如何找到时间列的平均值并使用r对其进行分组？

d7v8vwbk 于 2023-10-13 发布在其他

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我有一个数据框，其中包含一个名为ride_length的列，该列已经是hh：mm：ss格式。我想计算该列的平均值，并将其分为两类：member和casual（位于member_casual列）。
我已经尝试过使用Lubridate库的管道：

df %>%
  group_by(member_casual) %>%
  seconds_to_period(mean(period_to_seconds(hms(ride_length))))

即使我的论点与网上找到的其他例子相同，我仍然得到这样的信息：
秒到周期的误差（.，mean（period_to_seconds（hms（ride_length）：未使用的参数（mean（period_to_seconds（hms（ride_length）
我也尝试了一条更长的路：

df$nride_length <- difftime(strptime(df$ride_length,"%H:%M:%S"),
                     strptime("00:00:00","%H:%M:%S"),
                     units="mins")
df.means <- aggregate(df$nride_length,by=list(df$member_casual),mean)
df.means$ride_length <- format(.POSIXct(df.means$x,tz="GMT"), "%H:%M:%S")
df.means

但结果还是有问题：
Group.1 x ride_length 1休闲NA mins 2成员NA mins
我也试着总结一下：

df %>%
  group_by(member_casual) %>%
  summarise(length_mean = seconds_to_period(mean(period_to_seconds(hms(ride_length)))))

但这表明：

# A tibble: 2 × 2
  member_casual length_mean
  <chr>         <Period>   
1 casual        NA         
2 member        NA         
Warning message:
There were 2 warnings in `summarise()`.
The first warning was:
ℹ In argument: `length_mean =
  seconds_to_period(mean(period_to_seconds(hms(ride_length))))`.
ℹ In group 1: `member_casual = "casual"`.
Caused by warning in `.parse_hms()`:
! Some strings failed to parse, or all strings are NAs
ℹ Run dplyr::last_dplyr_warnings() to see the 1 remaining warning.

请帮

来源：https://stackoverflow.com/questions/77278084/how-to-find-the-mean-of-a-time-column-and-group-it-using-r

2条答案

按热度按时间

ndh0cuux1#

您可以单独使用aggregate()。像使用方差分析那样指定分组。我改变了数据。框架一点，所以有三个“成员”和“休闲”。

dtf <- structure(list(rideable_type=c("electric_bike",
  "classic_bike", "classic_bike", "electric_bike",
  "classic_bike", "classic_bike"), day_of_week=c(1, 1, 1, 6, 7,
  2), ride_length=structure(c(990, 810, 576, 296, 686, 294),
  class=c("hms", "difftime"), units="secs"),
  member_casual=c("member", "member", "member", "casual",
  "casual", "casual"), nride_length=structure(c(16.5, 13.5, 9.6,
  4.93, 11.43, 4.9), class="difftime", units="mins")),
  row.names=c(NA, -6L), class=c("tbl_df", "tbl", "data.frame"))
    
aggregate(ride_length ~ member_casual, data=dtf, mean)
  #   member_casual    ride_length
  # 1        casual 425.33333 secs
  # 2        member 792.00000 secs

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zy1mlcev2#

假设您的数据如下：

x <- c("09:10:01", "10:10:02", "09:40:03","07:10:16", "09:20:02", "08:52:10")
df <- data.frame(member_casual=c(rep('A',3),rep('B',3)),
                   ride_length=hms(x),stringsAsFactors = F)
df
  member_casual ride_length
1             A   9H 10M 1S
2             A  10H 10M 2S
3             A   9H 40M 3S
4             B  7H 10M 16S
5             B   9H 20M 2S
6             B  8H 52M 10S

我试过你上面试过的代码，它对我很有效。

df %>%
   group_by(member_casual) %>%
   summarise(mean=seconds_to_period(mean(period_to_seconds(ride_length))))
# A tibble: 2 × 2
  member_casual mean                    
  <chr>         <Period>                
1 A             9H 40M 2S               
2 B             8H 27M 29.3333333333321S

所以请确认你的数据有正确的格式，特别是名为'ride_length'的列，单独运行hms(df$ride_length)并检查它是否成功运行。

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如何找到时间列的平均值并使用r对其进行分组？

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