我的数据看起来像这样（为了简单起见，我删除了其他几列。）
Index Date Time Humid Temp id
93 4/3/16 12:00:00 AM 63.8 46.7 RSOSW
94 4/3/16 12:15:00 AM 60.3 47.8 RSOSW
95 4/3/16 12:30:00 AM 64.4 46.2 RSOSW
96 4/3/16 12:45:00 AM 60.4 46.8 RSOSW
97 4/3/16 1:00:00 AM 61.3 46.6 RSOSW
98 4/3/16 1:15:00 AM 68.5 44.3 RSOSW
99 4/3/16 1:30:00 AM 70.5 43.4 RSOSW
100 4/3/16 1:45:00 AM 75.1 41.8 RSOSW
101 4/3/16 2:00:00 AM 74.9 41.3 RSOSW
102 4/3/16 2:15:00 AM 73.6 41.1 RSOSW
103 4/3/16 2:30:00 AM 72.8 41.2 RSOSW
104 4/3/16 2:45:00 AM 71.1 41.2 RSOSW
93 4/3/16 12:00:00 AM 64.9 47.8 RSOSE
94 4/3/16 12:15:00 AM 61.2 48.9 RSOSE
95 4/3/16 12:30:00 AM 63.3 45.3 RSOSE
96 4/3/16 12:45:00 AM 62.6 42.3 RSOSE
97 4/3/16 1:00:00 AM 60.9 49.9 RSOSE
98 4/3/16 1:15:00 AM 67.3 45.3 RSOSE
99 4/3/16 1:30:00 AM 72.1 42.1 RSOSE
100 4/3/16 1:45:00 AM 79.0 40.5 RSOSE
101 4/3/16 2:00:00 AM 73.4 42.3 RSOSE
102 4/3/16 2:15:00 AM 73.6 40.1 RSOSE
103 4/3/16 2:30:00 AM 71.9 46.5 RSOSE
104 4/3/16 2:45:00 AM 70.6 45.4 RSOSE
我想通过id获得每小时平均温度和湿度。我正在寻找的结果是:(我想在每条记录中保留其他简单的数据删除列。）
Date Hour Humid Temp id
4/3/16 00 62.225 46.875 RSOSW
4/3/16 01 68.85 44.025 RSOSW
4/3/16 02 73.1 41.2 RSOSW
4/3/16 00 63 46.075 RSOSE
4/3/16 01 69.825 44.45 RSOSE
4/3/16 02 72.375 43.575 RSOSE
更新
Index Date Time Humid Temp serialnum id farm location
93 4/3/16 12:00:00 AM 63.8 46.7 1310014696 RSOSW_16 River School Outside
94 4/3/16 12:15:00 AM 60.3 47.8 1310014696 RSOSW_16 River School Outside
95 4/3/16 12:30:00 AM 64.4 46.2 1310014696 RSOSW_16 River School Outside
96 4/3/16 12:45:00 AM 60.4 46.8 1310014696 RSOSW_16 River School Outside
97 4/3/16 1:00:00 AM 61.3 46.6 1310014696 RSOSW_16 River School Outside
98 4/3/16 1:15:00 AM 68.5 44.3 1310014696 RSOSW_16 River School Outside
serialnum，id，farm和location都是字符。
提前致谢。
答案 0 :(得分：3)
library(lubridate)
df[,2] <- mdy_hms(df[,2])
df %>% mutate(hour = hour(df[,2])) %>%
group_by(id, hour) %>% summarise_at(vars(Humid, Temp), mean)
结果如下
Source: local data frame [6 x 4]
Groups: id [?]
id hour Humid Temp
<fctr> <int> <dbl> <dbl>
1 RSOSE 0 63.000 46.075
2 RSOSE 1 69.825 44.450
3 RSOSE 2 72.375 43.575
4 RSOSW 0 62.225 46.875
5 RSOSW 1 68.850 44.025
6 RSOSW 2 73.100 41.200
如果您希望保持列不变，并使用您计算的方法替换值，则可以
df %>% mutate(hour = hour(df[,2])) %>%
group_by(id, hour) %>% mutate_at(vars(Humid, Temp), mean) %>% head
它将导致
Source: local data frame [6 x 6]
Groups: id, hour [2]
Index datetime Humid Temp id hour
<int> <time> <dbl> <dbl> <fctr> <int>
1 93 2016-04-03 00:00:00 62.225 46.875 RSOSW 0
2 94 2016-04-03 00:15:00 62.225 46.875 RSOSW 0
3 95 2016-04-03 00:30:00 62.225 46.875 RSOSW 0
4 96 2016-04-03 00:45:00 62.225 46.875 RSOSW 0
5 97 2016-04-03 01:00:00 68.850 44.025 RSOSW 1
6 98 2016-04-03 01:15:00 68.850 44.025 RSOSW 1
df <- read.table(text =
"93 4/3/16 12:00:00 AM 63.8 46.7 RSOSW
94 4/3/16 12:15:00 AM 60.3 47.8 RSOSW
95 4/3/16 12:30:00 AM 64.4 46.2 RSOSW
96 4/3/16 12:45:00 AM 60.4 46.8 RSOSW
97 4/3/16 1:00:00 AM 61.3 46.6 RSOSW
98 4/3/16 1:15:00 AM 68.5 44.3 RSOSW
99 4/3/16 1:30:00 AM 70.5 43.4 RSOSW
100 4/3/16 1:45:00 AM 75.1 41.8 RSOSW
101 4/3/16 2:00:00 AM 74.9 41.3 RSOSW
102 4/3/16 2:15:00 AM 73.6 41.1 RSOSW
103 4/3/16 2:30:00 AM 72.8 41.2 RSOSW
104 4/3/16 2:45:00 AM 71.1 41.2 RSOSW
93 4/3/16 12:00:00 AM 64.9 47.8 RSOSE
94 4/3/16 12:15:00 AM 61.2 48.9 RSOSE
95 4/3/16 12:30:00 AM 63.3 45.3 RSOSE
96 4/3/16 12:45:00 AM 62.6 42.3 RSOSE
97 4/3/16 1:00:00 AM 60.9 49.9 RSOSE
98 4/3/16 1:15:00 AM 67.3 45.3 RSOSE
99 4/3/16 1:30:00 AM 72.1 42.1 RSOSE
100 4/3/16 1:45:00 AM 79.0 40.5 RSOSE
101 4/3/16 2:00:00 AM 73.4 42.3 RSOSE
102 4/3/16 2:15:00 AM 73.6 40.1 RSOSE
103 4/3/16 2:30:00 AM 71.9 46.5 RSOSE
104 4/3/16 2:45:00 AM 70.6 45.4 RSOSE")
df[,2] <- paste(df[,2], df[,3], df[,4])
df <- df[,c(-3,-4)]
names(df) <- c("Index", "datetime", "Humid", "Temp", "id")