提问人：Ahsk 提问时间：7/29/2023 更新时间：7/30/2023 访问量：37
如何在 R 中仅选择每周的最后一小时天气数据？
How to select only the last hour of weather data from each week in R?
问：
我有一个天气数据集，每隔 15 分钟收集一次观测数据，持续数周。我只想提取天气数据，而忽略其余的。last houreach week
例如，在一周中，我只想保留单元格 147 到 150 的行（最后四行 - 每行代表 15 分钟：4*15 = 1 小时）并忽略其余的天气数据。15
可重复的示例
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,
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0,
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0,
0,
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1,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0.200000000000045,
0,
0,
0,
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0,
0,
0,
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0,
0,
0,
0,
0,
0,
0,
0,
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0,
0,
0,
0,
0,
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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
7/29/2023
#1
由于您没有“时间”变量，因此您要求提供“每周的最后 4 行”，在这种情况下：
library(dplyr)
slice_tail(df, n=4, by=week)
# # A tibble: 104 × 2
# week precipitation
# <fct> <dbl>
# 1 1 0
# 2 1 0
# 3 1 0
# 4 1 0
# 5 2 0
# 6 2 0
# 7 2 0
# 8 2 0
# 9 3 0
# 10 3 0
# # ℹ 94 more rows
# # ℹ Use `print(n = ...)` to see more rows
1赞
AndS.
7/30/2023
#2
下面是一个基本的 R 选项：
split(df, df$week) |>
Map(f = \(x) tail(x, 4)) |>
do.call(what = rbind.data.frame)
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