The focus of this lab is on data outliers, data preparation, and data modeling. This lab requires the use of Microsoft Excel, R, and ERDplus.
Remember to always set your working directory to the source file location. Go to ‘Session’, scroll down to ‘Set Working Directory’, and click ‘To Source File Location’. Read carefully the below and follow the instructions to complete the tasks and answer any questions. Submit your work to RPubs as detailed in previous notes.
For your assignment you may be using different data sets than what is included in this worksheet demo. Make sure to read carefully the instructions on Sakai.
First, we must calculate the mean, standard deviation, maximum, and minimum for the Age column using R.
In R, we must read in the file again, extract the column and find the values that are asked for.
#Read File
mydata = read.csv(file="data/creditrisk.csv")
head(mydata)
## Loan.Purpose Checking Savings Months.Customer Months.Employed Gender
## 1 Small Appliance 0 739 13 12 M
## 2 Furniture 0 1230 25 0 M
## 3 New Car 0 389 19 119 M
## 4 Furniture 638 347 13 14 M
## 5 Education 963 4754 40 45 M
## 6 Furniture 2827 0 11 13 M
## Marital.Status Age Housing Years Job Credit.Risk
## 1 Single 23 Own 3 Unskilled Low
## 2 Divorced 32 Own 1 Skilled High
## 3 Single 38 Own 4 Management High
## 4 Single 36 Own 2 Unskilled High
## 5 Single 31 Rent 3 Skilled Low
## 6 Married 25 Own 1 Skilled Low
#Name the extracted variable
age = mydata$Age
#Calculate the average age below. Refer to Worksheet 2 for the correct command.
meanAge=mean(age)
meanAge
## [1] 34.39765
#Calculate standard deviation of age below. Refer to Worksheet 2 for the correct command.
stdAge=sd(age)
stdAge
## [1] 11.04513
#Calculate the maximum of age below. The command to find the maximum is max(variable) where variable is the extracted variable.
maxAge=max(age)
maxAge
## [1] 73
#Calculate the minimum of age below. The command to find the minimum is min(variable) where variable is the extracted variable.
minAge=min(age)
minAge
## [1] 18
Next, use the formula from class to detect any outliers. An outlier is value that “lies outside” most of the other values in a set of data. A common way to estimate the upper and lower threshold is to take the mean (+ or -) 3 * standard deviation. Try using this formula to find the upper and lower limit for age.
#Use the formula above to calculate the upper and lower threshold
upper=meanAge+3*stdAge
upper
## [1] 67.53302
lower=meanAge-3*stdAge
lower
## [1] 1.262269
Another similar method to find the upper and lower thresholds discussed in introductory statistics courses involves finding the interquartile range. Follow along below to see how we first calculate the interquartile range..
quantile(age)
## 0% 25% 50% 75% 100%
## 18 26 32 41 73
lowerq = quantile(age)[2]
upperq = quantile(age)[4]
iqr = upperq - lowerq
The formula below calculates the threshold. The threshold is the boundaries that determine if a value is an outlier. If the value falls above the upper threshold or below the lower threshold, it is an outlier.
Below is the upper threshold:
upperthreshold = (iqr * 1.5) + upperq
upperthreshold
## 75%
## 63.5
Below is the lower threshold:
lowerthreshold = lowerq - (iqr * 1.5)
lowerthreshold
## 25%
## 3.5
Are there any outliers? How many? It can also be useful to visualize the data using a box and whisker plot. The boxplot below supports the IQR we found of 15 and upper and lower threshold.
boxplot(age)
Next, we must read the ‘creditriskorg.csv’ file into R. This is the original dataset and contains missing values.
newdata = read.csv(file="data/creditriskorg.csv",skip=1,header=TRUE,sep=",")
head(newdata)
## Small.Appliance X.. X.739.00 X13 X12 M Single X23 Own X3
## 1 Furniture $- $1,230.00 25 0 M Divorced 32 Own 1
## 2 New Car $- $389.00 19 119 M Single 38 Own 4
## 3 Furniture $638.00 $347.00 13 14 M Single 36 Own 2
## 4 Education $963.00 $4,754.00 40 45 M Single 31 Rent 3
## 5 Furniture $2,827.00 $- 11 13 M Married 25 Own 1
## 6 New Car $- $229.00 13 16 M Married 26 Own 3
## Unskilled Low X X.1
## 1 Skilled High NA Age
## 2 Management High NA
## 3 Unskilled High NA
## 4 Skilled Low NA
## 5 Skilled Low NA
## 6 Unskilled Low NA
We observe that the column names are shifted down below because of the empty line. So, we must make sure to use the command skip and set the header to true.
newdata=read.csv(file="data/creditriskorg.csv")
head(newdata)
## Loan.Purpose Checking Savings Months.Customer Months.Employed
## 1 Small Appliance $- $739.00 13 12
## 2 Furniture $- $1,230.00 25 0
## 3 New Car $- $389.00 19 119
## 4 Furniture $638.00 $347.00 13 14
## 5 Education $963.00 $4,754.00 40 45
## 6 Furniture $2,827.00 $- 11 13
## Gender Marital.Status Age Housing Years Job Credit.Risk X X.1
## 1 M Single 23 Own 3 Unskilled Low NA
## 2 M Divorced 32 Own 1 Skilled High NA Age
## 3 M Single 38 Own 4 Management High NA
## 4 M Single 36 Own 2 Unskilled High NA
## 5 M Single 31 Rent 3 Skilled Low NA
## 6 M Married 25 Own 1 Skilled Low NA
To calculate the mean for Checking in R, follow Worksheet 2. Extract the Checking column first and then find the average using the function built in R. What happens when we try to use the function?
checking = newdata$Checking
meanChecking=mean(checking)
## Warning in mean.default(checking): argument is not numeric or logical:
## returning NA
meanChecking
## [1] NA
To resolve the error, we must understand where it is coming from and correct for. There are missing values in the csv file, which is quite common as most datasets are not perfect. Additionally, there are commas within the excel spreadsheet, and R does not recognize that ‘1,234’ is equivalent to ‘1234’. Lastly, there are ‘$’ symbols throughout the file which is not a numerical symbol either.
The sub function replaces these symbols with something else. So, in order to remove the comma in the number “1,234”, we must substitute it with just an empty space.
As shown on the worksheet, type and copy the exact commands to find the mean with the NA values removed.
checking = newdata$Checking
#substitute comma with blank in all of checking. Below are examples using a hypothetical variable name 'new'.
# Example new = sub(",","",new)
checking=sub(",","",checking)
#substitute dollar sign with blank in all of checking
# Example new = sub("\\$","",new)
checking=sub("\\$","",checking)
#Convert values to numeric to remove any NA
# Example new = as.numeric(new)
checking=as.numeric(checking)
## Warning: NAs introduced by coercion
#Calculate mean of checking with NA removed
# Example mean(new,na.rm=TRUE)
mean(checking,na.rm=TRUE)
## [1] 2559.805
meanChecking
## [1] NA
What are some other ways to clean this data? How about Excel? How does Excel treat the missing values and the “$” symbols?
Now, we will look at Chicago taxi data. Go and explore the interactive dashboard and read the description of the data.
Chicago Taxi Dashboard: https://data.cityofchicago.org/Transportation/Taxi-Trips-Dashboard/spcw-brbq
Chicago Taxi Data Description: http://digital.cityofchicago.org/index.php/chicago-taxi-data-released/
– Open in RStudio or Excel the taxi trips sample csv file located in the data folder. Note the size of the file, the number of columns and of rows here. Identify the unique entities, and fields in the data. The size of the file is 39.7 MB. There is 24 columns and 1000000 rows including header. The unique entity is trip id. The rest of the headers are the fields.
mydata = read.csv(file="data/Taxi_Trips_Sample.csv")
head(mydata)
## Trip.ID
## 1 3e7d6d8ccf1425ae1dcd584f5c3ca303cf6362ed
## 2 3e7d6e5c4e87f01a475c8200b33777e85497da89
## 3 3e7d6e69c1d6755d9e7484a453cd93a3ee9fed4c
## 4 3e7d6efe43222b0ebc698583916674c648dd4520
## 5 3e7d6f001e9bcda8478a489cb53293d26328ac85
## 6 3e7d6f2a03527d63dc01b95e829fdfdd706102da
## Taxi.ID
## 1 b47c583b142d75b42882975eaab19c6cb98d82686016576cce6e305b1b99eb16aacfb9a21ff61c84873a6c3dde282756c162c538c8b69554fd8f811f3a8f60a2
## 2 bc1c0381e3bca623e6c04f3410f7b67201a9fc85c6b66d0f420a88099d38448f9b9874e246da49cf2ef32ea3d027eec9c5b484fe77dbfc033c389b5576ac66bd
## 3 f529487ccf3a5d538cd246342379d54314e90dc6c573f94a72f5c2238189c5131e12c1e493c71ccdaed6751d13f53fa1d8b51a1591c48891dc6beb3f9df3a18f
## 4 0f831bff43d83f396f2e4950126c6137dcdb60fb4c8580ffe860203747a83a789b22f2f9e4fbdd0dd8ed8c310366d8935228ddbcadf708fb9691ca5dd1b6c802
## 5 e5274d6c103515af3ce705182d0bbbea7ca077a6f23b1736254f2de8ba3e1687dd77f5fb541b7f00b1ebc24cfde54caf5a9562f046a0559acbfe1e7159e17c1a
## 6 329d9f0b72ce0fea6c2cc7ea3347924c11d98702e5cc39eacf6252038304bb019457c905f38066a2f9aa4b8732ac30ea22d8740e177a314ef7b0327ad5766cee
## Trip.Start.Timestamp Trip.End.Timestamp Trip.Seconds Trip.Miles
## 1 08/19/2014 02:45:00 PM 08/19/2014 02:45:00 PM 480 0.15
## 2 09/23/2013 05:15:00 PM 09/23/2013 05:15:00 PM 420 0.00
## 3 02/16/2014 10:15:00 PM 02/16/2014 10:30:00 PM 420 1.70
## 4 05/10/2013 08:00:00 PM 05/10/2013 08:45:00 PM 2340 13.80
## 5 02/21/2016 07:15:00 PM 02/21/2016 07:15:00 PM 300 0.70
## 6 12/10/2015 05:30:00 PM 12/10/2015 06:00:00 PM 1020 1.40
## Pickup.Census.Tract Dropoff.Census.Tract Pickup.Community.Area
## 1 17031280100 17031839100 28
## 2 17031081800 17031281900 8
## 3 NA NA 6
## 4 17031980000 17031060400 76
## 5 17031081500 17031081500 8
## 6 NA NA NA
## Dropoff.Community.Area Fare Tips Tolls Extras Trip.Total Payment.Type
## 1 32 $7.05 $0.00 $0.00 $1.50 $8.55 Cash
## 2 28 $6.05 $0.00 $0.00 $0.00 $6.05 Cash
## 3 4 $7.05 $0.00 $0.00 $0.00 $7.05 Cash
## 4 6 $31.25 $0.00 $0.00 $3.00 $34.25 Cash
## 5 8 $5.50 $0.00 $0.00 $0.00 $5.50 Cash
## 6 NA $9.25 $0.00 $0.00 $1.00 $10.25 Cash
## Company Pickup.Centroid.Latitude
## 1 41.88530
## 2 Taxi Affiliation Services 41.89322
## 3 Taxi Affiliation Services 41.94423
## 4 41.97907
## 5 41.89251
## 6 NA
## Pickup.Centroid.Longitude Pickup.Centroid.Location
## 1 -87.64281 POINT (-87.642808 41.8853)
## 2 -87.63784 POINT (-87.637844 41.893216)
## 3 -87.65600 POINT (-87.655998 41.944227)
## 4 -87.90304 POINT (-87.90304 41.979071)
## 5 -87.62621 POINT (-87.626215 41.892508)
## 6 NA
## Dropoff.Centroid.Latitude Dropoff.Centroid.Longitude
## 1 41.88099 -87.63275
## 2 41.87926 -87.64265
## 3 41.97517 -87.68752
## 4 41.95067 -87.66654
## 5 41.89251 -87.62621
## 6 NA NA
## Dropoff.Centroid..Location Community.Areas
## 1 POINT (-87.632746 41.880994) 29
## 2 POINT (-87.642649 41.879255) 37
## 3 POINT (-87.687516 41.975171) 57
## 4 POINT (-87.666536 41.950673) 75
## 5 POINT (-87.626215 41.892508) 37
## 6 NA
– Define a relational business logic integrity check for the column field ‘Trip Seconds’.
The trip start time and end time equals to the trip seconds which means the difference between start time and end time cannot be greater than the trip second.
– Using https://erdplus.com/#/standalone draw a star like schema using at least the following tables: