R语言如何将每隔15分钟录得的天气数据转换为每小时的数据？

ht4b089n 于 2023-03-20 发布在其他

关注(0)|答案(5)|浏览(219)

我们的气象站每隔15分钟记录一次温度数据。我想计算每小时min最小值、每日max最小值和每日mean温度。我如何用R编程语言（最好是dplyr语法）来完成这项工作？
我的数据如下所示

date temperature
1   2014-05-26 11:45:00        25.0
2   2014-05-26 12:00:00        25.2
3   2014-05-26 12:15:00        25.3
4   2014-05-26 12:30:00        25.1
5   2014-05-26 12:45:00        25.4
[...]
96  2014-05-27 11:30:00        26.3
97  2014-05-27 11:45:00        25.7
98  2014-05-27 12:00:00        24.9
99  2014-05-27 12:15:00        24.9
100 2014-05-27 12:30:00        26.0

下面是一个可复制的例子。dput函数改变了日期格式，mdy_hm不能正确解析它。我想这就是为什么有些答案对我不起作用的原因。

df <- structure(list(date = structure(c(1401104700, 1401105600, 1401106500, 
1401107400, 1401108300, 1401109200, 1401110100, 1401111000, 1401111900, 
1401112800, 1401113700, 1401114600, 1401115500, 1401116400, 1401117300, 
1401118200, 1401119100, 1401120000, 1401120900, 1401121800, 1401122700, 
1401123600, 1401124500, 1401125400, 1401126300, 1401127200, 1401128100, 
1401129000, 1401129900, 1401130800, 1401131700, 1401132600, 1401133500, 
1401134400, 1401135300, 1401136200, 1401137100, 1401138000, 1401138900, 
1401139800, 1401140700, 1401141600, 1401142500, 1401143400, 1401144300, 
1401145200, 1401146100, 1401147000, 1401147900, 1401148800, 1401149700, 
1401150600, 1401151500, 1401152400, 1401153300, 1401154200, 1401155100, 
1401156000, 1401156900, 1401157800, 1401158700, 1401159600, 1401160500, 
1401161400, 1401162300, 1401163200, 1401164100, 1401165000, 1401165900, 
1401166800, 1401167700, 1401168600, 1401169500, 1401170400, 1401171300, 
1401172200, 1401173100, 1401174000, 1401174900, 1401175800, 1401176700, 
1401177600, 1401178500, 1401179400, 1401180300, 1401181200, 1401182100, 
1401183000, 1401183900, 1401184800, 1401185700, 1401186600, 1401187500, 
1401188400, 1401189300, 1401190200, 1401191100, 1401192000, 1401192900, 
1401193800), tzone = "UTC", class = c("POSIXct", "POSIXt")), 
    temperature = c(25, 25.2, 25.3, 25.1, 25.4, 26, 25.9, 25.6, 
    26.8, 27.8, 26.8, 26, 26, 26.3, 27, 27, 26.2, 25.8, 24.9, 
    25.1, 26.3, 25.6, 25.3, 25.2, 25.1, 24.8, 24.7, 24, 23, 22.7, 
    22.5, 22.5, 22.2, 21.9, 21.5, 21.1, 20.8, 20.5, 20.3, 20.3, 
    20.2, 20, 19.8, 19.6, 19.2, 19.1, 19.1, 18.9, 18.8, 18.6, 
    18.3, 18.2, 18.2, 18.2, 18.1, 17.9, 17.8, 17.7, 17.8, 18, 
    18.1, 18, 18.1, 18.6, 18.7, 18.5, 18.3, 18.1, 18.1, 18.6, 
    18.8, 18.6, 18.6, 18.3, 18.2, 18, 17.8, 18, 18.2, 18.9, 19.8, 
    19.6, 19.5, 19.7, 20.2, 21.5, 22.4, 23, 24, 23.3, 23.2, 23.7, 
    24.5, 24.8, 24.9, 26.3, 25.7, 24.9, 24.9, 26)), row.names = c(NA, 
-100L), class = c("tbl_df", "tbl", "data.frame"))

df$date = as.POSIXct(df$date)

我使用下面的代码将15分钟记录的数据转换为 daily，但我想更改为 hourly。

library(dplyr)
library(lubridate)
df %>%
   group_by(date = as.Date(date, "%m/%d/%Y")) %>%       # how to group by hour?
    summarise(min_temp = min(temperature, na.rm = TRUE),
              max_temp = max(temperature, na.rm = TRUE),
              mean_temp = mean(temperature, na.rm = TRUE)) %>%
              ungroup()

来源：https://stackoverflow.com/questions/75777105/how-to-convert-weather-data-recorded-at-15-minutes-interval-to-hourly-data

5条答案

按热度按时间

mo49yndu1#

使用round_date获得按小时分组的方法。使用玩具数据。
(...要确保您的 date 属于date类，请用途：df$date <- as.POSIXct(df$date, format="%m/%d/%Y %I:%M %p")）
编辑，使用过帐的df

library(dplyr)
library(lubridate)

df %>% 
  group_by(grp = cumsum(format(round_date(date, "hour"), "%H:%M") == 
                                 format(date, "%H:%M"))) %>% 
  summarize(date = unique(format(date, "%m/%d/%Y %I:00 %p")), 
            max_temperature = max(temperature), 
            min_temerature = min(temperature), 
            mean_temperature = mean(temperature)) %>% 
  select(-grp)
# A tibble: 26 × 4
   date                max_temperature min_temerature mean_temperature
   <chr>                         <dbl>          <dbl>            <dbl>
 1 05/26/2014 11:00 AM            25             25               25  
 2 05/26/2014 12:00 PM            25.4           25.1             25.2
 3 05/26/2014 01:00 PM            26.8           25.6             26.1
 4 05/26/2014 02:00 PM            27.8           26               26.6
 5 05/26/2014 03:00 PM            27             26.2             26.6
 6 05/26/2014 04:00 PM            26.3           24.9             25.5
 7 05/26/2014 05:00 PM            25.6           25.1             25.3
 8 05/26/2014 06:00 PM            24.8           23               24.1
 9 05/26/2014 07:00 PM            22.7           22.2             22.5
10 05/26/2014 08:00 PM            21.9           20.8             21.3
# … with 16 more rows
# ℹ Use `print(n = ...)` to see more rows

如果您确实需要所有小时的汇总表，请添加一个额外的summarize

df %>% 
  group_by(grp = cumsum(format(round_date(date, "hour"), "%H:%M") == 
                                 format(date, "%H:%M"))) %>% 
  summarize(Hour = unique(format(date, "%H:00")), 
            max_temperature = max(temperature), 
            min_temperature = min(temperature), 
            mean_temperature = mean(temperature)) %>% 
  select(-grp) %>% 
  summarize(across(ends_with("temperature"), mean), .by = Hour) %>%
  arrange(Hour)
# A tibble: 24 × 4
   Hour  max_temperature min_temperature mean_temperature
   <chr>           <dbl>           <dbl>            <dbl>
 1 00:00            18.6            18.2             18.3
 2 01:00            18.2            17.8             18  
 3 02:00            18.1            17.7             17.9
 4 03:00            18.7            18               18.4
 5 04:00            18.5            18.1             18.2
 6 05:00            18.8            18.6             18.7
 7 06:00            18.3            17.8             18.1
 8 07:00            19.8            18               18.7
 9 08:00            20.2            19.5             19.8
10 09:00            24              21.5             22.7
# … with 14 more rows
# ℹ Use `print(n = ...)` to see more rows

赞(0）回复(0）举报 2023-03-20

pengsaosao2#

使用lubridate::hour添加额外的分组变量，例如

> '4/20/2017 7:30 PM' |> 
+     strptime(format = '%m/%d/%Y %I:%M %p', tz = 'UTC') |> 
+     lubridate::hour()
[1] 19

(make确保时区正确，以防万一）

赞(0）回复(0）举报 2023-03-20

s3fp2yjn3#

下面是一个解决方案：
1.使用lubridate将日期转换为日期时间对象。
1.从datetime对象中提取小时。
1.按小时分组并总结。

library(dplyr)
library(lubridate)
library(stringr)

df <- tibble(
  date = c('4/20/2017 1:40 PM', '4/20/2017 1:45 PM',
           '4/20/2017 6:45 PM', '4/20/2017 7:45 PM'),
  temperature = c(14.5, 14.8, 19, 21.0)
)

# create date time, hour
df <- df %>%
  mutate(
    datetime = parse_date_time(df$date, '%m/%d/%Y %H:%M %p'),
    hour = lubridate::hour(datetime))

# group by hour
df %>%
  group_by(hour) %>%
  summarise(min_temp = min(temperature, na.rm = TRUE),
            max_temp = max(temperature, na.rm = TRUE),
            mean_temp = mean(temperature, na.rm = TRUE))

退货：

# A tibble: 3 × 4
   hour min_temp max_temp mean_temp
  <int>    <dbl>    <dbl>     <dbl>
1    13     14.5     14.8      14.6
2    18     19       19        19  
3    19     21       21        21

赞(0）回复(0）举报 2023-03-20

anauzrmj4#

我们在这里首先用mdy_hm()函数转换为日期时间格式。然后2.在计算下一行的差值后，我们应用cumsum() 3.我们用ceiling(diff/60)创建小时，然后使用现有代码：

library(lubridate)
library(dplyr)
df %>% 
  mutate(date = mdy_hm(date),
         diff = cumsum(as.numeric(difftime(date,lag(date, default = date[1]), units = "mins"))),
         hour = ceiling(diff / 60)) %>%
  group_by(hour) %>%
  summarise(min_temp = min(temperature, na.rm = TRUE),
            max_temp = max(temperature, na.rm = TRUE),
            mean_temp = mean(temperature, na.rm = TRUE)) %>%
  ungroup()

  hour min_temp max_temp mean_temp
  <dbl>    <dbl>    <dbl>     <dbl>
1     0     14.5     14.5      14.5
2     1     14.8     19.1      16.8
3     2     20.2     21.8      21.3
4     3     22.4     25.6      24.1
5     4     24.7     26.8      25.8
6     5     24.1     25.3      24.8
7     6     23       26.3      24.5
8     7     24.6     26.1      25.4
9     8     24.7     27.7      26.2

资料

df <- structure(list(date = c("4/20/2017 1:30 PM", "4/20/2017 1:45 PM", 
"4/20/2017 2:00 PM", "4/20/2017 2:15 PM", "4/20/2017 2:30 PM", 
"4/20/2017 2:45 PM", "4/20/2017 3:00 PM", "4/20/2017 3:15 PM", 
"4/20/2017 3:30 PM", "4/20/2017 3:45 PM", "4/20/2017 4:00 PM", 
"4/20/2017 4:15 PM", "4/20/2017 4:30 PM", "4/20/2017 4:45 PM", 
"4/20/2017 5:00 PM", "4/20/2017 5:15 PM", "4/20/2017 5:30 PM", 
"4/20/2017 5:45 PM", "4/20/2017 6:00 PM", "4/20/2017 6:15 PM", 
"4/20/2017 6:30 PM", "4/20/2017 6:45 PM", "4/20/2017 7:00 PM", 
"4/20/2017 7:15 PM", "4/20/2017 7:30 PM", "4/20/2017 7:45 PM", 
"4/20/2017 8:00 PM", "4/20/2017 8:15 PM", "4/20/2017 8:30 PM", 
"4/20/2017 8:45 PM", "4/20/2017 9:00 PM", "4/20/2017 9:15 PM", 
"4/20/2017 9:30 PM"), temperature = c(14.5, 14.8, 15.8, 17.3, 
19.1, 20.2, 21.4, 21.8, 21.7, 22.4, 23.2, 25.3, 25.6, 26.5, 26.8, 
24.7, 25.2, 25, 25.3, 24.7, 24.1, 23, 23.1, 25.6, 26.3, 26.1, 
25.8, 25.2, 24.6, 24.7, 25.6, 26.8, 27.7)), class = "data.frame", row.names = c(NA, 
-33L))

赞(0）回复(0）举报 2023-03-20

7vux5j2d5#

使用非tidyvert方法，我们可以使用by，将"POSIXct"转换为"Date"作为INDICES=，它将数据按日期拆分成一个类似列表的对象，然后我们可以在strftime和%H上使用aggregate，以获得每小时的min最大值。最后，只有data.frame，h_mins的min，以及每天的max和min，和rbind的事情。

by(df, as.Date(df$date), \(x) {
  h_mins <- aggregate(temperature ~ strftime(x$date, '%H'), x, min)[, 'temperature']
  data.frame(min_temp=min(h_mins), max_temp=max(x$temperature), mean_temp=mean(x$temperature))
}) |> do.call(what=rbind)
#            min_temp max_temp mean_temp
# 2017-04-20 12.09854 19.94601  17.01504
# 2017-04-21 12.09854 19.94601  17.01504
# 2017-04-22 12.09854 19.94601  17.01504

**注：**我使用下面的数据来获得更多的可用天数。这也适用于您的dput，但您需要df$date <- as.POSIXct(df$date)，因为您有"POSIXct"。

数据：*

df <- data.frame(
  date=seq.POSIXt(as.POSIXct('2017-04-20', tz='GMT'), as.POSIXct('2017-04-22 23:45', tz='GMT'), by='15 mins'),
  temperature=dnorm(seq.int(-1, 1, length.out=96))*.5e2
)

赞(0）回复(0）举报 2023-03-20

我来回答

R语言如何将每隔15分钟录得的天气数据转换为每小时的数据？

5条答案

相关问题

热门标签

最新问答

R语言 如何将每隔15分钟录得的天气数据转换为每小时的数据？

5条答案

相关问题

热门标签

最新问答

R语言如何将每隔15分钟录得的天气数据转换为每小时的数据？