Loading and preprocessing the data

1.Load the data

activity <- read.csv("C:/Users/kwak/Desktop/activity.csv")
head(activity)
  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

2.Process/transform the data into a format suitable for your analysis

activity_sum<-ddply(activity, .(date), summarize, total_steps = sum(steps, na.rm=T))
head(activity_sum)
        date total_steps
1 2012-10-01           0
2 2012-10-02         126
3 2012-10-03       11352
4 2012-10-04       12116
5 2012-10-05       13294
6 2012-10-06       15420

What is mean total number of steps taken per day?

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

ggplot(activity_sum, aes(total_steps))+geom_histogram(binwidth=1000, na.rm=TRUE)+xlab("Average Steps Taken per Day")+ylab("Frequency")

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

summary(activity_sum)
         date     total_steps   
 2012-10-01: 1   Min.   :    0  
 2012-10-02: 1   1st Qu.: 6778  
 2012-10-03: 1   Median :10395  
 2012-10-04: 1   Mean   : 9354  
 2012-10-05: 1   3rd Qu.:12811  
 2012-10-06: 1   Max.   :21194  
 (Other)   :55

What is the average dailiy activity pattern?

1.Make a time sereies plot(i.e. type=“1”) of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all days (y-axis)

activity_avg<-ddply(activity, .(interval), summarize, avg_steps = mean(steps, na.rm=T))
ggplot(activity_avg, aes(x=interval, y=avg_steps)) + geom_line()+xlab("interval")+ylab("Averge Steps")

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

activity_avg[which.max(activity_avg$avg_steps),]
    interval avg_steps
104      835  206.1698

Imputing missing values

1.Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)

count(is.na(activity$steps))
      x  freq
1 FALSE 15264
2  TRUE  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.

activity_upd<-activity
for (i in 1:nrow(activity_upd)){
  compare <- activity_upd$steps[i]
  if(is.na(compare)){
    for(j in 1:nrow(activity_avg)){
      if(activity_upd$interval[i] == activity_avg$interval[j]){
        activity_upd$steps[i] <- activity_avg$avg_steps[j]
        break
      }
    }
  }  
}
head(activity_upd)
      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

3.Create a new dataset that is equal to the original dataset but with the missing data filled in.

activity_upd_sum<-ddply(activity_upd, .(date), summarize, total_steps = sum(steps, na.rm=T))
head(activity_upd_sum)
        date total_steps
1 2012-10-01    10766.19
2 2012-10-02      126.00
3 2012-10-03    11352.00
4 2012-10-04    12116.00
5 2012-10-05    13294.00
6 2012-10-06    15420.00

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?

g1=ggplot(activity_sum, aes(total_steps))+geom_histogram(binwidth=1000, na.rm=TRUE)+xlab("Average Steps Taken per Day")+ylab("Frequency")+labs(title="original")
g2=ggplot(activity_upd_sum, aes(total_steps))+geom_histogram(binwidth=1000, na.rm=TRUE)+xlab("Average Steps Taken per Day")+ylab("Frequency")+labs(title="revision")
grid.arrange(g1, g2, ncol=2)

summary(activity_sum)
         date     total_steps   
 2012-10-01: 1   Min.   :    0  
 2012-10-02: 1   1st Qu.: 6778  
 2012-10-03: 1   Median :10395  
 2012-10-04: 1   Mean   : 9354  
 2012-10-05: 1   3rd Qu.:12811  
 2012-10-06: 1   Max.   :21194  
 (Other)   :55                  
summary(activity_upd_sum) 
         date     total_steps   
 2012-10-01: 1   Min.   :   41  
 2012-10-02: 1   1st Qu.: 9819  
 2012-10-03: 1   Median :10766  
 2012-10-04: 1   Mean   :10766  
 2012-10-05: 1   3rd Qu.:12811  
 2012-10-06: 1   Max.   :21194  
 (Other)   :55                  
#imputation의 결과로 median value와 mean value가 상승하였고 이를 히스토그램에서 시각적으로 확인할 수 있다.

Are there differences in activity patterns between weekdays and weekends?

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.

activity_week<-activity_upd
activity_week$date<-as.Date(activity_week$date) 
activity_week$week<-weekdays(activity_week$date)
activity_week$weekend <- ifelse(activity_week$week == "토요일" | activity_week$week == "일요일", "weekend","weekday")
activity_week_avg<-ddply(activity_week, .(interval, weekend), summarize, total_steps = mean(steps, na.rm=T)) 
head(activity_week_avg)
  interval weekend total_steps
1        0 weekday  2.25115304
2        0 weekend  0.21462264
3        5 weekday  0.44528302
4        5 weekend  0.04245283
5       10 weekday  0.17316562
6       10 weekend  0.01650943

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). The plot should look something like the following, which was creating using simulated data:

ggplot(activity_week_avg, aes(color=weekend, x=interval, y=total_steps)) + geom_line()+xlab("interval")+ylab("Averge Steps Taken")+facet_grid( . ~ weekend)