Summing rows by month in R

Question

So I have a data frame that has a date column, an hour column and a series of other numerical columns. Each row in the data frame is 1 hour of 1 day for an entire year.

Answer 1

Another base R solution

# to sum by date
rowsum(dat[-1], dat$Date)
#           Hour Melbourne Southern Flagstaff
#2009-05-01   21         0      496       715

# or by month and year
rowsum(dat[-1], format(dat$Date, "%b-%y") )
#       Hour Melbourne Southern Flagstaff
#May-09   21         0      496       715

Answer 2

I create the data set by

data <- read.table( text="   Date    Hour    Melbourne   Southern    Flagstaff
                       1   2009-05-01  0   0   5   17
                       2   2009-05-01  2   0   2   1
                       3   2009-05-01  1   0   11  0
                       4   2009-05-01  3   0   3   8
                       5   2009-05-01  4   0   1   0
                       6   2009-05-01  5   0   49  79
                       7   2009-05-01  6   0   425 610",
                    header=TRUE,stringsAsFactors=FALSE)

You can do the summation with the function aggregate:

byday <- aggregate(cbind(Melbourne,Southern,Flagstaff)~Date,
             data=data,FUN=sum)
library(lubridate)
bymonth <- aggregate(cbind(Melbourne,Southern,Flagstaff)~month(Date),
             data=data,FUN=sum)

Look at ?aggregate to understand the function better. Starting with the last argument (because that makes explaining easier) the arguments do the following:

• FUN is the function that should be used for the aggregation. I use sum to sum up the values, but i could also be mean, max or some function you wrote yourself.
• data is used to indicate that data frame that I want to aggregate.
• The first argument tells the function what exactly I want to aggregate. On the left side of the ~, I indicate the variables I want to aggregate. If there is more than one, they are combined with cbind. On the right hand side is the variable by which the data should be split. Putting Date means that aggregate will sum up the variables for each distinct value of Date.

For the aggregation by month, I used the function month from the package lubridate. It does what one expects: it returns a numeric value indicating the month for a given date. Maybe you first need to install the package by install.packages("lubridate").

If you prefer not to use lubridate, you could do the following instead:

data <- transform(data,month=as.numeric(format(as.Date(Date),"%m")))
bymonth <- aggregate(cbind(Melbourne,Southern,Flagstaff)~month,
                     data=data,FUN=sum)

Here I added a new column to data that contains the month and then aggregated by that column.

Answer 3

This could be another way to do this using data.table

library(data.table)
# Edited as per Arun's comment
out = setDT(data)[, lapply(.SD, sum), by=Date] 

#>out
#         Date Hour Melbourne Southern Flagstaff
#1: 2009-05-01   21         0      496       715

or by using dplyr

library(dplyr)
out = data %>% group_by(Date) %>% summarise_each(funs(sum))

#>out
#Source: local data frame [1 x 5]
#        Date Hour Melbourne Southern Flagstaff
#1 2009-05-01   21         0      496       715

Answer 4

I'd use dplyr::summarize and group_by, with a sum for each of your numeric columns:

summarize(group_by(df, Date), m_count = sum(Melbourne), s_count = sum(Southern), f_count = sum(Flagstaff)