1

我试图每 10 分钟计算一列的几何平均值。

我的样本数据是..

TimeDate    diam    ratio
2016-05-11 8:25 134.491 1.83074
2016-05-11 8:25 117.777 1.34712
2016-05-11 8:25 104.27  0.927635
2016-05-11 8:25 204.085 1.43079
2016-05-11 8:25 96.8011 0.991716
2016-05-11 8:25 119.152 1.09884
2016-05-11 8:25 113.871 0.932493
2016-05-11 8:26 150.468 0.710525
2016-05-11 8:26 116.576 1.11207
2016-05-11 8:26 192.257 1.61558
2016-05-11 8:26 128.071 0.756608
2016-05-11 8:26 177.667 0.73309
2016-05-11 8:27 97.7377 0.862858
2016-05-11 8:27 98.3195 1.00681
2016-05-11 8:27 91.3603 0.95051
2016-05-11 8:27 152.95  0.842145
2016-05-11 8:27 133.125 1.28365
2016-05-11 8:27 95.2516 0.573588

我已经尝试使用 dplyr 函数，但下面的代码不会每 10 分钟产生一次值，而是产生一个几何平均值和一个几何 sd 值。

mydata$TimeDate <- as.POSIXct(strptime(mydata$TimeDate, format = "%Y-%m-%d %H:%M","GMT"))

mydata %>%
    group_by(by10 = cut(TimeDate, breaks="10 min")) %>%
   summarize(Geo.Mean=exp(mean(log(diam))),
          Geo.SD=exp(sd(log(diam))))

数据格式本身是可以的，因为下面的聚合函数很好，尽管它不会创建几何平均值。

aggregate(mydata["diam"], 
               list(TimeDate=cut(mydata$TimeDate, "10 mins")),
                median, na.rm=T)
4

1 回答 1

2

一种选择是使用lubridate::floor_date函数为每 10 分钟全天候创建组。20-30 分钟之间的所有数据将被分组为第 20 分钟，依此类推。

library(dplyr)
library(lubridate)

mydata %>% mutate(TimeDate = as.POSIXct(TimeDate, format = "%Y-%m-%d %H:%M")) %>%
  group_by(Diff_10 = floor_date(TimeDate, "10minute")) %>%
  summarise(Geo.Mean=exp(mean(log(diam))),
            Geo.SD=exp(sd(log(diam))))

# # A tibble: 1 x 3
#     Diff_10             Geo.Mean Geo.SD
#     <dttm>                 <dbl>  <dbl>
#   1 2016-05-11 08:20:00      125   1.28

#Result with modified data
# # A tibble: 6 x 3
#   Diff_10             Geo.Mean Geo.SD
#   <dttm>                 <dbl>  <dbl>
# 1 2016-05-11 08:20:00    118     1.14
# 2 2016-05-11 08:30:00    141     1.69
# 3 2016-05-11 08:40:00    127     1.16
# 4 2016-05-11 08:50:00    150     1.28
# 5 2016-05-11 09:10:00     98.0   1.00
# 6 2016-05-11 09:20:00    115     1.29

cut如果从开始时间开始每 10 分钟对数据进行分组，则可以使用。在OP中，组将是2016-05-11 08:25，2016-05-11 08:35依此类推。

修改OP的数据：

mydata <- read.table(text = 
"TimeDate    diam    ratio
'2016-05-11 8:25' 134.491 1.83074
'2016-05-11 8:25' 117.777 1.34712
'2016-05-11 8:25' 104.27  0.927635
'2016-05-11 8:35' 204.085 1.43079
'2016-05-11 8:35' 96.8011 0.991716
'2016-05-11 8:42' 119.152 1.09884
'2016-05-11 8:45' 113.871 0.932493
'2016-05-11 8:46' 150.468 0.710525
'2016-05-11 8:56' 116.576 1.11207
'2016-05-11 8:56' 192.257 1.61558
'2016-05-11 8:56' 128.071 0.756608
'2016-05-11 8:59' 177.667 0.73309
'2016-05-11 9:17' 97.7377 0.862858
'2016-05-11 9:17' 98.3195 1.00681
'2016-05-11 9:27' 91.3603 0.95051
'2016-05-11 9:27' 152.95  0.842145
'2016-05-11 9:27' 133.125 1.28365
'2016-05-11 9:27' 95.2516 0.573588",
header = TRUE, stringsAsFactors = FALSE)
于 2018-04-11T06:54:15.267 回答