by Vasilis Daroglou
github repo with RMarkdown source code:
https://github.com/VDaroglou/RepData_PeerAssessment1

Prepare the Environment

First of all, you NEED make sure to SET the WORKING DIRECTORY to your LOCAL REPO FOLDER.
Throughout this report you can always find the code that I used to generate my output presents here. When writing code chunks in the R markdown document, always use echo = TRUE so that someone else will be able to read the code. This assignment will be evaluated via peer assessment so it’s essential that my peer evaluators be able to review my code and my analysis together.
I also change the locale to English in order to ensure the same output for peers with other system languages.

library(knitr)
## Warning: package 'knitr' was built under R version 3.1.3
opts_chunk$set(echo = TRUE)
Sys.setlocale("LC_TIME","English")

Loading and preprocessing the data

  1. The data is read as a data variable type data frame.
  2. steps column is converted to numeric type.(missing values are coded as NA)
  3. date column is converted to Date type.
  4. interval column is converted to numeric type.
unzip("activity.zip")
activity <- read.csv("activity.csv",
                     header=TRUE,
                     na.strings="NA",
                     colClasses=c("numeric", "Date", "numeric"))

Now, we can proceed with the data pre-examination of its str.

str(activity)
## 'data.frame':    17568 obs. of  3 variables:
##  $ steps   : num  NA NA NA NA NA NA NA NA NA NA ...
##  $ date    : Date, format: "2012-10-01" "2012-10-01" ...
##  $ interval: num  0 5 10 15 20 25 30 35 40 45 ...

What is mean total number of steps taken per day?

For this part of the assignment, we can ignore the missing values in the data set.

We initiate with a pre-calculation of steps aggregation by day:

steps_taken_per_day <- aggregate(steps ~ date, activity, sum)
colnames(steps_taken_per_day) <- c("date", "steps")

The following histogram displays the total number of steps taken per day, ignoring missing values.

hist(steps_taken_per_day$steps,
     breaks = 20,
     col = "blue",
     main = "Total Number of Steps Taken per Day",
     xlab = "Number of Steps per Day",
     ylab = "Count")

Finally, we calculate and report the mean and median of the total number of steps taken per day:

mean_steps = mean(steps_taken_per_day$steps, na.rm=TRUE)
median_steps = median(steps_taken_per_day$steps, na.rm=TRUE)

The mean is 10766.19 and the median is 10765.

What is the average daily activity pattern?

First, we calculate the aggregation of steps by intervals of 5-minutes and give names for columns of the result of this aggregation.

steps_per_interval <- aggregate(steps ~ interval, activity, mean)
colnames(steps_per_interval) <- c("interval", "steps")

Then, we present below a plot with the time series of the average number of steps taken (averaged across all days) versus the 5-minute intervals:

plot(steps_per_interval, type = "l", 
     col = "dark blue",
     main = "Average Daily Activity Pattern",
     xlab = "Interval",
     ylab = "Number of steps")

Then, we calculate the 5-minute interval with the maximum number of steps, on average across all the days in the dataset:

steps_per_interval$interval[which.max(steps_per_interval$steps)]
## [1] 835

Therefore, the 5-minute interval between 08:35 and 08:40 has the maximum number of steps, on average across all days.

Imputing missing values

Total number of missing values in the dataset

First we calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)

na_count = sum(is.na(activity))
na_count
## [1] 2304

So there are 2304 missing values in our dataset.

Strategy for filling in all of the missing values in the dataset

In this part of the assignment, we devise a strategy for filling in all of the missing values in the dataset. The strategy is not sophisticated.
We choose to use the mean for the 5-minute interval as fillers for missing values.

Creation of new dataset with the missing data filled in

So we start creating a new dataseta that is equal to the original dataset but with the missing data filled in with the mean for the 5-minute interval.

imputed_activity <- merge(activity,
                          steps_per_interval, 
                          by="interval", 
                          suffixes=c("",".interval.mean"))
nas <- is.na(imputed_activity$steps)
imputed_activity$steps[nas] <- imputed_activity$steps.interval.mean[nas]
imputed_activity <- imputed_activity[,1:3]

This is the str of the new dataset.

str(imputed_activity)
## 'data.frame':    17568 obs. of  3 variables:
##  $ interval: num  0 0 0 0 0 0 0 0 0 0 ...
##  $ steps   : num  1.72 0 0 0 0 ...
##  $ date    : Date, format: "2012-10-01" "2012-11-23" ...

And the count of its missing values. We should get a “0”.

sum(is.na(imputed_activity))
## [1] 0

Histogram of the total number of steps taken each day.(missing data filled in)

We initiate with a pre-calculation of steps aggregation by day:

full_steps_taken_per_day <- aggregate(steps ~ date, imputed_activity, sum)
colnames(full_steps_taken_per_day) <- c("date", "steps")

The following histogram displays the total number of steps taken per day, ignoring missing values.

hist(full_steps_taken_per_day$steps,
     breaks = 20,
     col = "blue",
     main = "Total Number of Steps Taken per Day
     (populated missing values)",
     xlab = "Number of Steps per Day",
     ylab = "Count")

Calculate and report the mean and median total number of steps taken per day

Then, we calculate and report the mean and median of the total number of steps taken per day:

mean_full_steps = mean(full_steps_taken_per_day$steps)
median_full_steps = median(full_steps_taken_per_day$steps)

The mean is 10766.19 and the median is 10766.19.

What is the impact of imputing missing data on the estimates of the total daily number of steps?

So now we wil compare these values with the values from the estimates from the first part of the assignment.

  • The mean before populating missing values is 10766.19.

  • The mean after populating missing values is 10766.19.

  • The median before populating missing values is 10765.

  • The median after populating missing values is 10766.19.

So we can observe, that while the mean value remains unchanged, the median value has shifted and matches the mean.
It seems that the impact of imputing missing values has increase our median, but has not affected negatively our predictions.

Are there differences in activity patterns between weekdays and weekends?

For this part of the assignment we are going to Use the dataset with the filled-in missing values.
At first, we will reate a new factor variable in the dataset with two levels – “weekday” and “weekend” indicating whether a given date is a weekday or weekend day. We will define the “daytype”" function for this purpose.

daytype <- function(date) {
    if (weekdays(as.Date(date)) %in% c("Saturday", "Sunday")) {
        "weekend"
    } else {
        "weekday"
    }
}
imputed_activity$daytype <- as.factor(sapply(imputed_activity$date, daytype))

Then, we make a panel plot containing a time series plot (i.e. type = “l”) of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all weekday days or weekend days (y-axis).

daytype_interval_steps <- aggregate(data = imputed_activity,
                                  steps ~ daytype + interval,
                                  FUN = mean)

library(lattice)

xyplot(steps ~ interval | daytype,
       data = daytype_interval_steps,
       xlab = "Interval",
       ylab = "Number of steps",
       col = "dark blue",
       type = "l",
       layout =c(1, 2))

We can observe from the graph above that activities on the weekdays have the greatest peak from all steps intervals. But, we can also observe that weekends steps have more peaks over hundred steps than weekdays. We can relate this to the fact that in weekends a person has more free time to do more intesive activities along the day than in weekdays, probably because of work.