What do your rules look like? editrules 1.8-x answers with the help of igraph

We (Edwin de Jonge and me) have recently updated our editrules package. The most important new features include (beta) support for categorical data. However, in this
post I’m going to show some visualizations we included, made possible by Gabor Csardi’s awesome igraph package.

Make sure you run

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install.packages(c('igraph','editrules'))

before trying the code below.

First, let’s load editrules’ built-in editset:

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> data(edits)
> edits
  name          edit               description
1   b1   t == ct + p             total balance
2   b2 ct == ch + cp              cost balance
3   s1    p <= 0.6*t             profit sanity
4   s2   cp <= 0.3*t     personnel cost sanity
5   s3   ch <= 0.3*t       housing cost sanity
6   p1          t >0       turnover positivity
7   p2        ch > 0   housing cost positivity
8   p3        cp > 0 personnel cost positivity
9   p4        ct > 0     total cost positivity

Here, edits is a data.frame with a “name” column, naming the rules, an “edit” column, with a character representation of the edit rules and a “description” column. The variables have the following meaning: t: turnover, ct total cost, p profit, ch housing costs and cp personnel cost. The rules demand balance accounts to add up (e.g.cost + profit equals turnover) and demand some sanity checks (e.g. profit can not exceed 60% of turnover).

The sanity checks here are completely fictional. To do anything useful with these rules, turn them into an editmatrix.

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> (E <- editmatrix(edits))
Edit matrix:
   ct  p    t ch cp Ops CONSTANT
b1 -1 -1  1.0  0  0  ==        0
b2  1  0  0.0 -1 -1  ==        0
s1  0  1 -0.6  0  0  <=        0
s2  0  0 -0.3  0  1  <=        0
s3  0  0 -0.3  1  0  <=        0
p1  0  0 -1.0  0  0   <        0
p2  0  0  0.0 -1  0   <        0
p3  0  0  0.0  0 -1   <        0
p4 -1  0  0.0  0  0   <        0
 
Edit rules:
b1 : t == ct + p [ total balance ]
b2 : ct == ch + cp [ cost balance ]
s1 : p <= 0.6*t [ profit sanity ]
s2 : cp <= 0.3*t [ personnel cost sanity ]
s3 : ch <= 0.3*t [ housing cost sanity ]
p1 : 0 < t [  turnover positivity ]
p2 : 0 < ch [ housing cost positivity ]
p3 : 0 < cp [ personnel cost positivity ]
p4 : 0 < ct [ total cost positivity ]

Although the matrix representation and the textual representation have their merits, it is hard to see which rules are (indirectly) related via shared variables. This may be visualized by plotting the rules in a graph, where each variable and edit is a node, and a variable node is connected with an editnode if the variable occurs in the edits. Just do

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plot(E)

to get the plot below.

The round, blue nodes represent variables and the square nodes represent edit rules.
You can see at a glance that everything is connected, so the editmatrix does not block into submatrices. If you want to leave out the variables, and just see how the edits are connected, use

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plot(E,nodetype='rules')

to get

Here, (slightly) thicker lines indicate that more variables are shared. Plotting connections between variables can be done with

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plot(E,nodetype='vars')

Which you can try for yourself. We can do cooler stuff. For example, lets define a faulty record and detect which rules it violates.

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> r <- c(ct = 100, ch = 30, cp = 70, p=30,t=130 ) 
> (v <- violatedEdits(E,r))
      edit
record    b1    b2    s1   s2    s3    p1    p2    p3    p4
     1 FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE

So only edit s2 is violated. We can visualize this as well.

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plot(E,violated=v)

gives

Edit graph with violated edits colored

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adapt <- errorLocalizer(E,r)$searchBest()$adapt
plot(E,
    violated=v,
    adapt=adapt
)

Edit graph, with violated edits and variables to adapt colored red

If you don’t like the colors or want to play with the igraph objects yourself, see the as.igraph or adjacency functions.

Oh, and if you wander which are the possible values to use for p and t, just substitute all the other values in the editmatrix:

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> substValue(E,names(r)[!adapt],r[!adapt])
Edit matrix:
   ct  p    t ch cp Ops CONSTANT
b1  0 -1  1.0  0  0  ==      100
s1  0  1 -0.6  0  0  <=        0
s2  0  0 -0.3  0  0  <=      -70
s3  0  0 -0.3  0  0  <=      -30
p1  0  0 -1.0  0  0   <        0
 
Edit rules:
b1 : t == p + 100 
s1 : p <= 0.6*t 
s2 : 70 <= 0.3*t 
s3 : 30 <= 0.3*t 
p1 : 0 < t

The solution set to the above system of equations is the set of possible values for t and p.