PA1_template.md

title	Reproducible Research: Peer Assessment 1
output	html_document keep_md true

Loading and preprocessing the data

library(stringi)
library(lubridate)
library(ggplot2)
if (!file.exists("activity.csv")) unzip("activity.zip")
activity <- read.csv("activity.csv", colClasses = c("integer", "Date", "character"))
activity$time <- parse_date_time(stri_pad_left(activity$interval, 4, 0), "hm")
str(activity)

## 'data.frame':	17568 obs. of  4 variables:
##  $ steps   : int  NA NA NA NA NA NA NA NA NA NA ...
##  $ date    : Date, format: "2012-10-01" "2012-10-01" ...
##  $ interval: chr  "0" "5" "10" "15" ...
##  $ time    : POSIXct, format: "0000-01-01 00:00:00" "0000-01-01 00:05:00" ...

What is mean total number of steps taken per day?

total_steps_per_day <- aggregate(steps ~ date, activity, sum)
hist(total_steps_per_day$steps)

Calculating the mean and median

mean(total_steps_per_day$steps)

## [1] 10766.19

median(total_steps_per_day$steps)

## [1] 10765

What is the average daily activity pattern?

mean_steps_per_interval <- aggregate(steps ~ time, activity, mean)

Calculating which 5-minute interval has the maximum of the mean steps

max_steps <- mean_steps_per_interval[which.max(mean_steps_per_interval$steps),]
max_label <- paste(round(max_steps$steps, 1),
                   "steps maximum at", format(max_steps$time, "%H:%M"))
max_label

## [1] "206.2 steps maximum at 08:35"

Plotting the mean steps for each 5-minute interval

with(mean_steps_per_interval, {
  plot(time, steps, type = "l")
  text(max_steps, max_label, pos = 4)
})

Imputing missing values

Calculating the total number of rows with missing values

steps <- activity$steps
sum(is.na(steps))

## [1] 2304

Filling in missing values with the mean for the 5-minute interval

for (i in which(is.na(steps))) {
  steps[i] <- subset(mean_steps_per_interval, time == activity$time[i])$steps
}
filled_in_total_steps_per_day <- aggregate(steps ~ date,
                                           cbind(steps, subset(activity, select = 'date')), sum)
hist(filled_in_total_steps_per_day$steps)

Mean and median with the missing values filled in

mean(filled_in_total_steps_per_day$steps)

## [1] 10766.19

median(filled_in_total_steps_per_day$steps)

## [1] 10766.19

Using the means of the 5-minute intervals to fill in missing values did not affect the mean total steps per day, but it increased the median so that it became equal to the mean.

Are there differences in activity patterns between weekdays and weekends?

Assigning a factor to each observation identifying weekdays and weekends

weekpart <- rep("weekday", length(activity$date))
weekpart[which(weekdays(activity$date) %in% c("Saturday", "Sunday"))] <- "weekend"
activity$weekpart <- as.factor(weekpart)

Plotting the average daily activity for weekdays and weekends

qplot(time, steps, geom = "line", facets = weekpart ~ .,
      data = aggregate(steps ~ time + weekpart, activity, mean))

The weekday and weekend activity patterns appear similar, but more steps were taken in the morning on weekdays, whereas more steps were taken later in the day on weekends.