Plot value over hour of day with xts/zoo R

问题

I do have a time-series looking like this (minute values):

   "timestamp", value 
    "2012-04-09 05:03:00",2
    "2012-04-09 05:04:00",4
    "2012-04-09 05:05:00",5
    "2012-04-09 05:06:00",0
    "2012-04-09 05:07:00",0
    "2012-04-09 05:08:00",3
    "2012-04-09 05:09:00",0
    "2012-04-09 05:10:00",1

Is there an easy way to plot these values over the hour of the day: X-Axis from 1 to 24 hours (or 0 and 23). So - all values between 5:00 and 5:59 over the 5th hour etc. It doesn't depend wich date, I am just interested in the houers of the day. Thank you!

Additional Question: Can I plot this as a boxplot? Right now it's plot(df$hh,df$coredata.._data..)in zoo format.

回答1:

Using xts package, you can use .indexhour Generic functions to format the index of an xts object to hours.

For example , I read the data,

dat <- read.zoo(text='timestamp, value 
    "2012-04-09 05:03:00",2
    "2012-04-09 05:04:00",4
    "2012-04-09 05:05:00",5
    "2012-04-09 05:06:00",0
    "2012-04-09 05:07:00",0
    "2012-04-09 05:08:00",3
    "2012-04-09 05:09:00",0
    "2012-04-09 05:10:00",1',header=TRUE,tz='',index=0:1,sep=',')

transform(dat, hh = .indexhour(dat))
                    coredata.._data.. hh
2012-04-09 05:03:00                 2  5
2012-04-09 05:04:00                 4  5
2012-04-09 05:05:00                 5  5
2012-04-09 05:06:00                 0  5
2012-04-09 05:07:00                 0  5
2012-04-09 05:08:00                 3  5
2012-04-09 05:09:00                 0  5
2012-04-09 05:10:00                 1  5

回答2:

Read the data creating zoo object z and then create hour variable and plot:

library(zoo)

# read in data
Lines <- ' "timestamp", value 
    "2012-04-09 05:03:00",2
    "2012-04-09 05:04:00",4
    "2012-04-09 05:05:00",5
    "2012-04-09 05:06:00",0
    "2012-04-09 05:07:00",0
    "2012-04-09 05:08:00",3
    "2012-04-09 05:09:00",0
    "2012-04-09 05:10:00",1
'
z <- read.zoo(text = Lines, header = TRUE, sep = ",", tz = "")

# plot
hour <- as.POSIXlt(time(z))$hour
plot(hour, z)

回答3:

You could get a vector of the hours by converting those times to POSIXlt classes and then extracting the "hours" component from the resulting list for each one, using sapply to return a vector. You can then add this to the original dataframe:

df <- read.table( text = "timestamp, value 
    2012-04-09 05:03:00,2
    2012-04-09 05:04:00,4
    2012-04-09 05:05:00,5
    2012-04-09 05:06:00,0
    2012-04-09 05:07:00,0
    2012-04-09 05:08:00,3
    2012-04-09 05:09:00,0
    2012-04-09 05:10:00,1" , h = T , sep = "," )

# Convert to POSIX class
df$timestamp <- as.POSIXlt(df$timestamp)

# Each entry in timestamp has some hidden attributes
attributes(df$timestamp[1])
$names
# [1] "sec"   "min"   "hour"  "mday"  "mon"   "year"  "wday"  "yday"  "isdst"

hours <- sapply( seq_len( nrow( df ) ) , function(x) df$timestamp[x]$hour )

df$hours <- hours

df
#               timestamp value hours
#   1 2012-04-09 05:03:00     2     5
#   2 2012-04-09 05:04:00     4     5
#   3 2012-04-09 05:05:00     5     5
#   4 2012-04-09 05:06:00     0     5
#   5 2012-04-09 05:07:00     0     5
#   6 2012-04-09 05:08:00     3     5
#   7 2012-04-09 05:09:00     0     5
#   8 2012-04-09 05:10:00     1     5

来源：https://stackoverflow.com/questions/16059965/plot-value-over-hour-of-day-with-xts-zoo-r

标签

time-series

xts

zoo