Reproducible Research: Peer Assessment 1
Marcelo Gomes Marques
Saturday, July 19, 2014
Created with knitr in R x64 3.1.0 Patched on Windows 7 64 bits SO
github repo with RMarkdown source code: https://github.com/mgmarques/RepData_PeerAssessment1
Executive Summary
This document presents the results of the Reproducible Research’s Peer Assessment 1 in a report using a single R markdown document that can be processed by knitr and be transformed into an HTML file.
Through this report you can see that activities on weekdays mostly follow a work related routine, where we find some more intensity activity in little a free time that the employ can made some sport.
An important consideration is the fact of our data presents as a t-student distribution (see both histograms), it means that the impact of imputing missing values with the mean has a good impact on our predictions without a significant distortion in the distribution of the data.
Prepare the Environment
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..
First, we set echo equal a TRUE and results equal a ‘hold’ as global options for this document.
library(knitr)
opts_chunk$set(echo = TRUE, results = 'hold')Load data.table, xtable and ggplot2 libraries
library(data.table)
library(xtable)
library(ggplot2) # we shall use ggplot2 for plotting figuresGetting and Cleaning Data
Loading and preprocessing the data
This assignment 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.
This assignment instructions request to show any code that is needed to loading and preprocessing the data, like to:
- Load the data (i.e. > read.csv())
- Process/transform the data (if necessary) into a format suitable for your analysis
In summary, I proceed with:
- Initiate with functions definitions and calls libraries
- Check if the file source is on expected, IE, the uncompressed file of activities is on the directory of this assessment
2.1. Check if the zip source file is present
2.1.1. uncompressed the file
2.1.2. If not the data file is downloaded and uncompressed - The data is read as a data variable type data frame.
- The interval column is converted to factor type.
- The date column is converted to Date type.
- The data is examined by using summary and str well formatted on this document.
Function definitions and load libraries code segment
First we define a function to check if the files exists in the path defined by file_path. If don’t exists stops execution of the current expression, and executes an error action
check_file_exist <- function(file_path)
{
if (!file.exists(file_path))
stop("The ", file_path, " not found!") else TRUE
}Next, we use data set and data_dir to define the file_path, call the check_file_exist function to check, if the file exist send a message to user for waiting the load of file. Finally returned with data set load to data variable.
load_data <- function(data_dir , fileURL, fileSource)
{
# Dataset check and load
source_path <- paste(data_dir, "/", fileSource , sep="")
txt_file <- paste(data_dir, "/","activity.csv", sep="")
if (!file.exists(txt_file)) {
if (!file.exists(source_path)) {
message(paste("Please Wait! Load", fileURL, "..."));
download.file(fileURL, destfile=source_path);
}
else {
message(paste("Please Wait! Unzip", source_path, " file..."));
unzip(source_path, exdir = data_dir);
}
}
message(paste("Please Wait! Load", txt_file, " to dataset..."));
data <- read.csv(txt_file,
header=TRUE, na.strings="NA",
colClasses=c("numeric", "character", "numeric"))
data$interval <- factor(data$interval)
data$date <- as.Date(data$date, format="%Y-%m-%d")
data
}Assign the directory that all data set was unzip and confirm its exists
Maybe you need to change this data_dir variable to yours peeress directory, because the line code that ask you inform where the data directory is find, use readline function, not function at markdown documents.
data_dir <- "C:/Users/Marcelo/RepData_PeerAssessment1";Check if the “./RepData_PeerAssessment1” directory exists, if doesn’t ask to user the path of his data directory. If user inform a invalid directory path stops execution of the current expression and executes an error action.
if (!file.exists(data_dir)){
# data_dir <- readline(prompt = "Please, inform your data directory path: ")
data_dir <-"./RepData_PeerAssessment1" ## simulate a valid data entry
if (!file.exists(data_dir)){
stop("You inform a invalid directory path")
}
}Here its rely the point at the all data load and preparation is call and run to assign the tidy data a data frame variable named tidy.
Here, we initiate the main variables and run data load and preparation process for the activate data.
fileURL <- "https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
fileSource <-"activity.zip"
source_path <- paste(data_dir, "/", fileSource , sep="")
txt_file <- paste(data_dir, "/","activity.csv", sep="")
if (!file.exists(txt_file)) {
if (!file.exists(source_path)) {
message(paste("Please Wait! Load", fileURL, "..."));
download.file(fileURL, destfile=source_path);
}
else {
message(paste("Please Wait! Unzip", source_path, " file..."));
unzip(source_path, exdir = data_dir);
}
}
message(paste("Please Wait! Load", txt_file, " to dataset..."));## Please Wait! Load C:/Users/Marcelo/RepData_PeerAssessment1/activity.csv to dataset... tidy <- read.csv(txt_file,
header=TRUE, sep=",",
colClasses=c("numeric", "character", "numeric"))
tidy$interval <- factor(tidy$interval)
tidy$date <- as.Date(tidy$date, format="%Y-%m-%d")Now, we can proceed with the data pre-examination of its str and …
str(tidy)## '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: Factor w/ 288 levels "0","5","10","15",..: 1 2 3 4 5 6 7 8 9 10 ...…summary methods.
xt <- xtable(summary(tidy))
print(xt, type = "html")| steps | date | interval | |
|---|---|---|---|
| 1 | Min. : 0.0 | Min. :2012-10-01 | 0 : 61 |
| 2 | 1st Qu.: 0.0 | 1st Qu.:2012-10-16 | 5 : 61 |
| 3 | Median : 0.0 | Median :2012-10-31 | 10 : 61 |
| 4 | Mean : 37.4 | Mean :2012-10-31 | 15 : 61 |
| 5 | 3rd Qu.: 12.0 | 3rd Qu.:2012-11-15 | 20 : 61 |
| 6 | Max. :806.0 | Max. :2012-11-30 | 25 : 61 |
| 7 | NA’s :2304 | (Other):17202 |
Questions from this Peer Assessment:
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, tidy, sum)
colnames(steps_taken_per_day) <- c("date", "steps")- At following, we present a histogram of the total number of steps taken each day, plotted with a bin interval of 1000 steps per Day.
ggplot(steps_taken_per_day, aes(x = steps)) +
geom_histogram(fill = "darkblue", binwidth = 1000) +
labs(title="Histogram of Steps Taken per Day",
x = "Number of Steps per Day", y = "Number of times (Count)") +
theme_bw() 
- Finally, we presents the mean and median of total number of steps taken per day:
mean_steps = round(mean(steps_taken_per_day$steps, na.rm=TRUE), 2)
median_steps = round(median(steps_taken_per_day$steps, na.rm=TRUE), 2)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, coerce the interval to integer and give names for colons of the result of this aggregation.
steps_per_interval <- aggregate(tidy$steps,
by = list(interval = tidy$interval),
FUN=mean, na.rm=TRUE)
# convert to integers for plotting
steps_per_interval$interval <-
as.integer(levels(steps_per_interval$interval)[steps_per_interval$interval])
colnames(steps_per_interval) <- c("interval", "steps")Then, we present below a plot with the time series of the average number of steps taken (averaged across all days) versus the 5-minute intervals:
ggplot(steps_per_interval, aes(x=interval, y=steps)) +
geom_line(color="darkblue", size=1) +
labs(title="Average Daily Activity Pattern", x="Interval", y="Number of steps") +
theme_bw() + theme(legend.position = "bottom")
Now, we find the 5-minute interval with the containing the maximum number of steps:
max_step_interval <- steps_per_interval[which.max(
steps_per_interval$steps),]$intervalThe 835th 5-minute interval containing the maximum number of steps.
Imputing missing values:
Total number of missing values in the dataset:
We already been saw the total number of NA through summary of tidy data. We proceed with the new instruction below for show how can you get only this result.
sum(is.na(tidy$steps))## [1] 2304Strategy for filling in all of the missing values and create a new dataset:
To populate missing values, we choose to replace them with the mean value at the same interval across days. In most of the cases the median is a better centrality measure than mean, but in our case the total median is not much far away from total mean, and probably we can make the mean and median meets.
fill_na <- function(data, defaults) {
na_indices <- which(is.na(data$steps))
na_replacements <- unlist(lapply(na_indices, FUN=function(idx){
interval = data[idx,]$interval
defaults[defaults$interval == interval,]$steps
}))
fill_steps <- data$steps
fill_steps[na_indices] <- na_replacements
fill_steps
}
data_fill <- data.frame(
steps = fill_na(tidy, steps_per_interval),
date = tidy$date,
interval = tidy$interval)A histogram of the total number of steps taken each day
Here is a histogram of the daily total number of steps taken, plotted with a bin interval of 1000 steps, after the populate missing values.
full_steps_per_day <- aggregate(steps ~ date, data_fill, sum)
colnames(full_steps_per_day) <- c("date", "steps")
ggplot(full_steps_per_day, aes(x=steps)) +
geom_histogram(fill="darkblue", binwidth=1000) +
labs(title="Histogram of Full Steps Taken per Day",
x="Number of Steps after populate missing values",
y="Count") +
theme_bw() 
Calculate and report the mean and median total number of steps taken per day
full_mean_steps = round(mean(full_steps_per_day$steps), 2)
full_median_steps = round(median(full_steps_per_day$steps), 2)- Mean after populate missing values is 10766.19
Median populate missing values is 10766.19
- Mean before populate missing values is 10766.19
Median before missing values is 10765
What is the impact of imputing missing data on the estimates of the total daily number of steps?
As you can see, comparing with the calculations done in the first section of this document, we observe that while the mean value remains unchanged, the median value has shifted and virtual matches to the mean.
Since our data has shown a t-student distribution (see both histograms), it seems that the impact of imputing missing values has increase our peak, but it’s not affect negatively our predictions.
Are there differences in activity patterns between weekdays and weekends?
We do this comparison with the table with filled-in missing values.
1. Augment the table with a column that indicates the day of the week
2. Subset the table into two parts - weekends (Saturday and Sunday) and weekdays (Monday through Friday).
3. Tabulate the average steps per interval for each data set.
4. Plot the two data sets side by side for comparison.
weekdays_steps <- function(data) {
weekdays_steps <- aggregate(data$steps, by=list(interval = data$interval),
FUN=mean, na.rm=T)
# convert to integers for plotting
weekdays_steps$interval <-
as.integer(levels(weekdays_steps$interval)[weekdays_steps$interval])
colnames(weekdays_steps) <- c("interval", "steps")
weekdays_steps
}
data_by_weekdays <- function(data) {
data$weekday <-
as.factor(weekdays(data$date)) # weekdays in portuguese
weekend_data <- subset(data, weekday %in% c("sábado","domingo"))
weekday_data <- subset(data, !weekday %in% c("sábado","domingo"))
weekend_steps <- weekdays_steps(weekend_data)
weekday_steps <- weekdays_steps(weekday_data)
weekend_steps$dayofweek <- rep("weekend", nrow(weekend_steps))
weekday_steps$dayofweek <- rep("weekday", nrow(weekday_steps))
data_by_weekdays <- rbind(weekend_steps, weekday_steps)
data_by_weekdays$dayofweek <- as.factor(data_by_weekdays$dayofweek)
data_by_weekdays
}
data_weekdays <- data_by_weekdays(data_fill)Below you can see the panel plot comparing the average number of steps taken per 5-minute interval across weekdays and weekends:
ggplot(data_weekdays, aes(x=interval, y=steps)) +
geom_line(color="steelblue", size=1) +
facet_wrap(~ dayofweek, nrow=2, ncol=1) +
labs(x="Interval", y="Number of steps") +
theme_bw()
Conclusion:
We can see at the graph above that activity on the weekday has the greatest peak from all steps intervals. But, we can see too that weekends activities has more peaks over a hundred than weekday. This could be due to the fact that activities on weekdays mostly follow a work related routine, where we find some more intensity activity in little a free time that the employ can made some sport. In the other hand, at weekend we can see better distribution of effort along the time.