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‎PA1_template.Rmd

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---
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output: html_document
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output:
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md_document:
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variant: markdown_github
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---
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# Reproducible Research: Peer Assessment 1
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‎PA1_template.md

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Reproducible Research: Peer Assessment 1
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========================================
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### 1. Load the data from course website; (i.e. read.csv())
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activity<- read.csv("activity.csv")
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### 2. Process/transform the data (if necessary) into a format suitable for your analysis
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activity$date <- as.Date(activity$date, format = "%Y-%m-%d")
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activity$interval <- as.factor(activity$interval)
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What is mean total number of steps taken per day?
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-------------------------------------------------
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### 1.Calculate the total number of steps taken per day.
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steps_daily <- aggregate(steps ~ date, activity, sum)
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colnames(steps_daily) <- c("date","steps")
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head(steps_daily)
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## date steps
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## 1 2012-10-02 126
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## 2 2012-10-03 11352
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## 3 2012-10-04 12116
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## 4 2012-10-05 13294
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## 5 2012-10-06 15420
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## 6 2012-10-07 11015
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### 2. Make a histogram of the total number of steps taken each day.
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library(ggplot2)
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## Warning: package 'ggplot2' was built under R version 3.2.5
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ggplot(steps_daily, aes(x = steps)) +
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geom_histogram(fill = "green", binwidth = 1500) +
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labs(title="Summary of Steps Taken per Day (Oct-Nov 2012)",
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x = "Total Number of Steps per Day", y = "Daily Frequency (intervals)" ) +theme_bw()
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![](PA1_template_files/figure-markdown_strict/unnamed-chunk-4-1.png)
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### 3. Calculate and report the mean and median of the total number of steps taken per day
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mean_steps_daily <- mean(steps_daily$steps, na.rm=TRUE)
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median_steps_daily <- median(steps_daily$steps, na.rm=TRUE)
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Print:
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mean_steps_daily
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## [1] 10766.19
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median_steps_daily
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## [1] 10765
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The average (mean) steps taken per day is 10766.19 while the median is
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10765.
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What is the average daily activity pattern?
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-------------------------------------------
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Prior to plotting, convert the steps taken by the 5-minute intervals
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into integers
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interval_steps <- aggregate(activity$steps,
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by = list(interval = activity$interval),
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FUN=mean, na.rm=TRUE)
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interval_steps$interval <- as.integer(levels(interval_steps$interval)[interval_steps$interval])
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as.integer(levels(interval_steps$interval)[interval_steps$interval])
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## integer(0)
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colnames(interval_steps) <- c("interval", "steps")
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### 1. Make a time series plot (i.e. type = "l") of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all days (y-axis)
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ggplot(interval_steps, aes(x=interval, y=steps)) +
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geom_line(color="green", size=1.5) +
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labs(title="Average Daily Activity Pattern (Oct-Nov 2012)", x="Interval (5-Minutes)", y="Average Number of steps") +
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theme_bw()
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![](PA1_template_files/figure-markdown_strict/unnamed-chunk-9-1.png)
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### 2. Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?
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max_interval_steps <- interval_steps[which.max(interval_steps$steps),]
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print(max_interval_steps)
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## interval steps
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## 104 835 206.1698
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The 835th interval has the most steps taken at 206 steps.
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Imputing Missing Values
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-----------------------
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To address possible bias caused by missing values (NA), identify the
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number of missing values and device a strategy to fill out these values.
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Create a new (modified) dataset that integrates the new values.
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### 1. Total number of missing values:
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Calculate and report the total number of missing values in the dataset
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(i.e. the total number of rows with NAs)
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missing_steps <- sum(is.na(activity$steps))
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print(missing_steps)
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## [1] 2304
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There are a total of 2304 missing values.
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### 2. Fill the missing values and create a new dataset with the proxy values
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Devise a strategy for filling in all of the missing values in the
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dataset. The strategy does not need to be sophisticated. For example,
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you could use the mean/median for that day, or the mean for that
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5-minute interval, etc.
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na_steps_proxy <- function(data, pervalue) {
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na_list <- which(is.na(data$steps))
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na_proxy <- unlist(lapply(na_list, FUN=function(idx){
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interval = data[idx,]$interval
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pervalue[pervalue$interval == interval,]$steps
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}))
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proxy_na_steps <- data$steps
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proxy_na_steps[na_list] <- na_proxy
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proxy_na_steps
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}
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activity_proxy <- data.frame(
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steps = na_steps_proxy(activity, interval_steps),
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date = activity$date,
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interval = activity$interval)
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Check if there still missing values in the new dataset: activity\_proxy.
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missing_steps_new <- sum(is.na(activity_proxy$steps))
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print(missing_steps_new)
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## [1] 0
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str(activity_proxy)
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## 'data.frame': 17568 obs. of 3 variables:
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## $ steps : num 1.717 0.3396 0.1321 0.1509 0.0755 ...
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## $ date : Date, format: "2012-10-01" "2012-10-01" ...
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## $ interval: Factor w/ 288 levels "0","5","10","15",..: 1 2 3 4 5 6 7 8 9 10 ...
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### 3. Plot a histogram of the total number of steps taken each day
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Plot a historgram of the total number of steps taken per day with the
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proxy values
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steps_daily_proxy<- aggregate(steps ~ date, activity_proxy, sum)
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colnames(steps_daily_proxy) <- c("date","steps")
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ggplot(steps_daily_proxy, aes(x = steps)) +
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geom_histogram(fill = "green", binwidth = 1500) +
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labs(title="Summary of Steps Taken per Day (Oct-Nov 2012)",
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x = "Total Number of Steps per Day", y = "Daily Frequency (intervals)") + theme_bw()
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![](PA1_template_files/figure-markdown_strict/unnamed-chunk-14-1.png)
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Calculate the new mean and median of the data set with proxy values
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mean_steps_daily_proxy <- mean(steps_daily_proxy$steps, na.rm=TRUE)
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median_steps_daily_proxy <- median(steps_daily_proxy$steps, na.rm=TRUE)
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print(mean_steps_daily_proxy)
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## [1] 10766.19
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print (median_steps_daily_proxy)
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## [1] 10766.19
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Note that the new mean and median are now equal at 10766.189; compared
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to the previous dataset with missing values. However, it is important to
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note that the value of the mean has not changed thus replacing the
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missing values has no impact on the estimates of computing the total
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daily number of steps.
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Are there differences in activity patterns between weekdays and weekends?
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-------------------------------------------------------------------------
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### 1.) Using the new dataset, add a new column indicating what type of day the observations were taken (Weekday or weekend)
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activity_proxy$dateType <- ifelse(as.POSIXlt(activity_proxy$date)$wday %in% c(0,6), 'weekend', 'weekday')
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head(activity_proxy)
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## steps date interval dateType
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## 1 1.7169811 2012-10-01 0 weekday
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## 2 0.3396226 2012-10-01 5 weekday
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## 3 0.1320755 2012-10-01 10 weekday
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## 4 0.1509434 2012-10-01 15 weekday
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## 5 0.0754717 2012-10-01 20 weekday
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## 6 2.0943396 2012-10-01 25 weekday
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### 2.) Plot a graph that would show if there is any difference in the activity by type of day.
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activitybydaytype<- aggregate(steps ~ interval + dateType, activity_proxy, mean)
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library(lattice)
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xyplot(steps ~ interval|dateType, data = activitybydaytype,
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type = "l", layout = c(1,2),
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grid = TRUE,
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xlab="5-Minute Time Interval", ylab = "Average Number of Steps Taken",
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main= "Average Steps Taken in Weekdays vs Weekends")
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![](PA1_template_files/figure-markdown_strict/unnamed-chunk-17-1.png)
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The graph shows that activity during weekdays has the highest peak but
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weekends have more frequent peaks.

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