Reproducible Research: Peer Assessment 1

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

Loading and preprocessing the data

download.file("https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip",destfile = "rawdata.zip")
unzip("rawdata.zip")
data<- read.csv("activity.csv")

What is mean total number of steps taken per day?

Make a histogram of the total number of steps taken each day.

data<- read.csv("activity.csv")
dat <- select(data, steps, date) %>%
    group_by(date) %>% 
    summarize(total_steps= sum(steps, na.rm = T)) 

    with(dat, hist(total_steps, main="Total number of steps taken each day"))

Report the mean and median total number of steps taken per day.

The mean and median total number of steps taken per day:

select(dat, total_steps) %>%
summarise(mean= mean(total_steps), median= median(total_steps)) %>% as.data.frame

##      mean median
## 1 9354.23  10395

What is the average daily activity pattern?

Make a time series plot of the 5-minute interval and the average number of steps taken, averaged across all days.

library(lubridate)  
dat <- group_by(data, interval) %>%
    summarize(mean_steps= mean(steps, na.rm = T)) %>%
    with(plot(interval, mean_steps,main = "Average Number of Steps Taken Each Day", type = 'l', xlab = "5-min Interval", ylab = "Average Number of Steps"))

The 5-minute interval contains the maximum number of steps

    group_by(data, interval) %>%
    summarize(mean_steps= mean(steps, na.rm = T)) %>%
    filter(mean_steps==max(mean_steps), na.rm = T) %>%
    select(the_interval_maximize_steps= interval) %>% as.data.frame

##   the_interval_maximize_steps
## 1                         835

Imputing missing values

The total number of rows with missing data is:

sum(is.na(data$steps))

## [1] 2304

The strategy for filling in all of the missing values is to replace the missing values with the mean for that 5-minute interval.

## mean steps table for intervals
dat <- group_by(data, interval) %>%
    summarize(mean_steps= mean(steps, na.rm = T)) %>% as.data.frame

## the replace function to replace the NAs with the means steps across days
replace_value <- function(steps,interval_value){
    if (!is.na(steps)) {steps}
    else {dat[dat$interval==interval_value,2]}
}  

## replace the NAs in the data
n= length(data$steps)
for (i in 1:n){ 
    data$steps[i]= replace_value(data$steps[i],data$interval[i])
}

analysis with the new data

dat <- select(data, steps, date) %>%
    group_by(date) %>% 
    summarize(total_steps= sum(steps))
    with(dat, hist(total_steps, main="Total number of steps taken each day"))

The mean and median total number of steps taken per day.

select(dat, total_steps) %>%
summarise(mean= mean(total_steps), median= median(total_steps)) %>% as.data.frame

##       mean   median
## 1 10766.19 10766.19

From the report we can see that both values are increased. and the histigram becomes more concentrated to the central.

Are there differences in activity patterns between weekdays and weekends?

create a new vairable indicating the type of days

data<- mutate(data, date= as.Date(date))
day_type<- function(date){
        if (weekdays(date)=="Sunday"|weekdays(date)=="Saturday") {"weekend"}
        else {"workday"}
}

dat <- mutate(data, type= sapply(date,day_type)) %>%
    mutate(type = as.factor(type))

From the following pictures we can see that the partterns are different.

plotdat <- select(dat, steps, interval, type) %>%
    group_by(interval,type) %>%   
    summarize(steps= mean(steps)) 
    
    xyplot(steps-mean(steps)~interval|type, layout= c(1,2), data = plotdat, type ="l")

    qplot(interval, steps, data = plotdat, facets = .~type, geom = c("point","smooth"))