Question

我有这个数据集

x<-data.frame(matrix(c("01-01-2010", "01-01-2010", "01-01-2010","01-01-2010"," 00:01"," 00:02"," 00:03"," 00:04", "12.2", "12.1", "13.1", "11.4", "12", "13", "5", "8","12", "4","7","9", "16.9", "17.5","18.8", "21.0"), ncol=6))
names(x)<-c("date","time","pressure","temperature","rain","windspeed")

        date     time pressure  temperature rain windspeed
1 01-01-2010   00:01     12.2          12   12      16.9
2 01-01-2010   00:02     12.1          13    4      17.5
3 01-01-2010   00:03     13.1           5    7      18.8
4 01-01-2010   00:04     11.4           8    9      21.0

这是我的数据集的简化版本。我的数据集从2010年1月1日00:01开始到2017年12月31日23:59。

我正在寻找

1）将平均压力，温度和风速转换为每小时数据。

2）将降雨汇总成每小时数据。

制作一个新的每小时时间戳以粘贴所有这些新数据很简单，我只需要知道什么是平均和汇总不同列的最佳方法，并且最多只能重复60行（60分钟即1个小时），直到2017/12/31 23:59

谢谢您的建议。

Answer 1

# sample data
x1 <- data.frame(matrix(c("01-01-2010", "01-01-2010", "01-01-2010","01-01- 
  2010","00:00:01","00:00:02","00:00:03","00:00:04", "12.2", "12.1", "13.1", "11.4", 
  "12", "13", "5", "8","12", "4","7","9", "16.9", "17.5","18.8", "21.0"), ncol=6))
x2 <- data.frame(matrix(c("01-01-2010", "01-01-2010", "01-01-2010","01-01- 
  2010","01:00:01","01:00:02","01:00:03","01:00:04", "12.2", "12.1", "13.1", "11.4", 
  "12", "13", "5", "8","12", "4","7","9", "16.9", "17.5","18.8", "21.0"), ncol=6))
x <- rbind(x1, x2)
names(x) <- c("date","time","pressure","temperature","rain","windspeed")
x[,3:6] <- apply(x[,3:6], 2, as.numeric)

# two separate aggregates 
aggregate(x[,c('pressure', 'temperature', 'windspeed')], by = list(paste0(x$date, 
  substring(x$time, 1, 2))), FUN = 'mean')
aggregate(x[,c('rain'), drop = FALSE], 
  by = list(paste0(x$date, substring(x$time, 1, 2))), FUN = 'sum')

#       Group.1 pressure temperature windspeed
#1 01-01-201000     12.2         9.5     18.55
#2 01-01-201001     12.2         9.5     18.55

#       Group.1 rain
#1 01-01-201000   32
#2 01-01-201001   32

Answer 2

我建议使用整洁的程序包和tibbletime来清楚，轻松地完成任务。我添加了一些清理代码，以便以所需的格式获取示例数据。

此方法是高度可重复和可解释的。 TibbleTime允许您在使用通用功能时对基于时间的数据进行大量汇总和滚动计算。

# The provided example data -----------------------------------------------
x<-data.frame(matrix(c("01-01-2010", "01-01-2010", "01-01-2010","01-01-2010"," 00:01"," 00:02"," 00:03"," 00:04", "12.2", "12.1", "13.1", "11.4", "12", "13", "5", "8","12", "4","7","9", "16.9", "17.5","18.8", "21.0"), ncol=6),
              stringsAsFactors = FALSE)
names(x)<-c("date","time","pressure","temperature","rain","windspeed")

# Load Libraries ----------------------------------------------------------
library(dplyr)
library(lubridate)
library(tibbletime)

# Fix column classes of data ----------------------------------------------
x <- x %>% 
  mutate_at(vars(pressure:windspeed),as.numeric)

# Convert to tibbletime object --------------------------------------------
x <- x %>%
  mutate(date_time = mdy_hm(paste0(date,time))) %>%
  as_tbl_time(index = date_time) %>%
  select(date_time,everything())

# Use tibbletime function to roll up hourly -------------------------------
x_hourly <- x %>%
  collapse_by('hourly',side = 'start') %>%
  group_by(date_time) %>%
  summarise(pressure = mean(pressure, na.rm = TRUE),
            temperature = mean(temperature, na.rm = TRUE),
            rain = sum(rain, na.rm = TRUE),
            windspeed = mean(windspeed, na.rm = TRUE))

结果：

> x_hourly
# A time tibble: 1 x 5
# Index: date_time
  date_time           pressure temperature  rain windspeed
  <dttm>                 <dbl>       <dbl> <dbl>     <dbl>
1 2010-01-01 00:01:00     12.2         9.5    32      18.6

将分钟数据集汇总或汇总为每日数据集，对R中的每60行将不同的函数应用于不同的列

2 个答案: