Analyzing_steps
Load libraries
library(ggplot2)
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union1. load data
my_data<-read.csv("~/MyProjects/representative projects/week2/activity.csv", header=TRUE)
head(my_data)## 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 25tail(my_data)## steps date interval
## 17563 NA 2012-11-30 2330
## 17564 NA 2012-11-30 2335
## 17565 NA 2012-11-30 2340
## 17566 NA 2012-11-30 2345
## 17567 NA 2012-11-30 2350
## 17568 NA 2012-11-30 2355str(my_data)## 'data.frame': 17568 obs. of 3 variables:
## $ steps : int NA NA NA NA NA NA NA NA NA NA ...
## $ date : Factor w/ 61 levels "2012-10-01","2012-10-02",..: 1 1 1 1 1 1 1 1 1 1 ...
## $ interval: int 0 5 10 15 20 25 30 35 40 45 ...2.histogram of number steps taken each day
Summarise data, removing missing numbers, formating dates and summing up total number of steps taken
my_data$newformat<-as.character(my_data$date)
my_data$newformat<-as.Date(my_data$newformat)
k<-my_data%>%
select(steps, newformat)%>%
filter(!is.na(steps))%>%
group_by(newformat)%>%
summarise(SumSteps=sum(steps))Putting the numbers on histogram
hist(k$SumSteps, main = "Distribution of Total Number of Steps",
xlab = "Number of Steps",
ylab = "Frequency",
breaks=10, col = "navy")
3.Mean and median number of steps taken each day
summary(k$SumSteps)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 41 8841 10760 10770 13290 211904. 5 minutes time series plot of the average number of steps taken
Takinf the means of each 5 minutes interval, exluding missing data
j<-my_data%>%
select(steps, interval)%>%
filter(!is.na(steps))%>%
group_by(interval)%>%
summarise(AvgSteps=mean(steps))
j## # A tibble: 288 x 2
## interval AvgSteps
## <int> <dbl>
## 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
## # ... with 278 more rowsPlotting the data, formating the intervals into the real time of he day
ggplot(j, aes(y=AvgSteps, x=strptime(sprintf("%04d", j$interval), format="%H%M")))+geom_line()+labs(x= "Time of the day") + labs(y="Average Steps taken")+labs(title="Pattern of a day")
Here we see the patterns of the day
5.The 5-minute interval that, on average, contains the maximum number of steps
j%>%
select(AvgSteps, interval)%>%
filter(AvgSteps==max(AvgSteps))## # A tibble: 1 x 2
## AvgSteps interval
## <dbl> <int>
## 1 206.1698 835Here we see that the average maximum number of steps was taken in the 835th inteval
6.Code to describe and show a strategy for imputing missing data
n.missing<-sum(is.na(my_data))#number of missing values in the datasetImputting missing data by mean number of steps taken a day.
StepsImputed<- inner_join(my_data, j, by="interval") %>%
mutate(steps=ifelse(is.na(steps), AvgSteps, steps)) %>%
select(date, interval, steps)
head(StepsImputed)## date interval steps
## 1 2012-10-01 0 1.7169811
## 2 2012-10-01 5 0.3396226
## 3 2012-10-01 10 0.1320755
## 4 2012-10-01 15 0.1509434
## 5 2012-10-01 20 0.0754717
## 6 2012-10-01 25 2.0943396Summary of the new data, wee see there is no NAs
summary(StepsImputed)## date interval steps
## 2012-10-01: 288 Min. : 0.0 Min. : 0.00
## 2012-10-02: 288 1st Qu.: 588.8 1st Qu.: 0.00
## 2012-10-03: 288 Median :1177.5 Median : 0.00
## 2012-10-04: 288 Mean :1177.5 Mean : 37.38
## 2012-10-05: 288 3rd Qu.:1766.2 3rd Qu.: 27.00
## 2012-10-06: 288 Max. :2355.0 Max. :806.00
## (Other) :15840it’s time to see new averages of steps done per day
z<-StepsImputed%>%
select(steps, date)%>%
group_by(date)%>%
summarise(SumSteps=sum(steps))
z ## # A tibble: 61 x 2
## date SumSteps
## <fctr> <dbl>
## 1 2012-10-01 10766.19
## 2 2012-10-02 126.00
## 3 2012-10-03 11352.00
## 4 2012-10-04 12116.00
## 5 2012-10-05 13294.00
## 6 2012-10-06 15420.00
## 7 2012-10-07 11015.00
## 8 2012-10-08 10766.19
## 9 2012-10-09 12811.00
## 10 2012-10-10 9900.00
## # ... with 51 more rowssummary(z$SumSteps)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 41 9819 10770 10770 12810 21190Wee see that the median increased by 10 steps/day.
#It’s time to plot
hist(z$SumSteps, main = "New Distibution of Steps per day",
xlab = "Steps per Day",
ylab = "Frequency ",
breaks=20, col = "grey")
8.Panel plot comparing the average number of steps taken per 5-minute interval across weekdays and weekends
Determine the weekdays and weekends, based on date column, with imputed data.
StepsImputed$date<-as.Date(StepsImputed$date)
StepsImputed <- StepsImputed %>%
mutate(Weekdays = ifelse(weekdays(date) %in% c("Saturday","Sunday"),
"weekend","weekday"))
head(StepsImputed)## date interval steps Weekdays
## 1 2012-10-01 0 1.7169811 weekday
## 2 2012-10-01 5 0.3396226 weekday
## 3 2012-10-01 10 0.1320755 weekday
## 4 2012-10-01 15 0.1509434 weekday
## 5 2012-10-01 20 0.0754717 weekday
## 6 2012-10-01 25 2.0943396 weekdayNumber of steps taken by the day of the week
a<-StepsImputed%>%
select(steps, interval, Weekdays)%>%
group_by(Weekdays, interval )%>%
summarise(AvgSteps=mean(steps))
a## # A tibble: 576 x 3
## # Groups: Weekdays [?]
## Weekdays interval AvgSteps
## <chr> <int> <dbl>
## 1 weekday 0 2.25115304
## 2 weekday 5 0.44528302
## 3 weekday 10 0.17316562
## 4 weekday 15 0.19790356
## 5 weekday 20 0.09895178
## 6 weekday 25 1.59035639
## 7 weekday 30 0.69266247
## 8 weekday 35 1.13794549
## 9 weekday 40 0.00000000
## 10 weekday 45 1.79622642
## # ... with 566 more rowsggplot(a, aes(y=AvgSteps, x=strptime(sprintf("%04d", a$interval), format="%H%M")))+geom_line(aes(col=Weekdays))+labs(x= "Time of the day") + labs(y="Average Steps taken")