`m = margin(tree,TBL,ResponseVarName)`

`m = margin(tree,TBL,Y)`

`m = margin(tree,X,Y)`

returns
the classification margins for the table of predictors `m`

= margin(`tree`

,`TBL`

,`ResponseVarName`

)`TBL`

and
class labels `TBL.ResponseVarName`

. For the definition,
see Margin.

The classification *margin* is the difference
between the classification *score* for the true
class and maximal classification score for the false classes. Margin
is a column vector with the same number of rows as in the matrix `X`

.

For trees, the *score* of a classification
of a leaf node is the posterior probability of the classification
at that node. The posterior probability of the classification at a
node is the number of training sequences that lead to that node with
the classification, divided by the number of training sequences that
lead to that node.

For example, consider classifying a predictor `X`

as `true`

when `X`

< `0.15`

or `X`

> `0.95`

, and `X`

is
false otherwise.

Generate 100 random points and classify them:

rng(0,'twister') % for reproducibility X = rand(100,1); Y = (abs(X - .55) > .4); tree = fitctree(X,Y); view(tree,'Mode','Graph')

Prune the tree:

tree1 = prune(tree,'Level',1); view(tree1,'Mode','Graph')

The pruned tree correctly classifies observations that are less
than 0.15 as `true`

. It also correctly classifies
observations from .15 to .94 as `false`

. However,
it incorrectly classifies observations that are greater than .94 as `false`

.
Therefore, the score for observations that are greater than .15 should
be about .05/.85=.06 for `true`

, and about .8/.85=.94
for `false`

.

Compute the prediction scores for the first 10 rows of `X`

:

[~,score] = predict(tree1,X(1:10)); [score X(1:10,:)]

ans = 0.9059 0.0941 0.8147 0.9059 0.0941 0.9058 0 1.0000 0.1270 0.9059 0.0941 0.9134 0.9059 0.0941 0.6324 0 1.0000 0.0975 0.9059 0.0941 0.2785 0.9059 0.0941 0.5469 0.9059 0.0941 0.9575 0.9059 0.0941 0.9649

Indeed, every value of `X`

(the right-most
column) that is less than 0.15 has associated scores (the left and
center columns) of `0`

and `1`

,
while the other values of `X`

have associated scores
of `0.91`

and `0.09`

. The difference
(score `0.09`

instead of the expected `.06`

)
is due to a statistical fluctuation: there are `8`

observations
in `X`

in the range `(.95,1)`

instead
of the expected `5`

observations.

Compute the classification margin for the Fisher iris data, trained on its first two columns of data, and view the last 10 entries.

```
load fisheriris
X = meas(:,1:2);
tree = fitctree(X,species);
M = margin(tree,X,species);
M(end-10:end)
```

ans = 0.1111 0.1111 0.1111 -0.2857 0.6364 0.6364 0.1111 0.7500 1.0000 0.6364 0.2000

The classification tree trained on all the data is better.

tree = fitctree(meas,species); M = margin(tree,meas,species); M(end-10:end)

ans = 0.9565 0.9565 0.9565 0.9565 0.9565 0.9565 0.9565 0.9565 0.9565 0.9565 0.9565

Was this topic helpful?