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Reproducible Research: Peer Assessment 1 |
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It is now possible to collect a large amount of data about personal movement using activity monitoring devices such as a Fitbit, Nike Fuelband, or Jawbone Up. These type of devices are part of the “quantified self” movement - a group of enthusiasts who take measurements about themselves regularly to improve their health, to find patterns in their behavior, or because they are tech geeks. But these data remain under-utilized both because the raw data are hard to obtain and there is a lack of statistical methods and software for processing and interpreting 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.
The data for this assignment can be downloaded from the course web site:
Dataset: [Activity monitoring data] (https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip)
The variables included in this dataset are:
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steps: Number of steps taking in a 5-minute interval (missing values are coded as NA)
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date: The date on which the measurement was taken in YYYY-MM-DD format
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interval: Identifier for the 5-minute interval in which measurement was taken
The dataset is stored in a comma-separated-value (CSV) file and there are a total of 17,568 observations in this dataset.
Setting global options and loading required libraries
library(knitr)
library(ggplot2)
library(data.table)
opts_chunk$set(echo = TRUE, results = 'hold')activity <- unzip("activity.zip")
act_data <- read.csv("activity.csv", header=TRUE, sep=",")
str(act_data)## 'data.frame': 17568 obs. of 3 variables:
## $ steps : int NA NA NA NA NA NA NA NA NA NA ...
## $ date : chr "2012-10-01" "2012-10-01" "2012-10-01" "2012-10-01" ...
## $ interval: int 0 5 10 15 20 25 30 35 40 45 ...
Convert some of the vectors to appropriate forms
act_data$date <- as.Date(act_data$date, format="%Y-%m-%d")
act_data$interval <- as.factor(act_data$interval)Post converting the column classes print the structure of the data
str(act_data)## 'data.frame': 17568 obs. of 3 variables:
## $ steps : int 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 ...
Print the header of the dataset
head(act_data, 10)## steps date interval
## 1 NA 2012-10-01 0
## 2 NA 2012-10-01 5
## 3 NA 2012-10-01 10
## 4 NA 2012-10-01 15
## 5 NA 2012-10-01 20
## 6 NA 2012-10-01 25
## 7 NA 2012-10-01 30
## 8 NA 2012-10-01 35
## 9 NA 2012-10-01 40
## 10 NA 2012-10-01 45
steps_per_day <- aggregate(steps ~ date, data=act_data, FUN=sum)
colnames(steps_per_day) <- c("date", "steps")Print the header of the data frame with steps aggregated per day
head(steps_per_day, 10)## date steps
## 1 2012-10-02 126
## 2 2012-10-03 11352
## 3 2012-10-04 12116
## 4 2012-10-05 13294
## 5 2012-10-06 15420
## 6 2012-10-07 11015
## 7 2012-10-09 12811
## 8 2012-10-10 9900
## 9 2012-10-11 10304
## 10 2012-10-12 17382
ggplot(steps_per_day, aes(x = steps)) +
geom_histogram(fill = "blue", binwidth = 1000) +
labs(title = "Histogram - Steps Taken Per Day", x = "Steps Per Day", y = "Frequency")
mean_steps_per_day <- mean(steps_per_day$steps)
mean_steps_per_day
median_steps_per_day <- median(steps_per_day$steps)
median_steps_per_day## [1] 10766.19
## [1] 10765
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)
steps_per_interval <- aggregate(steps ~ interval, data = act_data, FUN = mean, na.rm = TRUE)
steps_per_interval$interval <- as.integer(levels(steps_per_interval$interval)[steps_per_interval$interval])
colnames(steps_per_interval) <- c("interval", "steps")Print the header of the data frame with steps aggregated per interval
head(steps_per_interval, 10)## interval steps
## 1 0 1.7169811
## 2 5 0.3396226
## 3 10 0.1320755
## 4 15 0.1509434
## 5 20 0.0754717
## 6 25 2.0943396
## 7 30 0.5283019
## 8 35 0.8679245
## 9 40 0.0000000
## 10 45 1.4716981
Plot the timeseries graph
ggplot(steps_per_interval, aes(x = interval, y = steps)) +
geom_line(col = "blue", size = 1) +
labs(title = "Average Daily Activity Pattern", x = "Interval", y = "Steps")
2. Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?
max_interval <- steps_per_interval[which.max(steps_per_interval$steps),]
max_interval## interval steps
## 104 835 206.1698
1. Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)
missing_values <- sum(is.na(act_data$steps))
missing_values## [1] 2304
2. Devise a strategy for filling in all of the missing values in the dataset. The strategy does not need to be sophisticated. For example, you could use the mean/median for that day, or the mean for that 5-minute interval, etc.
To populate missing values, we choose to replace them with the mean value at the same interval across days.
new_act_data <- act_data
index_of_na <- which(is.na(new_act_data$steps))
for (i in index_of_na) {
new_act_data$steps[i] <- with(steps_per_interval, steps[interval = new_act_data$interval[i]])
}Print the top rows of newly created dataset
head(new_act_data, 10)## steps date interval
## 1 1.7169811 2012-10-01 0
## 2 0.3396226 2012-10-01 5
## 3 0.1320755 2012-10-01 10
## 4 0.1509434 2012-10-01 15
## 5 0.0754717 2012-10-01 20
## 6 2.0943396 2012-10-01 25
## 7 0.5283019 2012-10-01 30
## 8 0.8679245 2012-10-01 35
## 9 0.0000000 2012-10-01 40
## 10 1.4716981 2012-10-01 45
Given we have handled the missing values, let us check if the above strategy really worked out.
new_missing_values <- sum(is.na(new_act_data$steps))
new_missing_values## [1] 0
4. Make a histogram of the total number of steps taken each day and Calculate and report the mean and median total number of steps taken per day. Do these values differ from the estimates from the first part of the assignment? What is the impact of imputing missing data on the estimates of the total daily number of steps?
new_steps_per_day <- aggregate(steps ~ date, data = new_act_data, FUN=sum)
colnames(new_steps_per_day) <- c("date", "steps")
ggplot(new_steps_per_day, aes(x = steps)) +
geom_histogram(fill = "blue", binwidth = 1000) +
labs(title = "Histogram - Steps Taken Per Day", x = "Steps Per Day", y = "Frequency")
In order to find the impact of imputing the missing values, let us compute the mean and median of steps taken per day
new_mean_steps_per_day <- mean(new_steps_per_day$steps)
new_mean_steps_per_day
new_median_steps_per_day <- median(new_steps_per_day$steps)
new_median_steps_per_day## [1] 10766.19
## [1] 10766.19
1. Create a new factor variable in the dataset with two levels - “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.
Let us first add a factor variable to identify the given date as Weekday or Weekend
dt <- data.table(new_act_data)
dt[, weekday := ifelse(weekdays(date) %in% c("Saturday", "Sunday"), "Weekend", "Weekday")]
dt$weekday <- as.factor(dt$weekday)
dt$interval <- as.integer(levels(dt$interval)[dt$interval])
head(dt, 10)## steps date interval weekday
## 1: 1.7169811 2012-10-01 0 Weekday
## 2: 0.3396226 2012-10-01 5 Weekday
## 3: 0.1320755 2012-10-01 10 Weekday
## 4: 0.1509434 2012-10-01 15 Weekday
## 5: 0.0754717 2012-10-01 20 Weekday
## 6: 2.0943396 2012-10-01 25 Weekday
## 7: 0.5283019 2012-10-01 30 Weekday
## 8: 0.8679245 2012-10-01 35 Weekday
## 9: 0.0000000 2012-10-01 40 Weekday
## 10: 1.4716981 2012-10-01 45 Weekday
2. 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)
steps_per_weekday <- aggregate(steps ~ interval+weekday, data = dt, FUN = mean)
ggplot(steps_per_weekday, aes(x = interval, y = steps)) +
geom_line(col = "blue", size = 1) +
facet_wrap(~ weekday, nrow=2, ncol=1) +
labs(x = "Interval", y = "Number of Steps")