Reproducible Research: Peer Assessment 1

Introduction

This is an R Markdown document for the 1st Assignment in Reproducible Research course at Coursera. On this assignment I will create a single R markdown document that can be processed by knitr. To answer the different requirements, it will follow the same structure than the assignment instructions.

The only exception is that I did not Fork/Clone the GitHub repository created for this assignment, but I started from scratch. However the repository contains

1. Data
2. Readme
3. R markdown file (this)
4. figures folder with the figures shown on this report

Data

The data for this assignment can be downloaded from the course web site. But in fact, it has been also included in the repository.

• Dataset: Activity monitoring data [52K]

The variables included in this dataset are:

• steps: Number of steps taking in a 5-minute interval (missing values are coded as NA)

• date: The date on which the measurement was taken in YYYY-MM-DD format

• interval: Identifier for the 5-minute interval in which measurement was taken

Assignment

Loading and preprocessing the data

Show any code that is needed to

1. Load the data (i.e. read.csv())

activity<-read.csv("activity.csv", as.is = TRUE, dec=".", na.strings = "NA")

What is mean total number of steps taken per day?

For this part of the assignment, the missing values in the dataset are ignored.

1. Make a histogram of the total number of steps taken each day

hist_sum<-tapply(activity$steps, activity$date, sum)

hist(hist_sum, main="Histogram", xlab="Number of daily steps")

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

Mean

mean(hist_sum, na.rm=TRUE)

## [1] 10766.19

Median

median(hist_sum, na.rm=TRUE)

## [1] 10765

hist(hist_sum, main="Histogram", xlab="Number of daily steps")
abline(v=mean(hist_sum, na.rm=TRUE), col="red", lwd=2)
legend("topright", legend = c("mean"), col=c("red"), lty=c(1))

What is the average daily activity pattern?

1. Make 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 days (y-axis)

library(dplyr)

a_by_interval<-activity %>% 
               group_by(interval) %>% 
               summarise(avg_steps=mean(steps, na.rm=T))

with(a_by_interval, plot(avg_steps ~ interval, type="l"))

2. Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?

The average maximum number of steps is:

max(a_by_interval$avg_steps)

## [1] 206.1698

And it is achieved at interval:

pos_max_interval<-which.max(a_by_interval$avg_steps)
a_by_interval[pos_max_interval,"interval"][[1]]

## [1] 835

Imputing missing values

Note that there are a number of days/intervals where there are missing values (coded as NA). The presence of missing days may introduce bias into some calculations or summaries of the data.

1. Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)

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

## [1] 2304

2. Devise a strategy for filling in all of the missing values in the dataset. The strategy does not need to be sophisticated.

I decided to assign the mean for that 5-minute interval to each NA value.

3. Create a new dataset that is equal to the original dataset but with the missing data filled in.

new_activity<-merge(activity,a_by_interval, by="interval")
new_activity<-mutate(new_activity,
                     steps=coalesce(steps,as.integer(round(avg_steps))))

4. Make a histogram of the total number of steps taken each day and Calculate and report the mean and median total number of steps taken per day. Do these values differ from the estimates from the first part of the assignment? What is the impact of imputing missing data on the estimates of the total daily number of steps?

New Mean and Median are very similar to the previous one

new_hist_sum<-tapply(new_activity$steps, new_activity$date, sum)

mean(new_hist_sum, na.rm=TRUE)

## [1] 10765.64

median(new_hist_sum, na.rm=TRUE)

## [1] 10762

And the main impact on the Histogram is that the average interval increases. As it can be seen compared to the previous histogram.

par(mfrow=c(1,2))
hist(new_hist_sum, main="New Histogram", xlab="Number of daily steps")
abline(v=mean(new_hist_sum, na.rm=TRUE), col="red", lwd=2)
hist(hist_sum, ylim=c(0,35), main="Histogram", xlab="Number of daily steps")
abline(v=mean(hist_sum, na.rm=TRUE), col="red", lwd=2)

Are there differences in activity patterns between weekdays and weekends?

1. Create a new factor variable in the dataset with two levels – “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.

new_activity<-mutate(new_activity, weekday=weekdays(as.Date(new_activity$date)))
new_activity<-mutate(new_activity, weekend= if_else(weekday=="sábado" | weekday=="domingo", TRUE, FALSE))

1. 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). The plot should look something like the following, which was created using simulated data:

First of all we need to get the information on the average of steps by interval accross all days.

activity_weekend<-filter(new_activity, weekend==TRUE)
activity_weekday<-filter(new_activity, weekend==FALSE)

a_by_interval_weekday<-activity_weekday %>% 
        group_by(interval) %>% 
        summarise(avg_steps=mean(steps, na.rm=T))
a_by_interval_weekend<-activity_weekend %>% 
        group_by(interval) %>% 
        summarise(avg_steps=mean(steps, na.rm=T))

And then plot it to be able to compare the behaviour weekend vs weekday

par(mfrow=c(2,1))
with(a_by_interval_weekday, plot(avg_steps ~ interval, type="l"))
with(a_by_interval_weekend, plot(avg_steps ~ interval, type="l", ylim=c(0,250)))