Negative Payments in Local Spending Data

August 17, 2013

(This article was first published on OUseful.Info, the blog... » Rstats, and kindly contributed to R-bloggers)

In anticipation of a new R library from School of Data data diva @mihi_tr that will wrap the OpenSpending API and providing access to data directly from within R, I thought I’d start doodling around some ideas raised in Identifying Pieces in the Spending Data Jigsaw. In particular, common payment values, repayments/refunds and “balanced payments”, that is, multiple payments where the absolute value of a negative payment matches that of an above zero payment (so far example, -£269.72 and £269.72 would be balanced payments).

The data I’ve grabbed is Isle of Wight Council transparency (spending) data for the financial year 2012/2013. The data was pulled from the Isle of Wight Council website and cleaned using OpenRefine broadly according to the recipe described in Using OpenRefine to Clean Multiple Documents in the Same Way with a couple of additions: a new SupplierNameClustered column, originally created from the SupplierName column, but then cleaned to remove initials, (eg [SD]), direct debit prefixes (DD- etc) and then clustered using a variety of clustering algorithms; and a new unique index column, created with values '_'+row.index. You can find a copy of the data here.

To start with, let’s see how we can identify “popular” transaction amounts:

#We going to use some rearrangement routines...

#I'm loading in the data from a locally downloaded copy
iw <- read.csv("~/code/Rcode/eduDemos1/caseStudies/IWspending/IWccl2012_13_TransparencyTest4.csv")

#Display most common payment values

#     Var1 Freq
#1 1567.72 2177
#2 1930.32 1780
#3  1622.6 1347
#4 1998.08 1253
#5  1642.2 1227

This shows that the most common payment by value is for £1567.72. An obvious question to ask here is: does this correspond to some sort of “standard payment” or tariff? And if so, can we reconcile this against the extent of the delivery of a particular level of service, perhaps as disclosed elsewhere?

We can also generate a summary report to show what transactions correspond to this amount, or the most popular amounts.

#We can then get the rows corresponding to these common payments
  df[abs(df[[bcol]]) %in% commonAmounts,]

More usefully, we might generate “pivot table” style summaries of how the popular payments breakdown with respect to expenses area, supplier, or some other combination of factors. So for example, we can learn that there were 1261 payments of £1567.72 booked to the EF Residential Care services area, and SOMERSET CARE LTD [SB] received 231 payments of £1567.72. (Reporting by the clustered supplier name is often more useful…)

#R does pivot tables, sort of...
#We can also run summary statistics over those common payment rows.
zz1=aggregate(index ~ ServiceArea + Amount, data =cp.df, FUN="length")

#                       ServiceArea  Amount index
#1              EF Residential Care 1567.72  1261
#2             EMI Residential Care 1642.20   659
#3             EMI Residential Care 1930.32   645
#4              EF Residential Care 1930.32   561
#5              EF Residential Care 1622.60   530
#6 Elderly Frail Residential Income 1622.60   36

zz2=aggregate(index ~ ServiceArea +ExpensesType+ Amount, data =cp.df, FUN="length")
#                       ServiceArea                       ExpensesType  Amount
#1              EF Residential Care                Chgs from Ind Provs 1567.72
#2             EMI Residential Care                Chgs from Ind Provs 1930.32
#3             EMI Residential Care                Chgs from Ind Provs 1642.20
#4              EF Residential Care Charges from Independent Providers 1622.60

zz3=aggregate(index ~ SupplierName+ Amount, data =cp.df, FUN="length")
#                SupplierName  Amount index
#1     REDACTED PERSONAL DATA 1567.72   274
#2     SOMERSET CARE LTD [SB] 1567.72   231
#3 ISLAND HEALTHCARE LTD [SB] 1930.32   214

zz4=aggregate(index ~ SupplierNameClustered+ Amount, data =cp.df, FUN="length")

If I was a suspicious type, I suppose I might look for suppliers who received one of these popular amounts only once…

Let’s have a search for items declared as overpayments or refunds, perhaps as preliminary work in an investigation about why overpayments are being made…:

#Let's look for overpayments
  df=subset(df, grepl('((overpay)|(refund))', df[[fcol]],
  df[ order(df[,bcol]), ]

#View the largest "refund" transactions
#        Date                           ServiceArea         ExpensesType    Amount
#1 2013-03-28 Elderly Mentally Ill Residential Care     Provider Refunds -25094.16
#2 2012-07-18                   MH Residential Care Refund of Overpaymts -24599.12
#3 2013-03-25 Elderly Mentally Ill Residential Care     Provider Refunds -23163.84

#We can also generate a variety of other reports on the "refund" transactions
head(arrange(aggregate(index ~ SupplierName+ ExpensesType+Amount, data =rf, FUN="length"),-index))
#                 SupplierName     ExpensesType   Amount index
#1 THE ORCHARD HOUSE CARE HOME Provider Refunds  -434.84     8
#2 THE ORCHARD HOUSE CARE HOME Provider Refunds  -729.91     3
#3         SCIO HEALTHCARE LTD Provider Refunds  -434.84     3
##Which is to say: The Orchard House Care home made 8 refunds of £434.84 and 3 of £729.91

head(arrange(aggregate(index ~ SupplierName+ ExpensesType, data =rf, FUN="length"),-index))
#                 SupplierName     ExpensesType index
#1         SCIO HEALTHCARE LTD Provider Refunds    31
#2           SOMERSET CARE LTD Provider Refunds    31
#3 THE ORCHARD HOUSE CARE HOME Provider Refunds    22

Another area of investigation might be “balanced” payments. Here’s one approach for finding those, based around first identifying negative payments. Let’s also refine the approach in this instance for looking for balanced payments involving the supplier involved in the largest number of negative payments.

##Find which suppliers were involved in most number of negative payments
#Identify negative payments
#Count the number of each unique supplier
#Display the suppliers with the largest count of negative payments

#                    Var1 Freq
#4      SOMERSET CARE LTD   75

Let’s look for balanced payments around SOUTHERN ELECTRIC…

#Specific supplier search
#Limit transactions to just transactions involving this supplier
se=subset(iw, SupplierNameClustered=='SOUTHERN ELECTRIC')

##We can also run partial matching searches...
#sw=subset(iw,grepl('SOUTHERN', iw$SupplierNameClustered))

#Now let's search for balanced payments
  #Find the positive amounts
  positems=df[ df[[bcol]]>0, ]
  #Find the negative amounts
  negitems=df[ df[[bcol]]<=0, ]
  #Find the absolute unique negative amounts
  #Find the unique positive amounts
  #Find matching positive and negative amounts
  #Subset the data based on balanced positive and negative amounts
  #bals=subset(se,abs(Amount) %in% balitems)
  bals=df[abs(df[[bcol]]) %in% balitems,]
  #Group the data by sorting, largest absolute amounts first
  bals[ order(-abs(bals[,bcol])), ]


#                              Directorate       Date                ServiceArea ExpensesType   Amount
#20423               Economy & Environment 2012-07-11        Sandown Depot (East  Electricity -2770.35
#20424               Economy & Environment 2012-07-11        Sandown Depot (East  Electricity  2770.35
#52004 Chief Executive, Schools & Learning 2013-01-30 Haylands - Playstreet Lane  Electricity  2511.20
#52008 Chief Executive, Schools & Learning 2013-01-31 Haylands - Playstreet Lane  Electricity -2511.20

We can also use graphical techniques to try to spot balanced payments.

#Crude example plot to highlight matched +/1 amounts
se1a=subset(se,ServiceArea=='Ryde Town Hall' & Amount>0)
se1b=subset(se,ServiceArea=='Ryde Town Hall' & Amount<=0)

g=ggplot() + geom_point(data=se1a, aes(x=Date,y=Amount),pch=1,size=1)
g=g+geom_point(data=se1b, aes(x=Date,y=-Amount),size=3,col='red',pch=2)
g=g+ggtitle('Ryde Town Hall - Energy Payments to Southern Electric (red=-Amount)')+xlab(NULL)+ylab('Amount (£)')
g=g+theme(axis.text.x = element_text(angle = 45, hjust = 1))

In this first image, we see payments over time – the red markers are “minus” amounts on the negative payments. Notice that in some cases balanced payments seem to appear on the same day. If the y-value of a red and a black marker are the same, they are balanced in value. The x-axis is time. Where there is a period of equally spaced marks over x with the same y-value, this may represent a regular scheduled payment.

ryde - energy

g=ggplot(se1a)+geom_point( aes(x=Date,y=Amount),pch=1,size=1)
g=g+geom_point(data=se1b, aes(x=Date,y=-Amount),size=3,col='red',pch=2)
g=g+ggtitle('Ryde Town Hall - Energy Payments to Southern Electric (red=-Amount)')
g=g+xlab(NULL)+ylab('Amount (£)')
g=g+theme(axis.text.x = element_text(angle = 45, hjust = 1))

We can also facet out the payments by expense type.

ryde-energy facet

Graphical techniques can often help us spot patterns in the data that might be hard to spot just by looking at rows and columns worth of data. In many cases, it is worthwhile developing visual analysis skills to try out quick analyses “by eye” before trying to implement them as more comprehensive analysis scripts.

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