 20L, 19L, 10L, 7L, 5L, 
8L, 8L, 0L, 15L, 16L, 20L, 14L, 18L, 17L, 3L, 15L, 14L, 4L, 17L, 
16L, 11L, 12L, 10L, 7L, 0L, 15L, 3L, 12L, 17L, 6L, 4L, 16L, 4L, 
15L, 0L, 5L, 7L, 6L, 3L, 15L, 10L, 12L, 19L, 13L, 3L, 18L, 14L, 
11L, 18L, 15L, 17L, 19L, 1L, 18L, 16L, 14L, 2L, 3L, 2L, 16L, 
10L, 2L, 12L, 10L, 7L, 5L, 9L, 12L, 17L)), class = "data.frame", row.names = c(NA, 
-120L))

检查结果

# A tibble: 120 x 5
   Date       sum_criteria criteria2 Hour     Measure
   <chr>             <dbl>     <dbl> <chr>      <int>
 1 01-01-2020           17        NA 00:00:00       2
 2 01-01-2020           17        NA 01:00:00      14
 3 01-01-2020           17        NA 02:00:00       4
 4 01-01-2020           17        NA 03:00:00      19
 5 01-01-2020           17        NA 04:00:00       0
 6 01-01-2020           17        NA 05:00:00      15
 7 01-01-2020           17        NA 06:00:00      13
 8 01-01-2020           17        NA 07:00:00      17
 9 01-01-2020           17        NA 08:00:00       3
10 01-01-2020           17        NA 09:00:00      19
# ... with 110 more rows

【讨论】：

没有达到目的吗？
嗨@AnilGoyal - 我遇到了一些问题。我的原始数据集非常庞大，包含 70 个站点、365 天和每天 24 次测量。我无法总结并重新加入....但我正在努力寻找方法！我会及时通知你的。非常感谢您的帮助！！
这就是为什么总是建议使用样本数据的原因。尽管如此，在 group_by 中使用 site，在 right_join 中使用 arg；你将能够得到结果
您介意向我解释一下这到底是做什么的吗？为什么仅在第一项中> = 6的条件？ ifelse(lag(sum_criteria) >= 6 & lag(sum_criteria,2)
前两个日期显然是 NA。每个组中的下一行（摘要将在每个组中呈现一行）将为 1 或 0，具体取决于您的前两行是否都大于 5 的条件。