PA1_template.md

title	output
Project Markdown	html_document

Loading and preprocessing the data

if(!file.exists("getdata-projectfiles-UCI HAR Dataset.zip")) {
  Data <- tempfile()
    download.file("http://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip",Data)
    unzip(Data)
    unlink(Data)}
activity <- read.csv("activity.csv")

What is mean total number of steps taken per day

stepsday<-aggregate(steps~date,activity, sum)

hist(stepsday$steps, main="Steps per day", xlab="Steps", ylab = "Frecuency Day", col="grey")

Mean steps by day

mean_steps <- mean(stepsday$steps)

Median steps by day

median_steps <- median(stepsday$steps)

The mean is

mean(stepsday$steps)

The median is

median(stepsday$steps)

What is the average daily activity pattern?

steps_average<- aggregate(steps~interval, activity, mean)

plot(steps_average$interval,steps_average$steps, type ="l", col="blue", xlab = "Interval", ylab="Steps", main ="Average steps by interval")

Interval that contains the maximum number of steps On average across all the days in the dataset

max_steps_average <- steps_average[which.max(steps_average$steps),1]

#####The 5-minute interval, on average across all the days in the data set, containing the maximum number of steps is

steps_average[which.max(steps_average$steps),1]

Imputing missing values

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

number of missing values in the data

sum(!complete.cases(activity))

Devise a strategy for filling in all of the missing values in the dataset.

activity_imputed <- transform(activity,steps = ifelse(is.na(activity$steps), steps_average$steps[match(activity$interval,steps_average$interval)],activity$steps))

the strategy is to assign the average to the intervals for the steps that are equal to NA for the first date 2012-10-01

activity_imputed[as.character(activity_imputed$date) == "2012-10-01", 1] <- 0

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

stepsday2 <- aggregate(steps~date ,activity_imputed,sum)

Data vs imputed data through histrogram

stepsday_imputed <- aggregate(steps ~ date, activity_imputed, sum)
hist(stepsday_imputed$steps, main = "Total Steps by Day", col="blue", xlab="Steps")

#Histogram that show difference. 
hist(stepsday$steps, main = paste("Total Steps by Day"), col="grey", xlab="Steps", add=T)
legend("topright", c("Imputed", "Original Data"), col=c("blue", "grey"), lwd=10)

Values for imputed data

• Mean

mean_imput <- mean(stepsday_imputed$steps)
mean_imput

• Median

median_imput <- median(stepsday_imputed$steps)
median_imput

• Difference

The Diference in steps imputed vs steps original data is:

diff_steps <-( sum(stepsday_imputed$steps) - sum(stepsday$steps))
diff_steps

The Diference in mean imputed vs mean steps original data is:

mean_diff_steps <-( sum(mean_imput) - sum(mean_steps))
mean_diff_steps

The Diference in median imputed vs median steps original data is:

median_diff_steps <-( sum(median_imput) - sum(median_steps))
median_diff_steps

Differences in activity patterns between weekdays and weekends

#Define the name days (It's in spanish because I'm in Colombia)
weekdays <- c("lunes", "martes", "miercoles", "jueves","viernes")

activity_imputed$typeday=as.factor(ifelse(is.element(weekdays(as.Date(activity_imputed$date)),weekdays), "Weekday", "Weekend"))


steps_week_interval<- aggregate(steps~interval + typeday, activity_imputed, mean)

library(lattice)

xyplot(steps_week_interval$steps ~ steps_week_interval$interval|steps_week_interval$typeday, main="Average Steps per Day by Interval",xlab="Interval", ylab="Steps",layout=c(1,2), type="l")

head(steps_week_interval$type)