alfredhomere / repdata_peerassessment1

Reproducible Research_Peer Assessment1_Project1

RepData_PeerAssessment1:FitBit Data Analysis Alfred Homere NGANDAM MFONDOUM May, 2016

Context

This is the first project of the fith course of Data Analyst Especiallyzation intitle"Reproducible Research". Learners have to answer some questions by using data collected from Fitbit.

Purpose

The purpose of this project has been divided in three parts:

• loading and preprocessing data
• imputing missing values
• interpreting data to answer research questions

Data

The data for this assignment can be downloaded from the course web site:

• Dataset: Activity monitoring data [52K]

The variables included in this dataset are:

• steps: Number of steps taking in a 5-minute interval (missing values are coded as NA)

• date: The date on which the measurement was taken in YYYY-MM-DD format

• 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.

Loading and preprocessing the data

Download, unzip and load data into data frame data.

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

data <- read.csv("activity.csv")

What is mean total number of steps taken per day?

Sum steps by day, create Histogram, and calculate mean and median.

steps_by_day <- aggregate(steps ~ date, data, sum) hist(steps_by_day$steps, main = paste("Total Steps Each Day"), col="blue", xlab="Number of Steps")

2. Calculate and report the mean and median total number of steps taken per day

rmean <- mean(steps_by_day$steps) rmedian <- median(steps_by_day$steps)

The mean is 1.0766 × 104 and the median is 10765.

What is the average daily activity pattern?

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_by_interval <- aggregate(steps ~ interval, data, mean)

plot(steps_by_interval$interval,steps_by_interval$steps, type="l", xlab="Interval", ylab="Number of Steps",main="Average Number of Steps per Day by Interval") plot of chunk unnamed-chunk-3

2. Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?

max_interval <- steps_by_interval[which.max(steps_by_interval$steps),1] The 5-minute interval, on average across all the days in the data set, containing the maximum number of steps is 835.

Imputing missing values

Missing data needed to be imputed. Only a simple imputation approach was required for this assignment. Missing values were imputed by inserting the average for each interval. Thus, if interval 10 was missing on 10-02-2012, the average for that interval for all days (0.1320755), replaced the NA.

incomplete <- sum(!complete.cases(data)) imputed_data <- transform(data, steps = ifelse(is.na(data$steps), steps_by_interval$steps[match(data$interval, steps_by_interval$interval)], data$steps)) Zeroes were imputed for 10-01-2012 because it was the first day and would have been over 9,000 steps higher than the following day, which had only 126 steps. NAs then were assumed to be zeros to fit the rising trend of the data.

imputed_data[as.character(imputed_data$date) == "2012-10-01", 1] <- 0 Recount total steps by day and create Histogram.

steps_by_day_i <- aggregate(steps ~ date, imputed_data, sum) hist(steps_by_day_i$steps, main = paste("Total Steps Each Day"), col="blue", xlab="Number of Steps")

#Create Histogram to show difference. hist(steps_by_day$steps, main = paste("Total Steps Each Day"), col="red", xlab="Number of Steps", add=T) legend("topright", c("Imputed", "Non-imputed"), col=c("blue", "red"), lwd=10) plot of chunk unnamed-chunk-6

Calculate new mean and median for imputed data.

rmean.i <- mean(steps_by_day_i$steps) rmedian.i <- median(steps_by_day_i$steps) Calculate difference between imputed and non-imputed data.

mean_diff <- rmean.i - rmean med_diff <- rmedian.i - rmedian Calculate total difference.

total_diff <- sum(steps_by_day_i$steps) - sum(steps_by_day$steps) The imputed data mean is 1.059 × 104 The imputed data median is 1.0766 × 104 The difference between the non-imputed mean and imputed mean is -176.4949 The difference between the non-imputed mean and imputed mean is 1.1887 The difference between total number of steps between imputed and non-imputed data is 7.5363 × 104. Thus, there were 7.5363 × 104 more steps in the imputed data. Are there differences in activity patterns between weekdays and weekends?

Created a plot to compare and contrast number of steps between the week and weekend. There is a higher peak earlier on weekdays, and more overall activity on weekends.

weekdays <- c("Monday", "Tuesday", "Wednesday", "Thursday", "Friday") imputed_data$dow = as.factor(ifelse(is.element(weekdays(as.Date(imputed_data$date)),weekdays), "Weekday", "Weekend"))

steps_by_interval_i <- aggregate(steps ~ interval + dow, imputed_data, mean)

library(lattice)

xyplot(steps_by_interval_i$steps ~ steps_by_interval_i$interval|steps_by_interval_i$dow, main="Average Steps per Day by Interval",xlab="Interval", ylab="Steps",layout=c(1,2), type="l") plot of chunk unnamed-chunk-10