Overview

This project makes use of data from a personal activity monitoring device. This device collects data at 5 minute intervals through out the day. The data consists of two months of data from an anonymous individual collected during the months of October and November, 2012 and include the number of steps taken in 5 minute intervals each day.

The data for this assignment can be downloaded from https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip.

The variables included in this dataset are:

There are a total of 17,568 observations in this dataset.

We need to load the following libraries:

library(dplyr)
library(chron)
library(lattice)
library(ggplot2)

Loading and preprocessing the data

In the following script, we load the data using by the read.csv and the unzip functions, and save it into a variable called data. The zip file (“activity.zip”) is already in the repo.

data <- read.csv(unzip("activity.zip", "activity.csv"))
str(data)
## 'data.frame':    17568 obs. of  3 variables:
##  $ steps   : int  NA NA NA NA NA NA NA NA NA NA ...
##  $ date    : Factor w/ 61 levels "2012-10-01","2012-10-02",..: 1 1 1 1 1 1 1 1 1 1 ...
##  $ interval: int  0 5 10 15 20 25 30 35 40 45 ...

As we see, the date variable is a factor. In the follwoing script, we transform it to a Date variable.

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

Before going to the next step, let us see how many of the observations got missing values for each variables.

stepsNA <- mean(is.na(data$steps)) *100
dateNA <- mean(is.na(data$date)) *100
intervalNA <- mean(is.na(data$interval)) *100

So, the proportion of NA values w.r.t the whole data in steps, date, and interval are 13%, 0%, and 0%, respectively.

What is mean total number of steps taken per day?

We get “the total number of steps taken per day” by the following script.

stepsday <- data %>% group_by(date) %>% summarize(totalSteps = sum(steps, na.rm = TRUE))
stepsday

Here is the “histogram of the total number of steps taken each day”:

hist(stepsday$totalSteps, col = "green", breaks = 7, main = "Histogram of number of steps taken each day", xlab = "Total Steps")
rug(stepsday$totalSteps) 
abline(v = median(stepsday$totalSteps, na.rm = T), col = "magenta", lwd = 4)
abline(v = mean(stepsday$totalSteps, na.rm = T), lwd = 4)

In the following script, we calculate the “mean and median of the total number of steps taken per day”.

stepsmean <- mean(stepsday$totalSteps)
stepsmedian <- median(stepsday$totalSteps)

The mean and median of total number of steps taken per day are 9354.2295082 and 10395, respecitvely. The megenta and black vertical lines in the above histogram show them, respectively.

What is the average daily activity pattern?

The following script calculates the average number of steps, averaged across all days per interval. The result is shown in a plot.

avgstpinv <- data %>% group_by(interval) %>% summarize(avgerage_steps = mean(steps, na.rm = TRUE))
plot(avgstpinv, type = "l", main = "Average Steps per Interval")

The following script gets the interval, which contains the maximum number of steps, averaged across all the days.

maxinvstp <- avgstpinv$interval[which(avgstpinv$avgerage_steps == max(avgstpinv$avgerage_steps))]

So, the interval, which contains the maximum number of average steps is 835.

Imputing missing values

The total number of NA rows in data is calculated as follows:

nanum <- sum(is.na(data))
nanum
## [1] 2304

The following script replaces the NA values in steps with the mean of all steps taken in the corresponding interval. We take advantage of the data frame created in the former stages.

for(i in 1:dim(data)[1]){
        if(is.na(data$steps)[i]){
                data$steps[i] <- avgstpinv[avgstpinv$interval == data$interval[i],]$avgerage_steps
        }
}

As we see now, in the following script, there is no missing values in our data anymore.

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

Now, here is the updated “histogram of the total number of steps taken each day”. We first get the total number of steps taken per day and store the data in a data frame named stepsday2.

stepsday2 <- data %>% group_by(date) %>% summarize(totalSteps = sum(steps, na.rm = T))

hist(stepsday2$totalSteps, col = "green", main = "Histogram of number of steps taken each day", xlab = "Total Steps")
rug(stepsday2$totalSteps) 
abline(v = median(stepsday2$totalSteps), col = "magenta", lwd = 2, lty = 2)

#abline(v = mean(stepsday2$totalSteps), lwd = 2)
meann <- mean(stepsday2$totalSteps)
mediann <-median(stepsday2$totalSteps)

As we see in the plot, the median and mean of the total steps per day are now eqaul: mean == 1.076618910^{4} == median.

Are there differences in activity patterns between weekdays and weekends?

In the following script, we create a factor variable, named wday in data with two levels – “weekday” and “weekend”– indicating whether a given date is a weekday or weekend day.

data$wdays <- weekdays(data$date)
for(i in 1:dim(data)[1]){
        if(data$wdays[i] == "Saturday" | data$wdays[i] == "Sunday") {
                data$wdays[i] <- "weekend"
        }
        else{
                data$wdays[i] <- "weekday"
        }
}
data$wdays <- as.factor(data$wdays)

Now, in the following script, we make a panel plot containing a time series plot of the interval (x-axis) and the average number of steps taken, averaged across all weekday days or weekend days (y-axis).

avgstp <- data %>% group_by(wdays, interval) %>% summarize(avg_steps = mean(steps, na.rm = TRUE))
xyplot(avg_steps ~ interval | wdays, data = avgstp,  type = "l", ylab = "Number of steps", main = "Average number of Steps per interval")

  1. a.a.safilian@gmail.com↩