Summary of data for each year in R

问题

I have a data with two columns. In one column it is date and in another column it is flow data.

I was able to read the data as date and flow data. I used the following code:

creek <- read.csv("creek.csv")
library(ggplot2)
creek[1:10,]
colnames(creek) <- c("date","flow")
creek$date <- as.Date(creek$date, "%m/%d/%Y")

The link to my data is https://www.dropbox.com/s/eqpena3nk82x67e/creek.csv

Now, I want to find the summary of each year. I want to especially know mean, median, maximum etc.

Thanks.

Regards, Jdbaba

回答1:

Base R

Here are two methods from base R.

The first uses cut, split and lapply along with summary.

creekFlowSummary <- lapply(split(creek, cut(creek$date, "1 year")), 
                           function(x) summary(x[2]))

This creates a list. You can view the summaries of different years by accessing the corresponding list index or name.

creekFlowSummary[1]
# $`1999-01-01`
#       flow       
#  Min.   :0.3187  
#  1st Qu.:0.3965  
#  Median :0.4769  
#  Mean   :0.6366  
#  3rd Qu.:0.5885  
#  Max.   :7.2560  
# 
creekFlowSummary["2000-01-01"]
# $`2000-01-01`
#       flow       
#  Min.   :0.1370  
#  1st Qu.:0.1675  
#  Median :0.2081  
#  Mean   :0.2819  
#  3rd Qu.:0.2837  
#  Max.   :2.3800

The second uses aggregate:

aggregate(flow ~ cut(date, "1 year"), creek, summary)
#    cut(date, "1 year") flow.Min. flow.1st Qu. flow.Median flow.Mean flow.3rd Qu. flow.Max.
# 1           1999-01-01    0.3187       0.3965      0.4770    0.6366       0.5885    7.2560
# 2           2000-01-01    0.1370       0.1675      0.2081    0.2819       0.2837    2.3800
# 3           2001-01-01    0.1769       0.2062      0.2226    0.2950       0.2574    2.9220
# 4           2002-01-01    0.1279       0.1781      0.2119    0.5346       0.4966   14.3900
# 5           2003-01-01    0.3492       0.4761      0.7173    1.0350       1.0840   10.1500
# 6           2004-01-01    0.4178       0.5379      0.6524    0.9691       0.9020   11.7100
# 7           2005-01-01    0.4722       0.6094      0.7279    1.2340       1.0900   17.7200
# 8           2006-01-01    0.2651       0.3275      0.4282    0.5459       0.5758    3.3510
# 9           2007-01-01    0.2784       0.3557      0.4041    0.6331       0.6125    9.6290
# 10          2008-01-01    0.4131       0.5430      0.6477    0.8792       0.9540    4.5960
# 11          2009-01-01    0.3877       0.4572      0.5945    0.8465       0.8309    6.3830

Be careful with the aggregate solution though: All of the summary information is a single matrix. View str on the output to see what I mean.

`xts`

There are, of course other ways to do this. One way is to use the xts package.

First, convert your data to xts:

library(xts)
creekx <- xts(creek$flow, order.by=creek$date)

Then, use apply.yearly and whatever functions you are interested in.

Here is the yearly mean:

apply.yearly(creekx, mean)
#                 [,1]
# 1999-12-31 0.6365604
# 2000-12-31 0.2819057
# 2001-12-31 0.2950348
# 2002-12-31 0.5345666
# 2003-12-31 1.0351742
# 2004-12-31 0.9691180
# 2005-12-31 1.2338066
# 2006-12-31 0.5458652
# 2007-12-31 0.6331271
# 2008-12-31 0.8792396
# 2009-09-30 0.8465300

And the yearly maximum:

apply.yearly(creekx, max)
#              [,1]
# 1999-12-31  7.256
# 2000-12-31  2.380
# 2001-12-31  2.922
# 2002-12-31 14.390
# 2003-12-31 10.150
# 2004-12-31 11.710
# 2005-12-31 17.720
# 2006-12-31  3.351
# 2007-12-31  9.629
# 2008-12-31  4.596
# 2009-09-30  6.383

Or, put them together like this: apply.yearly(creekx, function(x) cbind(mean(x), sum(x), max(x)))

`data.table`

The data.table package may also be of interest for you, particularly if you are dealing with a lot of data. Here's a data.table approach. The key is to use as.IDate on your "date" column while you are reading your data in:

library(data.table)
DT <- data.table(date = as.IDate(creek$date), creek[-1])
DT[, list(mean = mean(flow),
          tot = sum(flow),
          max = max(flow)), 
   by = year(date)]
#     year      mean      tot    max
#  1: 1999 0.6365604 104.3959  7.256
#  2: 2000 0.2819057 103.1775  2.380
#  3: 2001 0.2950348 107.6877  2.922
#  4: 2002 0.5345666 195.1168 14.390
#  5: 2003 1.0351742 377.8386 10.150
#  6: 2004 0.9691180 354.6972 11.710
#  7: 2005 1.2338066 450.3394 17.720
#  8: 2006 0.5458652 199.2408  3.351
#  9: 2007 0.6331271 231.0914  9.629
# 10: 2008 0.8792396 321.8017  4.596
# 11: 2009 0.8465300 231.1027  6.383

回答2:

You can use the code of @ananda to create an additional column with the year, the line with cut. Assuming the column name is year, you can use ddply from the plyr package:

ddply(creek, .(year), summarise, mm = mean(flow), me = median(flow), ...etc)

回答3:

@Pauls code can be rewritten with the next iteration of plyr package, the dplyr package.

Create a year variable
Create a separate data frame that splits the original data frame by year
Compute summary statistics for each year in the data frame with the summarize()

creek <- mutate(creek, year = as.POSIXlt(date)$year + 1900)
years <- group_by(creek, year)
summarize(years, mm= mean(flow), 
          tot= sum(flow), 
          max = max(flow, na.rm = TRUE))

Here is a nice dplyr tutorial by Roger Peng.

回答4:

Base R Very basic response, but you can always use summary() and subset() for an overview of specific years. The benefit of this is that it takes seconds to prepare.

summary(subset(df,Year == '####'))

来源：https://stackoverflow.com/questions/14641874/summary-of-data-for-each-year-in-r

标签

aggregate

time-series

summary