我有一组10分钟移动平均线的天气数据，以1分钟的间隔显示。我想将其转换为平均1小时。
Date Direction Speed
1 2017-07-06 00:01:00 93 7.3
2 2017-07-06 00:02:00 92 7.4
3 2017-07-06 00:03:00 92 7.3
4 2017-07-06 00:04:00 91 7.4
5 2017-07-06 00:05:00 91 7.3
6 2017-07-06 00:06:00 91 7.3
7 2017-07-06 00:07:00 91 7.2
8 2017-07-06 00:08:00 90 7.1
9 2017-07-06 00:09:00 90 6.9
10 2017-07-06 00:10:00 91 6.7
...
(thousands of row of data in 1 min-interval
*上面的方向和速度是10分钟移动平均值
对于普通移动平均值内置函数，它们会遇到每个邻域值，例如：
rollmean(timeLine$Speed, 60, fill=FALSE, align = "right")
将在遇到n，n-1，n-2，n-3，...，n-59的每个值上产生滚动均值。
然而，由于我的原始数据已经是10分钟的平均值，我只需要取值n，n-10，n-20，n-30，n-40，n-50，以便将其转换为一小时的意思。
例如，如果我想要2001-07-06 10:00:00的每小时数据，我只需要对以下内容采取平均值：
有什么方法可以让我在R上顺利计算出来吗？
提前感谢您的帮助！
更新1：这是dput（head（timeLine，10））
structure(
list(
Date = structure(c(1499270460, 1499270520, 1499270580, 1499270640, 1499270700, 1499270760, 1499270820, 1499270880, 1499270940, 1499271000),
class = c("POSIXct", "POSIXt"), tzone = "Asia/Hong_Kong"),
Direction = c(93L, 92L, 92L, 91L, 91L, 91L, 91L, 90L, 90L, 91L),
Speed = c(7.3, 7.4, 7.3, 7.4, 7.3, 7.3, 7.2, 7.1, 6.9, 6.7)),
.Names = c("Date", "Direction", "Speed"),
row.names = c(NA, 10L),
class = "data.frame")
答案 0 :(得分：0)
lubridate包来轻松转换为日期时间格式：
library(tidyverse)
library(lubridate)
df = read.csv(text="
Date,Time,Direction,Speed
2001-07-04,09:01:00,310,4.0
2001-07-04,09:02:00,310,3.9
2001-07-04,09:03:00,310,3.9
2001-07-04,09:04:00,310,3.9
2001-07-04,09:05:00,300,3.9
2001-07-04,09:06:00,300,4.0
2001-07-04,09:07:00,300,3.9
2001-07-04,09:08:00,300,4.0
2001-07-04,09:09:00,300,4.0
2001-07-04,09:10:00,300,4.0
2001-07-04,09:11:00,290,4.0
2001-07-04,09:12:00,290,4.0
2001-07-04,09:13:00,290,4.0
2001-07-04,09:14:00,290,4.0
2001-07-04,09:15:00,290,4.0", sep=",", header = TRUE, row.names = NULL)
lagged_avg = function(col) {
lag_positions = c(0,10,20,30,40,50)
sum = 0
for (n in lag_positions) {
sum = sum + lag(col, n)
}
return(sum/6)
}
df = df %>%
mutate(datetime = ymd_hms(paste0(Date," ",Time))) %>%
mutate(lag = lagged_avg(Speed)) %>%
select(-Date, -Time)
答案 1 :(得分：0)
我会查看tibbletime package - 具体来说，collapse_by()功能很有帮助。以下应该可以工作（使用更多数据更容易测试）：
library(tidyverse)
library(lubridate)
library(tibbletime)
tbl_time(timeLine, index = Date) %>%
filter(minute(Date) %in% seq(0, 50, 10)) %>%
collapse_by("hour", clean = TRUE) %>%
group_by(Date) %>%
summarise_all(mean)
注意：根据您对小时数的看法，您可能希望将collapse_by行更改为collapse_by("hour", clean = TRUE, side = "start") - 默认情况下，它会使用side = "end"。
答案 2 :(得分：0)
解决方案是首先过滤0, 10, 20, 30, 40, 50th分钟数据。可以将minute的日期/时间除以10，并检查remainder是否等于0，以过滤0, 10, 20, 30, 40, 50th分钟数据的数据。每6次观察应用zoo::rollmean。以这种方式，将使用第10,20,30,40,40和0分钟数据计算每小时的平均值。最后过滤minute == 0（一小时）。
library(zoo)
library(lubridate)
library(tidyverse)
timeLine_mod %>% filter(minute(Date) %% 10 == 0) %>%
mutate(meanSpeed = rollmean(Speed, 6, fill = FALSE, align = "right")) %>%
filter(minute(Date) == 0)
# Date Direction Speed meanSpeed
# 1 2017-07-06 01:00:00 91 6.7 6.7
# 2 2017-07-06 02:00:00 91 6.7 6.7
# 3 2017-07-06 03:00:00 91 6.7 6.7
数据：由于OP仅提供10分钟的数据，这不足以计算每小时平均值。因此，我已将数据扩展到3小时：
timeLine <- structure(list(Date = structure(c(1499270460, 1499270520, 1499270580,
1499270640, 1499270700, 1499270760, 1499270820, 1499270880, 1499270940, 1499271000),
class = c("POSIXct", "POSIXt"), tzone = "Asia/Hong_Kong"),
Direction = c(93L, 92L, 92L, 91L, 91L, 91L, 91L, 90L, 90L, 91L),
Speed = c(7.3, 7.4, 7.3, 7.4, 7.3, 7.3, 7.2, 7.1, 6.9, 6.7)),
.Names = c("Date", "Direction", "Speed"), row.names = c(NA, 10L),
class = "data.frame")
#Extend data to cover 3 hours as
timeLine_mod <- timeLine %>% complete(Date = seq(min(Date),
min(Date)+60*60*3-60,by="1 min"))
#Repeat the value of Direction and Speed
timeLine_mod$Direction <- timeLine$Direction
timeLine_mod$Speed <- timeLine$Speed
答案 3 :(得分：0)
rollapplyr（最后r表示右对齐）允许使用width = list(offset_vector)来指定偏移：
transform(timeLine, avg = rollapplyr(Speed, list(seq(-50, 0, 10)), mean, fill = NA))