R语言如何平均正确的数据：温度-每天和每个站点的平均温度

uubf1zoe 于 2023-01-28 发布在其他

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我必须减去每天的平均温度到每一个温度值在白天也考虑到它来自车站，沿着9个月。我已经有了平均值，我的数据是这样的：

>>> df1
   Station     Date                Temperature
0  Station1    2022-05-1 9:30:00   7,4
1  Station1    2022-05-1 9:45:00   7,45
2  Station1    2022-05-1 10:00:00  8,2
3  Station1    2022-05-1 10:15:00  8,4
4  Station1    2022-05-1 10:30:00  8,9
5  Station1    2022-05-1 9:30:00   7,5
6  Station2    2022-05-1 9:45:00   7,56
7  Station2    2022-05-1 10:00:00  8,4
8  Station2    2022-05-1 10:15:00  8,7
9  Station2    2022-05-1 10:30:00  8,1
10 ...
>>> df2
   Station     Date        AverageTemperaturePerDayAndStation
0  Station1    2022-05-1   8
1  Station1    2022-05-2   8,3
2  Station1    2022-05-3   8,6
3  Station1    2022-05-4   8,4
4  Station1    2022-05-5   7,9
5  Station2    2022-05-1   6
6  Station2    2022-05-2   7,3
7  Station2    2022-05-3   8,6
8  Station2    2022-05-4   7,4
9  Station2    2022-05-5   6,9
10 ...

所以我想用R减去温度-每天和站点的平均温度，就像这样：

>>> df3
   Station     Date                CorrectedTemperature 
0  Station1    2022-05-1 9:30:00   7,4  - 8
1  Station1    2022-05-1 9:45:00   7,45 - 8
2  Station1    2022-05-1 10:00:00  8,2  - 8
3  Station1    2022-05-1 10:15:00  8,4  - 8
4  Station1    2022-05-1 10:30:00  8,9  - 8
5  Station1    2022-05-1 9:30:00   7,5  - 8
6  Station2    2022-05-1 9:45:00   7,56 - 6
7  Station2    2022-05-1 10:00:00  8,4  - 6
8  Station2    2022-05-1 10:15:00  8,7  - 6
9  Station2    2022-05-1 10:30:00  8,1  - 6
10 ...

r

来源：https://stackoverflow.com/questions/75258788/how-to-average-correct-data-temperature-average-temperature-per-day-and-stati

1条答案

按热度按时间

xzlaal3s1#

我谨提出以下建议：

#generate dataframes df1 and df2
df1 <- data.frame(Station =c(rep("Station1",5), rep("Station2",5)),
                  Date = c("2022-05-1 9:30:00","2022-05-1 9:45:00","2022-05-1 10:00:00","2022-05-1 10:15:00", "2022-05-1 10:30:00",
                           "2022-05-1 9:30:00","2022-05-1 9:45:00","2022-05-1 10:00:00","2022-05-1 10:15:00", "2022-05-1 10:30:00"),
                  Temperature = c(7.4, 7.45, 8.2, 8.4, 8.9, 7.5, 7.56, 8.4, 8.7, 8.1))
df2 <- data.frame(Station=c(rep("Station1",5), rep("Station2",5)),
                  Date = c("2022-05-01","2022-05-02","2022-05-03","2022-05-04","2022-05-05",
                           "2022-05-01","2022-05-02","2022-05-03","2022-05-04","2022-05-05"),
                  AverageTemperaturePerDayAndStation =c(8, 8.3, 8.6, 8.4, 7.9, 6, 7.3, 8.6, 7.4, 6.9))
#save Time and date in separate columns
df1$Time <- format(as.POSIXct(df1$Date), format = "%H:%M:%S")
df1$Date <- as.Date(df1$Date)
#change format of Date in df2 with as.Date
df2$Date <- as.Date(df2$Date)
#use left join (i.e. keep all entries from the first dataframe) and join by both Date and Station
df3 <- dplyr::left_join(df1, df2, by =c("Date","Station"))

然后，可以计算新列CorrectedTemperature。

df3$CorrectedTemperature <- df3$Temperature - df3$AverageTemperaturePerDayAndStation

希望这能帮上忙。

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R语言如何平均正确的数据：温度-每天和每个站点的平均温度

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R语言如何平均正确的数据：温度-每天和每个站点的平均温度