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Reproducible Research: Peer Assessment 1

We will use the dplyr package to simplify data operations, lubridate for date operations, and ggplot2 for graphics. If the dplyr, lubridate or ggplot2 package is not installed, please install it using install.packages('dplyr', 'lubridate', 'ggplot2')

  library(dplyr)
  library(lubridate)
  library(ggplot2)

Loading and preprocessing the data

Load the data from ./activity.zip:activity.csv into activity.data. Convert the date from a character string to a date object and add a column translating interval to time of day. Then examine the basic structure to ensure it loaded correctly.

  stopifnot(file.exists('./activity.zip'))
  
  activity.data <- unz('./activity.zip', 'activity.csv') %>%
    read.csv(colClasses = c('integer', 'character', 'integer')) %>%
    transmute(
      date = ymd(date),
      time = sprintf('%02i:%02i', interval %/% 100, interval %% 100 %% 60),
      steps)
  
  str(activity.data)
## 'data.frame':	17568 obs. of  3 variables:
##  $ date : POSIXct, format: "2012-10-01" "2012-10-01" ...
##  $ time : chr  "00:00" "00:05" "00:10" "00:15" ...
##  $ steps: int  NA NA NA NA NA NA NA NA NA NA ...

What is mean total number of steps taken per day?

Group the data by date, sum the number of steps and find the average.

  by.day <- activity.data %>%
    group_by(date) %>%
    summarize(steps = sum(steps, na.rm=TRUE))
  mean.daily.steps <- mean(by.day$steps)
  median.daily.steps <- median(by.day$steps)
  
  qplot(by.day$steps, xlab='steps', geom=c('histogram', 'rug'),
        main='Total Number of Steps Taken per Day')
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.


  print(c(mean = mean.daily.steps, median = median.daily.steps))
##     mean   median 
##  9354.23 10395.00

The mean number of steps per day is 9354.23 and the median number of steps per day is 10395.

What is the average daily activity pattern?

Find the mean of each 5 minute interval over the sample of days and graph the number of steps taken each interval on an averaged day.

  by.interval <- activity.data %>%
    group_by(time) %>%
    summarize(steps = mean(steps, na.rm = TRUE))
  
  max.step.interval <- by.interval %>% arrange(desc(steps)) %>% slice(1)
  x.breaks <- ceiling(seq(1, nrow(by.interval), length.out=7))
  x.ticks <- scale_x_continuous(breaks=x.breaks, labels=by.interval$time[x.breaks])
  
  qplot(seq_along(steps), steps, data=by.interval, geom=c('line'),
        main="Average Daily Activity Pattern", xlab='time of day') + 
    x.ticks

  print.AsIs(max.step.interval)
##    time    steps
## 1 08:35 206.1698

The interval with the maximum number of steps on average is the five minute period begining 08:35 in which an average of 206.17 steps are taken.

Imputing missing values

  sum(is.na(activity.data$steps))
## [1] 2304

There are 2304 missing step values in the activity data set.

Using the values calculated for each interval, impute the missing values.

  incomplete.cases <- activity.data %>%
    filter(is.na(activity.data$steps)) %>%
    left_join(by.interval, by='time') %>%
    transmute(date, time, steps = steps.y)
  imputed.data <- activity.data %>%
    filter(!is.na(activity.data$steps)) %>%
    bind_rows(incomplete.cases) %>%
    arrange(date, time)
  imputed.by.day <- imputed.data %>%
    group_by(date) %>%
    summarise(steps = sum(steps))
  imputed.mean.daily.steps <- mean(imputed.by.day$steps)
  imputed.median.daily.steps <- median(imputed.by.day$steps)
  
  qplot(imputed.by.day$steps, xlab='steps', geom=c('histogram', 'rug'),
        main='Total Number of Steps Taken per Day (Imputed Values)')
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.


  print(c(
    mean = imputed.mean.daily.steps,
    median = imputed.median.daily.steps))
##     mean   median 
## 10766.19 10766.19

With the imputed data, the mean is now 10766.19 steps and the median is now 10766.19 steps. The imputed mean is 1411.96 steps more than the collected data. The imputed median is 371.19 steps more than the collected data. There are fewer 0 values and more values near mean.

Are there differences in activity patterns between weekdays and weekends?

  day.types <- as.factor(c('weekend', 'weekday'))
  by.day.and.interval = activity.data %>%
    mutate(day.type = sapply(date, function(d)
      if (wday(d) %in% c(1, 7))
        day.types[[1]]
      else
        day.types[[2]])
      ) %>%
    group_by(time, day.type) %>%
    summarize(steps = sum(steps, na.rm = TRUE)) %>%
    ungroup
  
  qplot(as.integer(factor(time)), steps, data=by.day.and.interval,
        geom=c('line'), facets=day.type~., xlab='time',
        main='Differences in Activity Patterns') +
    x.ticks + theme(legend.position='none')