PA1_template.Rmd

Reproducible Research: Peer Assessment 1

Loading and preprocessing the data

Codes to load and preprocess the data:

directory <- ("activity")
file <- ("activity.csv")
activity <- read.csv(file.path(directory, file))
activity_na_omit <- na.omit(activity)
total_ac_na_omit <- aggregate(steps ~ date, activity_na_omit, sum)

What is mean total number of steps taken per day?

Code to make a histogram of the total number of steps taken each day:

hist(
        total_ac_na_omit$steps, 
        main = "Total of steps per day", 
        xlab="Steps per day", 
        ylab="Frequency", 
        col = "green", 
        breaks = 20
        )

Codes to calculate the mean and median total number of steps taken per day:

steps_mean <- mean(total_ac_na_omit$steps)
steps_median <- median(total_ac_na_omit$steps)
print(steps_mean)

## [1] 10766.19

print(steps_median)

## [1] 10765

What is the average daily activity pattern?

Code to preprocess the data:

total_ac_average <- aggregate(steps ~ interval, activity_na_omit, mean)

Code to create a time series plot:

library(ggplot2)
par(mar = c(1.5, 1.5, 1, 1))
ggplot(total_ac_average, aes(interval, steps)) +
        geom_line() +
        ggtitle("Time series plot of steps by interval") +
        xlab("Interval") +
        ylab("Number of steps")

Code to calculate the maximum number of steps:

max_number_steps <- which.max(total_ac_average$steps)
print(total_ac_average[max_number_steps,])

##     interval    steps
## 104      835 206.1698

Imputing missing values.

Code to calculate the total number of missing values in the dataset:

number_na <- sum(is.na(activity$steps))
print(number_na)

## [1] 2304

Code to fill in all of the missing values in the dataset. The chosen strategy is to use the mean itself to fill in all the NA:

fill_na <- function(steps, interval) {
        fill_ac <- NA
        if (!is.na(steps))
                fill_ac <- c(steps)
        else
                fill_ac <- (total_ac_average[total_ac_average$interval==interval, "steps"])
        return(fill_ac)
}

Codes to create a new dataset with the missing data filled in:

new_activity <- activity
new_activity$steps <- mapply(fill_na, new_activity$steps, new_activity$interval)

Codes to make a histogram of the total number of steps taken each day and to calculate the mean and median total number of steps taken per day:

total_ac_steps <- aggregate(steps ~ date, new_activity, sum)
hist(
        total_ac_steps$steps, 
        main = "Total number of steps per day", 
        xlab="Steps per day", 
        ylab="Frequency", 
        col = "orange", 
        breaks = 20
)

total_ac_steps_mean <- mean(total_ac_steps$steps) 
total_ac_steps_median <- median(total_ac_steps$steps) 
print(total_ac_steps_mean)

## [1] 10766.19

print(total_ac_steps_median)

## [1] 10766.19

Do these values differ from the estimates from the first part of the assignment?

They do differ, although slightly. The mean concerning the total number of steps taken per day - considering the missing values - is 10766.19, while the mean with all the NA filled in by the mean is also 10766.19. But the median is a bit different. While the median - considering the missing values - is 10765, the median with the NA filled in by the mean is 10766.19.

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

If we consider the mean, there was no impact at all, since I opted to use the mean itself to fill in all the NA. Although there was a difference at the medians, it was by 1.19. It seems to have no great impact.

Are there differences in activity patterns between weekdays and weekends?

Codes to create the new factor variable in the dataset with the levels “weekday” and “weekend”:

week_activity <- new_activity

week_activity$date <- as.Date(strptime(week_activity$date, format="%Y-%m-%d")) 

week_activity$day <-  factor(ifelse(as.POSIXlt(week_activity$date)$wday %in% c(0,6), 'weekend', 'weekday'))

Code to calculate the averaged number of steps taken by “weekday”" or “weekend”:

averaged_week_activity <- aggregate(steps ~ interval + day, data=week_activity, mean)

Code to plot a time series of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all weekday days or weekend days (y-axis):

library(ggplot2)
ggplot(averaged_week_activity, aes(interval, steps)) + 
        geom_line() + 
        facet_grid(day ~ .) +
        xlab("5-minute interval") + 
        ylab("avarage number of steps")