Reproducible Research: Peer Assessment 1
by Chan Chee-Foong on 14 Apr 2016
Libraries required
- lubridate
- dplyr
- ggplot2
- lattice
- knitr
- markdown
- rmarkdown
Loading and preprocessing the data
- Download the zip file if it is not found in the working directory and unzip the file for analysis.
datadir <- "./activity"
datafile <- "activity.csv"
zipfile <- "activity.zip"
datadirfile <- paste(datadir, datafile, sep="/")
zipdirfile <- paste(datadir, zipfile, sep="/")
if (!file.exists(datadirfile)) {
dir.create(datadir)
url <- "https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
download.file(url, destfile = zipdirfile, mode = 'wb')
unzip (zipdirfile, exdir = datadir)
}- Load the csv file in to myData dataframe. Update the date to the date class. Update the interval to factor class.
myData <- read.csv(datadirfile, header = TRUE, na.strings = 'NA')
myData$date <- ymd(myData$date)
myData$interval <- as.factor(myData$interval)- Check data loaded
str(myData)## 'data.frame': 17568 obs. of 3 variables:
## $ steps : int NA NA NA NA NA NA NA NA NA NA ...
## $ date : POSIXct, 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 ...summary(myData)## steps date interval
## Min. : 0.00 Min. :2012-10-01 0 : 61
## 1st Qu.: 0.00 1st Qu.:2012-10-16 5 : 61
## Median : 0.00 Median :2012-10-31 10 : 61
## Mean : 37.38 Mean :2012-10-31 15 : 61
## 3rd Qu.: 12.00 3rd Qu.:2012-11-15 20 : 61
## Max. :806.00 Max. :2012-11-30 25 : 61
## NA's :2304 (Other):17202head(myData)## 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 25What is mean total number of steps taken per day?
Use dplyr library to group and summarise data. Then calculate and show the mean and median.
- Notes: Could have used aggregate function here.
totalByDay <- myData[!is.na(myData$steps),] %>% group_by(date) %>% summarise(total = sum(steps))
head(totalByDay)## Source: local data frame [6 x 2]
##
## date total
## (time) (int)
## 1 2012-10-02 126
## 2 2012-10-03 11352
## 3 2012-10-04 12116
## 4 2012-10-05 13294
## 5 2012-10-06 15420
## 6 2012-10-07 11015hist(totalByDay[!is.na(totalByDay$total),]$total, breaks = 5, col = 'blue', xlab = 'Total number of steps', main = 'Total number of steps taken per day')
myMean <- round(mean(totalByDay[!is.na(totalByDay$total),]$total),0)
myMedian <- median(totalByDay[!is.na(totalByDay$total),]$total)
abline(v=myMean, lwd = 2, lty= 2, col = 'green')
abline(v=myMedian, lwd = 2, lty= 3, col = 'yellow')
myMean## [1] 10766myMedian## [1] 10765Mean and median of total number of steps taken per day are 10766 and 10765 respectively.
What is the average daily activity pattern?
Use dplyr library to group and summarise data. Then plot the mean number of steps every 5 minute interval throughout a day.
meanByInterval <- myData[!is.na(myData$steps),] %>% group_by(interval) %>% summarise(mean = mean(steps))
meanByInterval <- meanByInterval[!is.na(meanByInterval$mean),]
meanByInterval$interval <- as.numeric(as.character(meanByInterval$interval))
head(meanByInterval)## Source: local data frame [6 x 2]
##
## interval mean
## (dbl) (dbl)
## 1 0 1.7169811
## 2 5 0.3396226
## 3 10 0.1320755
## 4 15 0.1509434
## 5 20 0.0754717
## 6 25 2.0943396ggplot(data=meanByInterval, aes(x=interval, y=mean)) +
geom_line(size = 1, col = 'red') +
xlab("5 minute interval in a day") + ylab("Number of steps") +
ggtitle("Average number of steps at every 5 minute interval in a day")
myMax <- meanByInterval[meanByInterval$mean == max(meanByInterval$mean),]
myMax## Source: local data frame [1 x 2]
##
## interval mean
## (dbl) (dbl)
## 1 835 206.1698The 5 minute interval of 835 contains the maximum mean number of steps of 206
Imputing missing values
Study the NA values by considering the following:
- How many NA values are there?
- On which days are there NA values?
- Are these NA values occuring on days with values?
countNA <- nrow(myData[is.na(myData$steps),])
countNA## [1] 2304unique(myData[is.na(myData$steps),]$date)## [1] "2012-10-01 UTC" "2012-10-08 UTC" "2012-11-01 UTC" "2012-11-04 UTC"
## [5] "2012-11-09 UTC" "2012-11-10 UTC" "2012-11-14 UTC" "2012-11-30 UTC"myData[!is.na(myData$steps) & myData$date %in% unique(myData[is.na(myData$steps),]$date),]## [1] steps date interval
## <0 rows> (or 0-length row.names)- There is a total number of 2304 rows with NAs in the dataset.
- NA values occur only on 8 dates.
- On these 8 dates, all the steps values are NA.
Decision on imputing logic:
- The missing values are imputed based on the mean number of steps specific to weekday and interval.
- The means are calculated based only on data with steps values.
Steps:
1. Mean number of steps specific to weekday and interval are calculated and populated in meanByIntnDay
2. Merge with myData to form myNewData
3. Populate the missing NA values
4. Use dplyr library to group and summarise data.
5. Plot the histogram
myData$day <- wday(myData$date, label=TRUE)
meanByIntnDay <- myData[!is.na(myData$steps),] %>% group_by(interval, day) %>% summarise(mean = mean(steps))
myNewData <- merge(myData, meanByIntnDay,by = c('interval','day'))
myNewData[is.na(myNewData$steps),]$steps <- myNewData[is.na(myNewData$steps),]$mean
myNewData$interval <- as.numeric(as.character(myNewData$interval))
head(myNewData)## interval day steps date mean
## 1 0 Fri 0 2012-10-19 0
## 2 0 Fri 0 2012-10-05 0
## 3 0 Fri 0 2012-11-30 0
## 4 0 Fri 0 2012-10-12 0
## 5 0 Fri 0 2012-11-16 0
## 6 0 Fri 0 2012-10-26 0newTotalByDay <- myNewData[!is.na(myNewData$steps),] %>% group_by(date) %>% summarise(total = sum(steps))
head(newTotalByDay)## Source: local data frame [6 x 2]
##
## date total
## (time) (dbl)
## 1 2012-10-01 9974.857
## 2 2012-10-02 126.000
## 3 2012-10-03 11352.000
## 4 2012-10-04 12116.000
## 5 2012-10-05 13294.000
## 6 2012-10-06 15420.000hist(newTotalByDay[!is.na(newTotalByDay$total),]$total, breaks = 5, col = 'blue', xlab = 'Total number of steps', main = 'Total number of steps taken per day')
myNewMean <- round(mean(newTotalByDay[!is.na(newTotalByDay$total),]$total),0)
myNewMedian <- median(newTotalByDay[!is.na(newTotalByDay$total),]$total)
abline(v=myNewMean, lwd = 2, lty= 2, col = 'green')
abline(v=myNewMedian, lwd = 2, lty= 3, col = 'yellow')
myNewMean## [1] 10821myNewMedian## [1] 11015Mean and median of total number of steps taken per day are 10821 and 11015 respectively.
Before imputing the missing values, the mean and median are closed to each other. After imputing the missing values, the mean and median are quite different as observed on the histograms.
Are there differences in activity patterns between weekdays and weekends?
Additional field daytype was created to separate the data into 2 groups. Sat and Sun defined as weekends and the rest of the days as weekdays.
myNewData$daytype <- as.factor(ifelse(myNewData$day %in% c('Sat','Sun'),'Weekend','Weekday'))
head(myNewData)## interval day steps date mean daytype
## 1 0 Fri 0 2012-10-19 0 Weekday
## 2 0 Fri 0 2012-10-05 0 Weekday
## 3 0 Fri 0 2012-11-30 0 Weekday
## 4 0 Fri 0 2012-10-12 0 Weekday
## 5 0 Fri 0 2012-11-16 0 Weekday
## 6 0 Fri 0 2012-10-26 0 WeekdaymeanByIntnDT <- myNewData[!is.na(myNewData$steps),] %>% group_by(interval,daytype) %>% summarise(mean = mean(steps))
head(meanByIntnDT)## Source: local data frame [6 x 3]
## Groups: interval [3]
##
## interval daytype mean
## (dbl) (fctr) (dbl)
## 1 0 Weekday 2.310714
## 2 0 Weekend 0.000000
## 3 5 Weekday 0.450000
## 4 5 Weekend 0.000000
## 5 10 Weekday 0.175000
## 6 10 Weekend 0.000000with(meanByIntnDT,
xyplot(mean ~ interval | daytype, type= 'l',
layout = c(1,2),
lwd = 2,
main = 'Average number of steps taken in a day',
xlab = '5 minute interval in a day',
ylab = 'Number of steps',
panel = function(x,y,...) {
panel.xyplot(x,y,...)
panel.abline(h=100, col='red')
}
))
Noticed that there differences in activity patterns between weekdays and weekends. For weekdays, most activities occur earlier part of the day. For weekends, activities occur throughout the day.