如何从R中的aggregate（）函数创建表

Question

I have a data set that looks like this:

Type    Date    Lively  Count
sm      1/13/2010   10  10
sm      1/14/2010   10  20
sm      2/15/2010   20  30
am      4/16/2010   5   42
am      1/17/2010   10  34
am      3/18/2010   40  54
sm      1/19/2010   10  65
sm      4/20/2010   5   67
sm      3/21/2010   40  76

I'm trying to average out all the numeric parameters by month. So my resultant data set would ideally be:

Date     Lively Count
Jan 2010     10     32.25
Feb 2010     20     30.00
Mar 2010     40     65.00
Apr 2010      5     54.50

I'm very close to this, what I currently have is:

         Lively Count
Jan 2010     10     32.25
Feb 2010     20     30.00
Mar 2010     40     65.00
Apr 2010      5     54.50

As you can see I'm missing the title 'Date'. Here is my code:

library(zoo)
z <- zoo(data[3:4], as.Date(data[,2], "%m/%d/%Y"))
aggregate(z, as.yearmon, mean)

I don't know how to make a title for the left column ('Date'), and more importantly, I don't know how to make the output of aggregate() into a table (resultant data set).

Answer 1

You can use the list format to specify your names within aggregate .

To get your "date" values, you need to refer to the "index" of your zoo object

aggregate(list(Lively = z[, "Lively"], Count = z[, "Count"]), 
          list(Date = as.yearmon(index(z))), mean)
#       Date Lively Count
# 1 Jan 2010     10 32.25
# 2 Feb 2010     20 30.00
# 3 Mar 2010     40 65.00
# 4 Apr 2010      5 54.50

Alternatively, you can easily change your names if required. This will allow you to be able to use the much nicer formula method for aggregate .

x <- aggregate(. ~ as.yearmon(index(z)), z, mean)
names(x)[1] <- "Date"
x
#       Date Lively Count
# 1 Jan 2010     10 32.25
# 2 Feb 2010     20 30.00
# 3 Mar 2010     40 65.00
# 4 Apr 2010      5 54.50

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Note, however, that by doing so, you miss out on all the goodness that zoo has to offer by doing this. You've essentially taken a data.frame , converted it to a zoo object, and re-converted it to a data.frame .

Your aggregate(z, as.yearmon, mean) solution would be the way I would go about it, and refer to the date by using index() .

---

Update

If you're just doing this at a later stage for aesthetic reasons, you can keep working with zoo objects since it will give you a lot of flexibility that you might not get with base R functions, and then use cbind at the end.

Following from where you left off:

library(zoo)
z <- zoo(data[3:4], as.Date(data[,2], "%m/%d/%Y"))
x <- aggregate(z, as.yearmon, mean)
cbind(Date = index(x), 
      as.data.frame.matrix(x, row.names = NULL))
#       Date Lively Count
# 1 Jan 2010     10 32.25
# 5 Feb 2010     20 30.00
# 6 Mar 2010     40 65.00
# 8 Apr 2010      5 54.50