我有一个这样的数据集超过23天：
temp1 DateTime
1 448 2016-03-24 21:14:00
2 448 2016-03-24 21:14:01
3 448 2016-03-24 21:14:02
4 448 2016-03-24 21:14:03
5 448 2016-03-24 21:14:04
6 448 2016-03-24 21:14:05
1930724 500 2016-04-15 13:46:21
1930725 500 2016-04-15 13:46:22
1930726 500 2016-04-15 13:46:23
1930727 500 2016-04-15 13:46:24
1930728 500 2016-04-15 13:46:25
1930729 500 2016-04-15 13:46:26
我现在想要添加一个新列，为每个day写入DateTime - 值大于sunrise time且小于sunset time，所以我得到：
temp1 DateTime dayNight
1 448 2016-03-24 21:14:00 night
2 448 2016-03-24 21:14:01 night
3 448 2016-03-24 21:14:02 night
4 448 2016-03-24 21:14:03 night
5 448 2016-03-24 21:14:04 night
6 448 2016-03-24 21:14:05 night
1930724 500 2016-04-15 13:46:21 day
1930725 500 2016-04-15 13:46:22 day
1930726 500 2016-04-15 13:46:23 day
1930727 500 2016-04-15 13:46:24 day
1930728 500 2016-04-15 13:46:25 day
1930729 500 2016-04-15 13:46:26 day
我从函数中得到"2016-04-15 06:40:37 UTC"的日出和日落时间：
sunRise <- function(x){
sunrise.set(lat, long, x, timezone = "UTC", num.days = 1)[1,1]
}
和
sunSet <- function(x){
sunrise.set(lat, long, x, timezone = "UTC", num.days = 1)[1,2]
}
包含sunrise.set() - 来自包StreamMetabolism的{{1}}和lat <- 28.39675
现在的问题是那些太阳时代在数据框架的23天内发生了变化。
根据每一天的特定日出/设置时间，我有办法制作long <- -16.567131列吗？
有没有办法将每天的日出/设置时间作为一列添加到数据框？
答案 0 :(得分：2)
这是一种方法（基于我的评论）并编制了我可以提出的数据：
library(dplyr)
# You don't need this line if you already have DateTime in proper format
df$DateTime <- as.POSIXct(df$DateTime)
# Add a date column (with whatever timezone you want)
df$date <- as.Date(df$DateTime, tz = 'EST')
# Following generates the sunrise and sunset times for the two example dates
sunRise <- c(as.POSIXct('2016-04-15 06:40:37'), as.POSIXct('2016-03-24 06:55:00'))
sunSet <- c(as.POSIXct('2016-04-15 18:40:37'), as.POSIXct('2016-03-24 18:25:00'))
sun <- data.frame(date = as.Date(sunRise, tz = 'EST'), sunRise = sunRise, sunSet = sunSet)
sun
date sunRise sunSet
1 2016-04-15 2016-04-15 06:40:37 2016-04-15 18:40:37
2 2016-03-24 2016-03-24 06:55:00 2016-03-24 18:25:00
# Join the two tables and compute night/day
df <- inner_join(df, sun)
df$dayNight <- ifelse(df$DateTime > df$sunRise & df$DateTime < df$sunSet, 'day', 'night')
输出如下：
df
temp1 DateTime date sunRise sunSet dayNight
1 448 2016-03-24 21:14:00 2016-03-24 2016-03-24 06:55:00 2016-03-24 18:25:00 night
2 448 2016-03-24 21:14:01 2016-03-24 2016-03-24 06:55:00 2016-03-24 18:25:00 night
3 448 2016-03-24 21:14:02 2016-03-24 2016-03-24 06:55:00 2016-03-24 18:25:00 night
4 448 2016-03-24 21:14:03 2016-03-24 2016-03-24 06:55:00 2016-03-24 18:25:00 night
5 448 2016-03-24 21:14:04 2016-03-24 2016-03-24 06:55:00 2016-03-24 18:25:00 night
6 448 2016-03-24 21:14:05 2016-03-24 2016-03-24 06:55:00 2016-03-24 18:25:00 night
7 500 2016-04-15 13:46:21 2016-04-15 2016-04-15 06:40:37 2016-04-15 18:40:37 day
8 500 2016-04-15 13:46:22 2016-04-15 2016-04-15 06:40:37 2016-04-15 18:40:37 day
9 500 2016-04-15 13:46:23 2016-04-15 2016-04-15 06:40:37 2016-04-15 18:40:37 day
10 500 2016-04-15 13:46:24 2016-04-15 2016-04-15 06:40:37 2016-04-15 18:40:37 day
11 500 2016-04-15 13:46:25 2016-04-15 2016-04-15 06:40:37 2016-04-15 18:40:37 day
12 500 2016-04-15 13:46:26 2016-04-15 2016-04-15 06:40:37 2016-04-15 18:40:37 day
当然，您可以删除不必要的列并保留您想要的内容，如下所示：
df[, -c(3, 4, 5)]
temp1 DateTime dayNight
1 448 2016-03-24 21:14:00 night
2 448 2016-03-24 21:14:01 night
3 448 2016-03-24 21:14:02 night
4 448 2016-03-24 21:14:03 night
5 448 2016-03-24 21:14:04 night
6 448 2016-03-24 21:14:05 night
7 500 2016-04-15 13:46:21 day
8 500 2016-04-15 13:46:22 day
9 500 2016-04-15 13:46:23 day
10 500 2016-04-15 13:46:24 day
11 500 2016-04-15 13:46:25 day
12 500 2016-04-15 13:46:26 day