PA1_template.md

title	Reproducible Research: Peer Assessment 1
output	html_document keep_md true

Loading and preprocessing the data

Load the data.

data <- read.csv( "activity.csv" )

Clean the data by removing the rows with NA for steps.

cldata <- data[!is.na(data["steps"]),]

For easier typing later, assign each column to a variable.

steps     <- cldata[["steps"]]
d         <- cldata[["date"]]
interval  <- cldata[["interval"]]

What is mean total number of steps taken per day?

stepsPerDay <- tapply( steps , d , sum )

hist( 
  stepsPerDay , 
  main="Histogram of Total Steps per Day" ,
  xlab="Steps per Day"
)

mnStepsPerDay <- mean( stepsPerDay[ !is.na(stepsPerDay) ] )
mdStepsPerDay <- median( stepsPerDay[ !is.na(stepsPerDay) ] )

The mean of Total Steps per Day is 1.0766189 × 10⁴.
The median of Total Steps per Day is 10765.

What is the average daily activity pattern?

stepsPerInt <- tapply( steps , interval , mean )
plot( 
  interval[ 1:length(stepsPerInt) ] ,
  stepsPerInt ,
  type="l" ,
  main="Ave Steps per Interval" ,
  xlab="Interval" ,
  ylab="Number of steps"
)

maxInt <- names( which.max( stepsPerInt ) )

The 5-minute interval which on average contains the maximum number of steps is 835.

Imputing missing values

stepsNA <- is.na( data["steps"] )
totalNA <- sum( stepsNA )

The total number of missing values in the data set is 2304.

Missing values will be imputed by using the mean of the 5-minute interval.

imdata <- data
imdata["steps"][stepsNA] <- stepsPerInt[as.character(imdata["interval"][stepsNA])]

Create vars for each column with imputed data.

steps_     <- imdata[["steps"]]
d_         <- imdata[["date"]]
interval_  <- imdata[["interval"]]

Analyze the distribution which contains the imputed values.

stepsPerDay_ <- tapply( steps_ , d_ , sum )

hist( 
  stepsPerDay_ , 
  main="Histogram of Total Steps per Day\nUsing Imputed Values" ,
  xlab="Steps per Day"
)

mnStepsPerDay_ <- mean( stepsPerDay_ )
mdStepsPerDay_ <- median( stepsPerDay_ )

The mean of Total Steps per Day is 1.0766189 × 10⁴.
The median of Total Steps per Day is 1.0766189 × 10⁴.

Comparing the distribution containing data imputed by using the mean of the 5-minute interval with the distribution in which missing values were removed, the histograms appear to be the same. The means are the same, but the medians are slightly different.

Are there differences in activity patterns between weekdays and weekends?

Calculate the average steps per interval on weekends and weekdays.

weekday <- as.factor( 
  sapply(
    weekdays( strptime( imdata[,2] , "%Y-%m-%d" ) , abbreviate=T ) ,
    function(d) if(d=="Sat"|d=="Sun"){"weekend"} else{"weekday"}
  )
)
imdata <- data.frame( imdata , weekday )

stepsPerInt_wknd <- tapply( 
  imdata[,1][weekday=="weekend"] ,  # steps
  imdata[,3][weekday=="weekend"] ,  # interval
  mean
)
stepsPerInt_wkdy <- tapply( 
  imdata[,1][weekday=="weekday"] ,  # steps
  imdata[,3][weekday=="weekday"] ,  # interval
  mean
)

Plot the average steps per interval on weekends and weekdays.

stepsW    <- c( stepsPerInt_wknd , stepsPerInt_wkdy )
intervalW <- c( imdata[,3][1:(2*length(stepsPerInt_wknd))] )
weekdayW  <- c(
    rep( "weekend" , length(stepsPerInt_wknd) ) ,
    rep( "weekday" , length(stepsPerInt_wkdy) )
)

library(lattice)
xyplot(
  stepsW ~ intervalW | weekdayW ,
  type = "l" ,
  layout = c(1,2) ,
  main = "Ave Steps per Interval" ,
  xlab = "Interval" ,
  ylab = "Number of steps"
)