如果在 R 中满足条件，则连续两天（或 N）从列中更改值答案

【问题标题】：Change values from a column for two (or N) consecutive days if condition met in R如果在 R 中满足条件，则连续两天（或 N）从列中更改值
【发布时间】：2021-10-04 18:23:15
【问题描述】：

我有以下数据框：

df <- structure(list(DateTime = structure(c(1477978200, 1477980000, 1477981800, 1477983600, 1477985400, 1477987200, 1477989000, 1477990800, 1477992600, 1477994400, 1477996200, 1477998000, 1477999800, 1478001600, 1478003400, 1478005200, 1478007000, 1478008800, 1478010600, 1478012400, 1478014200, 1478016000, 1478017800, 1478019600, 1478021400, 1478023200, 1478025000, 1478026800, 1478028600, 1478030400, 1478032200, 1478034000, 1478035800, 1478037600, 1478039400, 1478041200, 1478043000, 1478044800, 1478046600, 1478048400, 1478050200, 1478052000, 1478053800, 1478055600, 1478057400, 1478059200, 1478061000, 1478062800, 1478064600, 1478066400, 1478068200, 1478070000, 1478071800, 1478073600, 1478075400, 1478077200, 1478079000, 1478080800, 1478082600, 1478084400, 1478086200, 1478088000, 1478089800, 1478091600, 1478093400, 1478095200, 1478097000, 1478098800, 1478100600, 1478102400, 1478104200, 1478106000, 1478107800, 1478109600, 1478111400, 1478113200, 1478115000, 1478116800, 1478118600, 1478120400), class = c("POSIXct", "POSIXt"), tzone = "America/Chicago"), Date = structure(c(17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17106, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107, 17107), class = "Date"), Rain_daily = c(8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), PET_daily = c(7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4), Mask = c(TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE)), row.names = c(NA, 80L), class = "data.frame")

如果某一天的“Rain_daily”大于“PET_daily”，我想连续两天将“Mask”列中的值更改为“FALSE”。例如，对于给定的数据框，“2016-11-01”和“2016-11-02”（如 8>7）天的 Mask 列中的值将替换为 FALSE。

我们怎样才能做到这一点？我想避免 for 循环。

【问题讨论】：

我的数据框非常大，有多个列。在这里，我只是放了一个示例数据框。值为每 30 分钟一次。 Rain_daily 和 PET_daily 在一天内保持不变。
条件应该适用于每一天（或日期）。例如，如果“2016-11-01”日的条件满足，则“2016-11-01”和“2016-11-02”这两天的值（对于列掩码）应更改为 FALSE。同样，如果条件也满足日期“2016-11-02”，则值（对于列掩码）应更改为“2016-11-02”和“2016-11-03”天的 FALSE。

标签： r dataframe dplyr lubridate

【解决方案1】：

使用来自dplyr 的case_when。

如果日期满足条件Rain_daily > PET_daily，则为这些 (df[Rain_daily > PET_daily, Date]) 和接下来的 (df[Rain_daily > PET_daily, Date]+1) 天转换为 Mask。在其他地方保留 Mask 原样 (TRUE ~ Mask)

library(data.table)
library(dplyr)

df <- df %>% as.data.table() # to simlify operations on df
df <- df %>% 
  mutate(Mask = case_when(Date %in% c(df[Rain_daily > PET_daily, Date],df[Rain_daily > PET_daily, Date]+1) ~ FALSE,
                          TRUE ~Mask ))

希望有帮助

【讨论】：

【解决方案2】：

In a comment，OP 提到了 我的数据框真的很大，有多个列。

因此，这是一个data.table 方法，通过引用分配仅Mask 列。这样可以避免复制整个数据集。

library(data.table)
setDT(df2)
df2[, Mask := !Date %in% CJ(.SD[Rain_daily > PET_daily, Date], 0:1, unique = TRUE)[, V1 + V2]]
df2

               DateTime       Date Rain_daily PET_daily  Mask
 1: 2015-01-01 00:00:00 2015-01-01          1         2  TRUE
 2: 2015-01-01 00:30:00 2015-01-01          1         2  TRUE
 3: 2015-01-01 01:00:00 2015-01-01          1         2  TRUE
 4: 2015-01-02 00:00:00 2015-01-02          3         2 FALSE
 5: 2015-01-02 00:30:00 2015-01-02          3         2 FALSE
 6: 2015-01-02 01:00:00 2015-01-02          3         2 FALSE
 7: 2015-01-03 00:00:00 2015-01-03          1         2 FALSE
 8: 2015-01-03 00:30:00 2015-01-03          1         2 FALSE
 9: 2015-01-03 01:00:00 2015-01-03          1         2 FALSE
10: 2015-01-04 00:00:00 2015-01-04          1         2  TRUE
11: 2015-01-04 00:30:00 2015-01-04          1         2  TRUE
12: 2015-01-04 01:00:00 2015-01-04          1         2  TRUE
13: 2015-01-05 00:00:00 2015-01-05          3         2 FALSE
14: 2015-01-05 00:30:00 2015-01-05          3         2 FALSE
15: 2015-01-05 01:00:00 2015-01-05          3         2 FALSE
16: 2015-01-06 00:00:00 2015-01-06          4         2 FALSE
17: 2015-01-06 00:30:00 2015-01-06          4         2 FALSE
18: 2015-01-06 01:00:00 2015-01-06          4         2 FALSE
19: 2015-01-07 00:00:00 2015-01-07          1         2 FALSE
20: 2015-01-07 00:30:00 2015-01-07          1         2 FALSE
21: 2015-01-07 01:00:00 2015-01-07          1         2 FALSE
22: 2015-01-08 00:00:00 2015-01-08          1         2  TRUE
23: 2015-01-08 00:30:00 2015-01-08          1         2  TRUE
24: 2015-01-08 01:00:00 2015-01-08          1         2  TRUE
               DateTime       Date Rain_daily PET_daily  Mask

请注意，此处使用了一个新的示例数据集，其中包含不同的日期来演示不同的用例，请参阅下面的部分。

新样本数据集

作为一个可重复的示例，每天只模拟三个时间戳，但在 8 个不同的日子。

library(data.table)
df2 <- fread("
      Date Rain_daily PET_daily
2015-01-01          1         2
2015-01-02          3         2
2015-01-03          1         2
2015-01-04          1         2
2015-01-05          3         2
2015-01-06          4         2
2015-01-07          1         2
2015-01-08          1         2
")[CJ(Date, DateTime = as.ITime(60 * 30 * 0:2))[
  , DateTime := as.POSIXct(Date) + DateTime], on = "Date"][
    , setcolorder(.SD, "DateTime")]
df2

               DateTime       Date Rain_daily PET_daily
 1: 2015-01-01 00:00:00 2015-01-01          1         2
 2: 2015-01-01 00:30:00 2015-01-01          1         2
 3: 2015-01-01 01:00:00 2015-01-01          1         2
 4: 2015-01-02 00:00:00 2015-01-02          3         2
 5: 2015-01-02 00:30:00 2015-01-02          3         2
 6: 2015-01-02 01:00:00 2015-01-02          3         2
 7: 2015-01-03 00:00:00 2015-01-03          1         2
 8: 2015-01-03 00:30:00 2015-01-03          1         2
 9: 2015-01-03 01:00:00 2015-01-03          1         2
10: 2015-01-04 00:00:00 2015-01-04          1         2
11: 2015-01-04 00:30:00 2015-01-04          1         2
12: 2015-01-04 01:00:00 2015-01-04          1         2
13: 2015-01-05 00:00:00 2015-01-05          3         2
14: 2015-01-05 00:30:00 2015-01-05          3         2
15: 2015-01-05 01:00:00 2015-01-05          3         2
16: 2015-01-06 00:00:00 2015-01-06          4         2
17: 2015-01-06 00:30:00 2015-01-06          4         2
18: 2015-01-06 01:00:00 2015-01-06          4         2
19: 2015-01-07 00:00:00 2015-01-07          1         2
20: 2015-01-07 00:30:00 2015-01-07          1         2
21: 2015-01-07 01:00:00 2015-01-07          1         2
22: 2015-01-08 00:00:00 2015-01-08          1         2
23: 2015-01-08 00:30:00 2015-01-08          1         2
24: 2015-01-08 01:00:00 2015-01-08          1         2
               DateTime       Date Rain_daily PET_daily

【讨论】：