在 R 中按月汇总行

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.

The data frame looks like this:

          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

The hours are out of order because this is subsetted from another data frame.

I would like to sum the values in the numerical columns by month and possibly by day. Does anyone know how I can do this?

Answer 1

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 2

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 3

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 4

我会使用 dplyr::summarize 和 group_by，并对每个数字列求和：

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