I need to convert my hourly data to daily data by taking the sum

I need to convert my hourly data to daily data by taking the sum



data set :


structure(list(time = structure(c(1506406740, 1506406770, 1506406860,
1506406890, 1506406920, 1506406950, 1506406980, 1506407010, 1506407040,
1506407070), class = c("POSIXct", "POSIXt"), tzone = "UTC"),
Column3 = c(131, 131, 131, 131, 131, 131, 131, 131, 131,
131), m_Pm = c(2.402842, 2.556558, 2.805165, 2.97428, 3.101824,
3.23984, 3.359587, 3.474448, 3.62753, 3.773597)), row.names = c(NA,
-10L), class = c("tbl_df", "tbl", "data.frame"))



It works alright till the first aggregate function . after that I am not able to get the correct answer after the sum function . I'm getting large values instead


library(dplyr)
library(ggplot2)

attach(data)
data %>% #filtering 131 and 132
select(time,Column3,m_Pm) %>%
filter(data,Column3=="131")
filter(data,Column3=="132")
data_131<-filter(data,Column3=="131")
data_132<-filter(data,Column3=="132")

#datehour column (dailyaverage)
data_131$datehour<-format(data_131$time,"%Y-%m-%d %H")
aggregate(m_Pm~datehour,data_131, mean)

#datecolumn

data_131$date1<-format(data_131$time,"%Y-%m-%d")
dE_131<-aggregate(m_Pm~date1,data_131,sum)
dE_131}



Since the data is too huge, I cannot post it here.
It is a 7 months data of every 30 seconds.






please share your dataset as well.

– Salman Lashkarara
Sep 10 '18 at 7:28






@SalmanLashkarara check edit

– fatma
Sep 10 '18 at 7:30






image is not good. you can run dput(head(YourDataSet,10)), and it gives you a sample of 10 rows

– Salman Lashkarara
Sep 10 '18 at 7:38


dput(head(YourDataSet,10))






@SalmanLashkarara done

– fatma
Sep 10 '18 at 7:47






Take a look at tibbletime package github.com/business-science/tibbletime

– Tung
Sep 10 '18 at 7:48


tibbletime




1 Answer
1


library(dplyr)
df %>% mutate(dayH= format(time,"%Y-%m-%d %H"), day= format(time,"%Y-%m-%d")) %>%
group_by(dayH) %>% mutate(th=mean(m_Pm)) %>% distinct(dayH, .keep_all = TRUE) %>%
group_by(day) %>% summarise(tday=sum(th))

# A tibble: 1 x 2
day tday
<chr> <dbl>
1 2017-09-26 15.7


df %>% mutate(dayH= format(time,"%Y-%m-%d %H"), day= format(time,"%Y-%m-%d"), month=format(time,"%Y-%m")) %>%
group_by(dayH) %>% mutate(th=mean(m_Pm)) %>% distinct(dayH, .keep_all = TRUE) %>%
group_by(day) %>% mutate(tday=sum(th)) %>% distinct(day, .keep_all = TRUE) %>%
group_by(month) %>% summarise(tmonth=sum(tday))






Error in mutate_impl(.data, dots) : Evaluation error: invalid 'trim' argument.

– fatma
Sep 10 '18 at 8:24






did you read my code ??

– fatma
Sep 10 '18 at 8:46






. but my sum function is wrong .

– fatma
Sep 10 '18 at 8:46






but i need to sum the hourly data to get total daily value

– fatma
Sep 10 '18 at 8:51






While this might answer the authors question, it lacks some explaining words and links to documentation. Raw code snippets are not very helpful without some phrases around it. You may also find how to write a good answer very helpful. Please edit your answer.

– hellow
Sep 10 '18 at 9:11



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