PA1_template.md

Reproducible Research: Peer Assessment 1

Loading and preprocessing the data

I started by unzipping the data and reading the csv file:

d <- read.csv('activity.csv')

Checking the summary and the 'head' of the data, everything seems fine:

summary(d)

##      steps                date          interval     
##  Min.   :  0.00   2012-10-01:  288   Min.   :   0.0  
##  1st Qu.:  0.00   2012-10-02:  288   1st Qu.: 588.8  
##  Median :  0.00   2012-10-03:  288   Median :1177.5  
##  Mean   : 37.38   2012-10-04:  288   Mean   :1177.5  
##  3rd Qu.: 12.00   2012-10-05:  288   3rd Qu.:1766.2  
##  Max.   :806.00   2012-10-06:  288   Max.   :2355.0  
##  NA's   :2304     (Other)   :15840

head(d)

##   steps       date interval
## 1    NA 2012-10-01        0
## 2    NA 2012-10-01        5
## 3    NA 2012-10-01       10
## 4    NA 2012-10-01       15
## 5    NA 2012-10-01       20
## 6    NA 2012-10-01       25

One could convert the date variable (which is numeric) to a Date variable, but it does not really matter.

What is mean total number of steps taken per day?

I start by summarizing the data frame by day with the ddply function.

library(plyr)
d_day <- ddply(d, .(date), summarize, total_steps = sum(steps))

Then, I plot the histogram:

hist(d_day$total_steps, main='Histogram of number of steps per day', xlab='steps per day')

Mean and median are (ignoring NA values):

mean(d_day$total_steps, na.rm=T)

## [1] 10766.19

median(d_day$total_steps, na.rm=T)

## [1] 10765

What is the average daily activity pattern?

I again apply the ddply function for averaging over all days for each interval, and then plot the result.

d_avgday <- ddply(d, .(interval), summarize, avg_steps = mean(steps, na.rm=T))
plot(d_avgday$interval, d_avgday$avg_steps, type='l', main='Average daily activity pattern', xlab='time of day', ylab='steps averaged over all days')

The maximum number of steps (on average over all days) occurs at: 835.

d_avgday$interval[which.max(d_avgday$avg_steps)]

## [1] 835

Imputing missing values

There are 2304 rows with missing values:

colSums(is.na(d))

##    steps     date interval 
##     2304        0        0

I will follow the strategy of using the average number of steps over all days for intervals with missing values. I am using a good-old for loop, there must be more elegant ways of dealing with this (like apply for vectors), but it works...

steps_filled <- numeric()
for(i in 1:nrow(d)) {
  s <- d[i,]$steps
  if(!is.na(s)) {
    steps_filled <- c(steps_filled, s)
  }
  else {
    steps_filled <- c(steps_filled, subset(d_avgday, interval==d[i,]$interval)$avg_steps)
  }
}

d_filled <- d
d_filled$steps <- steps_filled

I redo the ddply calculation like before, but now for the dataset with filled-in values.

d_day_filled <- ddply(d_filled, .(date), summarize, total_steps = sum(steps))

I again plot the histogram:

hist(d_day_filled$total_steps, main='Histogram of number of steps per day (imputed NA values)', xlab='steps per day')

Mean and median are:

mean(d_day_filled$total_steps, na.rm=T)

## [1] 10766.19

median(d_day_filled$total_steps, na.rm=T)

## [1] 10766.19

The differences in median and mean values are very small.

Are there differences in activity patterns between weekdays and weekends?

The plot indicates some differences in activity patterns between weekdays and weekends.Weekdays show a marked peak in activity between 8 and 9 AM, and less activity throughout the day. Weekends also show a morning peak, but it is less marked and builds up later (almost no activity until 8 AM) and more activity is visible throughout the day.

library(ggplot2)
d$weekday <- revalue(weekdays(as.Date(d$date, "%Y-%m-%d")), c('Saturday' = 'Weekend', 'Sunday' = 'Weekend', 'Monday' = 'Weekday', 'Tuesday' = 'Weekday', 'Wednesday' = 'Weekday', 'Thursday' = 'Weekday', 'Friday' = 'Weekday'))
d_avgday_2 <- ddply(d, .(interval, weekday), summarize, avg_steps = mean(steps, na.rm=T))
qplot(interval, avg_steps, data=d_avgday_2, geom="line", facets=weekday~.)