从给定列中的时间戳中提取特定小时答案

【问题标题】：extract specific hour from time stamp in given column从给定列中的时间戳中提取特定小时
【发布时间】：2019-11-29 17:32:16
【问题描述】：

timestamp
2001-06-12 14:05:00
2001-08-12 15:06:00
2001-09-11 18:05:00
2001-06-22 14:05:00
2001-06-02 14:05:00
2001-06-12 18:05:00
2001-06-12 14:05:00
2001-06-12 11:11:00
2001-06-12 18:59:00

我必须从上面的时间戳中找到18:00 hour（18:00:00 到 18:59:59）的频率。

输出应该如下所示：

Output      18th_hour      
Frequency   3

【问题讨论】：

标签： r string datetime timestamp data-manipulation

【解决方案1】：

您可以提取时间戳的小时部分并计算"18"th 小时的出现次数。

在基础 R 中，我们可以这样做

sum(format(as.POSIXct(df$timestamp), "%H") == "18")
#[1] 3

类似使用lubridate

library(lubridate)
sum(hour(ymd_hms(df$timestamp)) == 18)

数据

df <- structure(list(timestamp = structure(c(3L, 7L, 8L, 6L, 1L, 4L, 3L, 2L, 
5L), .Label = c("2001-06-02 14:05:00", "2001-06-12 11:11:00", 
"2001-06-12 14:05:00", "2001-06-12 18:05:00", "2001-06-12 18:59:00", 
"2001-06-22 14:05:00", "2001-08-12 15:06:00", "2001-09-11 18:05:00"
), class = "factor")), class = "data.frame", row.names = c(NA, -9L))

【讨论】：

【解决方案2】：

带有anytime 的选项将“时间戳”列转换为日期时间，提取hour 并获取18 个元素的sum

library(anytime)
sum(hour(anytime(df$timestamp)) == 18)
#[1] 3

或者使用strptime 来自base R

sum(strptime(df$timestamp, format = "%Y-%m-%d %H:%M:%S")$hour == 18)

或者使用正则表达式选项

sum(grepl(" 18", df$timestamp))
#[1] 3

数据

df <- structure(list(timestamp = c("2001-06-12 14:05:00", "2001-08-12 15:06:00", 
"2001-09-11 18:05:00", "2001-06-22 14:05:00", "2001-06-02 14:05:00", 
"2001-06-12 18:05:00", "2001-06-12 14:05:00", "2001-06-12 11:11:00", 
"2001-06-12 18:59:00")), class = "data.frame", row.names = c(NA, 
-9L))

【讨论】：