提问人：Ahsk 提问时间：4/11/2023 更新时间：4/11/2023 访问量：42
如何根据R中的条件选择某个时期？
How to select a certain period based on a condition in R?
问：
我进行了几次实验，持续了大约一周。对于每周，我希望保留降水开始到周末的周期/行。
我知道如何摒弃干燥的几周，但我不确定如何选择从下雨开始到潮湿周结束的时期。
我的数据如下所示
在 中，我只想保留单元格 144 到 150 中的行并丢弃其余的行。完整的可以丢弃，因为这是一个完全干燥的一周。week 15week 16
注意：如果降水开始时有多个干/湿时段，直到周末，这很好。关键是忽略每周降雨发生前的干旱期。感谢您的帮助。
这是一个可重现的例子
df <- structure(
list(
week = structure(
c(
1L,
1L,
1L,
1L,
1L,
1L,
1L,
1L,
1L,
1L,
2L,
2L,
2L,
2L,
2L,
2L,
2L,
2L,
2L,
2L,
3L,
3L,
3L,
3L,
3L,
3L,
3L,
3L,
3L,
3L,
4L,
4L,
4L,
4L,
4L,
4L,
4L,
4L,
4L,
4L,
5L,
5L,
5L,
5L,
5L,
5L,
5L,
5L,
5L,
5L,
6L,
6L,
6L,
6L,
6L,
6L,
6L,
6L,
6L,
6L,
7L,
7L,
7L,
7L,
7L,
7L,
7L,
7L,
7L,
7L,
8L,
8L,
8L,
8L,
8L,
8L,
8L,
8L,
8L,
8L,
9L,
9L,
9L,
9L,
9L,
9L,
9L,
9L,
9L,
9L,
10L,
10L,
10L,
10L,
10L,
10L,
10L,
10L,
10L,
10L,
11L,
11L,
11L,
11L,
11L,
11L,
11L,
11L,
11L,
11L,
12L,
12L,
12L,
12L,
12L,
12L,
12L,
12L,
12L,
12L,
13L,
13L,
13L,
13L,
13L,
13L,
13L,
13L,
13L,
13L,
14L,
14L,
14L,
14L,
14L,
14L,
14L,
14L,
14L,
14L,
15L,
15L,
15L,
15L,
15L,
15L,
15L,
15L,
15L,
15L,
16L,
16L,
16L,
16L,
16L,
16L,
16L,
16L,
16L,
16L,
17L,
17L,
17L,
17L,
17L,
17L,
17L,
17L,
17L,
17L,
18L,
18L,
18L,
18L,
18L,
18L,
18L,
18L,
18L,
18L,
19L,
19L,
19L,
19L,
19L,
19L,
19L,
19L,
19L,
19L,
20L,
20L,
20L,
20L,
20L,
20L,
20L,
20L,
20L,
20L,
21L,
21L,
21L,
21L,
21L,
21L,
21L,
21L,
21L,
21L,
22L,
22L,
22L,
22L,
22L,
22L,
22L,
22L,
22L,
22L,
23L,
23L,
23L,
23L,
23L,
23L,
23L,
23L,
23L,
23L,
24L,
24L,
24L,
24L,
24L,
24L,
24L,
24L,
24L,
24L,
25L,
25L,
25L,
25L,
25L,
25L,
25L,
25L,
25L,
25L,
26L,
26L,
26L,
26L,
26L,
26L,
26L,
26L,
26L,
26L
),
.Label = c(
"1",
"2",
"3",
"4",
"5",
"6",
"7",
"8",
"9",
"10",
"11",
"12",
"13",
"14",
"15",
"16",
"17",
"18",
"19",
"20",
"21",
"22",
"23",
"24",
"25",
"26"
),
class = "factor"
),
precipitation = c(
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,
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,
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,
0,
1.59999999999997,
0.800000000000011,
0.600000000000023,
0.199999999999989,
0.399999999999977,
0.400000000000034,
1,
0.799999999999955,
0.400000000000034,
0.800000000000011,
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,
9.60000000000002,
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,
0,
0,
1,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0.200000000000045,
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,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0.200000002980232,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0
)
),
class = c("tbl_df",
"tbl", "data.frame"),
row.names = c(NA,-260L)
)
答：
2赞
r2evans
4/11/2023
#1
我们可以为此使用：filter(cumany(.))
library(dplyr)
out <- df %>%
group_by(week) %>%
filter(cumany(precipitation > 0)) %>%
ungroup()
filter(out, week == "15")
# # A tibble: 7 × 2
# week precipitation
# <fct> <dbl>
# 1 15 9.60
# 2 15 0
# 3 15 0
# 4 15 0
# 5 15 0
# 6 15 0
# 7 15 0
评论
1赞
Konrad Rudolph
4/11/2023
很好，我本来打算用的，但这有点好。cumsum(precipitation) > 0
0赞
r2evans
4/11/2023
那是我答案的第一次迭代，直到我想起来：-）......我之前已经包含在 base-R 答案中（不知道），经常从某人那里得到一个提示，所以显然我对累积逻辑函数有一些内部包混淆cumanycumanydplyr::cumany
评论