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@DDakhno
DDakhno committed Apr 5, 2016
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45 changes: 29 additions & 16 deletions PA1_template.Rmd
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@@ -1,7 +1,7 @@
---
title: "Reproducible research : Course Project 1"
author: "D.Dakhno"
date: "2. April 2016"
date: "3. April 2016"
output: html_document
---

Expand Down Expand Up @@ -64,13 +64,11 @@ dev.off()
## Mean and median number of steps taken each day
Using the before summarized data print the mean and median number of steps taken each day
```{r, echo = TRUE}
mean(activity_summarized$SumSteps,na.rm = TRUE)
median(activity_summarized$SumSteps,na.rm = TRUE)
#For comparisons in future...
activity_summarized_before <- copy(activity_summarized)
activity_summarized$SumSteps <- NULL
activity_summarized
```
A lot of strange "0" as median values for the number of steps compared to the non-null means suggest, more than the half of time intervals were lazy.

##Time series plot of the average number of steps taken or What is the average daily activity pattern?
```{r, echo = TRUE}
activity_summarized_interval <- aggregate(steps ~ interval, activity, mean)
Expand All @@ -94,24 +92,30 @@ sapply(activity, function(x) sum(is.na(x)))
So, the total number of rows with NAs is 2304 (all missing values are in the column "steps")

###Devising a strategy for filling in all of the missing values in the dataset.
Out of all unsophisticated strategies we choose here setting up the mean for that 5-minute interval (over all the days) instead of the missing value
Out of all unsophisticated strategies we choose here setting up the mean for the same 5-minute interval (over all the days) instead of the missing value
```{r, echo = TRUE}
activity_summarized <- activity%>%group_by(interval)%>%summarize(MeanSteps = mean(steps, na.rm = TRUE))
activity_summarized$interval <- as.factor(activity_summarized$interval)
setkey(activity_summarized, interval)
#NB! Data tables are approached by reference, so copy() here to preserve the original data
activity_completed <- copy(activity)
#Calculating the mean values per interval, indexing (setkey)
activity_summarized <- aggregate(steps ~ interval, activity_completed, mean)
colnames(activity_summarized) <- c("interval","steps")
activity_summarized <- as.data.table(activity_summarized)
activity_summarized$interval <- as.factor(activity_summarized$interval)
setkey(activity_summarized,interval)
for (i in seq(along = activity_completed$interval)) {
#Setting the appropriate mean values at the place of NAs
for (i in seq(along = activity_completed$steps)) {
if (is.na(activity_completed[i,]$steps)) {
activity_completed[i,]$steps <- activity_summarized[as.character(activity_completed[i,]$interval)]$MeanSteps
intrv <- as.character(activity_completed[i,]$interval)
activity_completed[i,]$steps <- activity_summarized[as.character(intrv)]$steps
}
}
```
Compare
```{r, echo = TRUE}
activity
activity_completed
identical(activity,activity_completed)
summary(activity)
summary(activity_completed)
```

## Histogram of the total number of steps taken each day after missing values are imputed
Expand All @@ -124,15 +128,24 @@ hist(activity_summarized$SumSteps, main = "Total number of steps taken each day\

Do these values differ from the estimates from the first part of the assignment?
Anyway!

```{r, echo = TRUE}
activity_summarized_before
summary(activity_summarized_before)
summary(activity_summarized)
```

<img class=center src=./figure/Total_number_steps_with_NA.png width=672>

##What is the impact of imputing missing data on the estimates of the total daily number of steps?

The peak interval of 10000 to 15000 steps per day has been enhanced through the imputing missing data.

```{r, echo = TRUE}
par(mfrow=c(1,2))
hist(activity_summarized_before$SumSteps, main = "Total number of steps taken each day\n(NAs not treated)", xlab = "Total steps per day", col = "blue", ylim = c(0,40))
hist(activity_summarized$SumSteps, main = "Total number of steps\n(NAs substituted)", xlab = "Total steps per day", col = "blue", ylim = c(0,40))
```


Visually, the peak interval of 10000 to 15000 steps per day has been enhanced through the imputing missing data.

## Panel plot comparing the average number of steps taken per 5-minute interval across weekdays and weekends

Expand Down
136 changes: 62 additions & 74 deletions PA1_template.html
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168 changes: 92 additions & 76 deletions PA1_template.knit.md
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Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
---
title: "Reproducible research : Course Project 1"
author: "D.Dakhno"
date: "2. April 2016"
date: "3. April 2016"
output: html_document
---

Expand Down Expand Up @@ -116,47 +116,41 @@ dev.copy(png, file = "figure/Total_number_steps_with_NA.png")

```
## png
## 3
## 5
```

```r
dev.off()
```

```
## png
## 2
## RStudioGD
## 2
```

## Mean and median number of steps taken each day
Using the before summarized data print the mean and median number of steps taken each day

```r
#For comparisons in future...
activity_summarized_before <- copy(activity_summarized)
activity_summarized$SumSteps <- NULL
activity_summarized
mean(activity_summarized$SumSteps,na.rm = TRUE)
```

```
## Source: local data table [61 x 3]
##
## date MeanSteps MedianSteps
## (date) (dbl) (dbl)
## 1 2012-10-01 NA NA
## 2 2012-10-02 0.43750 0
## 3 2012-10-03 39.41667 0
## 4 2012-10-04 42.06944 0
## 5 2012-10-05 46.15972 0
## 6 2012-10-06 53.54167 0
## 7 2012-10-07 38.24653 0
## 8 2012-10-08 NA NA
## 9 2012-10-09 44.48264 0
## 10 2012-10-10 34.37500 0
## .. ... ... ...
```
A lot of strange "0" as median values for the number of steps compared to the non-null means suggest, more than the half of time intervals were lazy.
## [1] 10766.19
```

```r
median(activity_summarized$SumSteps,na.rm = TRUE)
```

```
## [1] 10765
```

```r
#For comparisons in future...
activity_summarized_before <- copy(activity_summarized)
```
##Time series plot of the average number of steps taken or What is the average daily activity pattern?

```r
Expand Down Expand Up @@ -194,59 +188,63 @@ sapply(activity, function(x) sum(is.na(x)))
So, the total number of rows with NAs is 2304 (all missing values are in the column "steps")

###Devising a strategy for filling in all of the missing values in the dataset.
Out of all unsophisticated strategies we choose here setting up the mean for that 5-minute interval (over all the days) instead of the missing value
Out of all unsophisticated strategies we choose here setting up the mean for the same 5-minute interval (over all the days) instead of the missing value

```r
activity_summarized <- activity%>%group_by(interval)%>%summarize(MeanSteps = mean(steps, na.rm = TRUE))
activity_summarized$interval <- as.factor(activity_summarized$interval)
setkey(activity_summarized, interval)
#NB! Data tables are approached by reference, so copy() here to preserve the original data
activity_completed <- copy(activity)
#Calculating the mean values per interval, indexing (setkey)
activity_summarized <- aggregate(steps ~ interval, activity_completed, mean)
colnames(activity_summarized) <- c("interval","steps")
activity_summarized <- as.data.table(activity_summarized)
activity_summarized$interval <- as.factor(activity_summarized$interval)
setkey(activity_summarized,interval)

for (i in seq(along = activity_completed$interval)) {
#Setting the appropriate mean values at the place of NAs
for (i in seq(along = activity_completed$steps)) {
if (is.na(activity_completed[i,]$steps)) {
activity_completed[i,]$steps <- activity_summarized[as.character(activity_completed[i,]$interval)]$MeanSteps
intrv <- as.character(activity_completed[i,]$interval)
activity_completed[i,]$steps <- activity_summarized[as.character(intrv)]$steps
}
}
```
Compare

```r
activity
identical(activity,activity_completed)
```

```
## 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
## ---
## 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 2355
## [1] FALSE
```

```r
activity_completed
summary(activity)
```

```
## 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
## ---
## 17564: 4.6981132 2012-11-30 2335
## 17565: 3.3018868 2012-11-30 2340
## 17566: 0.6415094 2012-11-30 2345
## 17567: 0.2264151 2012-11-30 2350
## 17568: 1.0754717 2012-11-30 2355
## steps date interval
## Min. : 0.00 Min. :2012-10-01 Min. : 0.0
## 1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
## Median : 0.00 Median :2012-10-31 Median :1177.5
## Mean : 37.38 Mean :2012-10-31 Mean :1177.5
## 3rd Qu.: 12.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
## Max. :806.00 Max. :2012-11-30 Max. :2355.0
## NA's :2304
```

```r
summary(activity_completed)
```

```
## steps date interval
## Min. : 0.00 Min. :2012-10-01 Min. : 0.0
## 1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
## Median : 0.00 Median :2012-10-31 Median :1177.5
## Mean : 37.38 Mean :2012-10-31 Mean :1177.5
## 3rd Qu.: 27.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
## Max. :806.00 Max. :2012-11-30 Max. :2355.0
```

## Histogram of the total number of steps taken each day after missing values are imputed
Expand Down Expand Up @@ -284,33 +282,51 @@ hist(activity_summarized$SumSteps, main = "Total number of steps taken each day\
Do these values differ from the estimates from the first part of the assignment?
Anyway!


```r
activity_summarized_before
summary(activity_summarized_before)
```

```
## Source: local data table [61 x 4]
##
## date SumSteps MeanSteps MedianSteps
## (date) (dbl) (dbl) (dbl)
## 1 2012-10-01 NA NA NA
## 2 2012-10-02 126 0.43750 0
## 3 2012-10-03 11352 39.41667 0
## 4 2012-10-04 12116 42.06944 0
## 5 2012-10-05 13294 46.15972 0
## 6 2012-10-06 15420 53.54167 0
## 7 2012-10-07 11015 38.24653 0
## 8 2012-10-08 NA NA NA
## 9 2012-10-09 12811 44.48264 0
## 10 2012-10-10 9900 34.37500 0
## .. ... ... ... ...
## date SumSteps MeanSteps MedianSteps
## Min. :2012-10-01 Min. : 41 Min. : 0.1424 Min. :0
## 1st Qu.:2012-10-16 1st Qu.: 8841 1st Qu.:30.6979 1st Qu.:0
## Median :2012-10-31 Median :10765 Median :37.3785 Median :0
## Mean :2012-10-31 Mean :10766 Mean :37.3826 Mean :0
## 3rd Qu.:2012-11-15 3rd Qu.:13294 3rd Qu.:46.1597 3rd Qu.:0
## Max. :2012-11-30 Max. :21194 Max. :73.5903 Max. :0
## NA's :8 NA's :8 NA's :8
```

```r
summary(activity_summarized)
```

```
## date SumSteps MeanSteps MedianSteps
## Min. :2012-10-01 Min. : 41 Min. : 0.1424 Min. : 0.000
## 1st Qu.:2012-10-16 1st Qu.: 9819 1st Qu.:34.0938 1st Qu.: 0.000
## Median :2012-10-31 Median :10766 Median :37.3826 Median : 0.000
## Mean :2012-10-31 Mean :10766 Mean :37.3826 Mean : 4.474
## 3rd Qu.:2012-11-15 3rd Qu.:12811 3rd Qu.:44.4826 3rd Qu.: 0.000
## Max. :2012-11-30 Max. :21194 Max. :73.5903 Max. :34.113
```

<img class=center src=./figure/Total_number_steps_with_NA.png width=672>

##What is the impact of imputing missing data on the estimates of the total daily number of steps?

The peak interval of 10000 to 15000 steps per day has been enhanced through the imputing missing data.


```r
par(mfrow=c(1,2))
hist(activity_summarized_before$SumSteps, main = "Total number of steps taken each day\n(NAs not treated)", xlab = "Total steps per day", col = "blue", ylim = c(0,40))
hist(activity_summarized$SumSteps, main = "Total number of steps\n(NAs substituted)", xlab = "Total steps per day", col = "blue", ylim = c(0,40))
```

<img src="PA1_template_files/figure-html/unnamed-chunk-13-1.png" title="" alt="" width="672" />


Visually, the peak interval of 10000 to 15000 steps per day has been enhanced through the imputing missing data.

## Panel plot comparing the average number of steps taken per 5-minute interval across weekdays and weekends

Expand All @@ -326,6 +342,6 @@ library(lattice)
xyplot(x ~ Group.2| Group.1, data = activity_summarized, type = "l", layout = c(1, 2), lty=1, xlab = "Interval", ylab = "Number of steps")
```

<img src="PA1_template_files/figure-html/unnamed-chunk-13-1.png" title="" alt="" width="672" />
<img src="PA1_template_files/figure-html/unnamed-chunk-14-1.png" title="" alt="" width="672" />

The daily pattern on weekends looks more uniform than on weekdays (definite morning peaks).
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