Integration of adaptive neural fuzzy inference system and fuzzy rough set theory with support vector regression to urban growth modelling

Abstract

Land change models are amongst the most widely developed tools for spatial decision support. Despite this progress, only a few models have been created thus far that simulate urban growth that incorporate two important aspects of uncertainty inherent to land use dynamics: fuzziness and roughness. Combining fuzziness and roughness into models will enhance the use of these tools for decision support. This study applied and evaluated a fuzzy-based approach to the feature selection effects on the accuracy of a land change model. Fuzzy rough set theory (FRST) was employed here as feature selection method and was integrated with a support vector regression (SVR) algorithm to simulate urban growth of Tabriz mega city in northwest Iran. In order to apply feature selection to a FRST algorithm, incoming data has been first fuzzified by an adaptive neural fuzzy inference system (ANFIS). To evaluate the application of FRST, SVR was used with and without FRST (SVR and SVR-FRST), while for performance evaluation logistic regression (LR) and kernelled LR (KLR) models were integrated with and without FRST (LR, LR-FRST, KLR, and KLR-FRST). The accuracy of the simulated maps of all models were evaluated by calculating the overall accuracy (OA), true positive rate (TPR), true negative rate (TNR), total operating characteristic (TOC) and their area under curve (AUC). The results showed that integrating FRST with the above-mentioned models enhanced the overall performances based on the above criteria. Among the above mentioned models, SVM-FRST and KLR-FRST yielded the best goodness of fit measures. Moreover, SVM-FRST with 83.6% OA, 41.6% TPR, and 90.4% TNR performs better than KLR-FRST with 82.4% OA, 37.4% TPR, and 89.8% TNR. However, KLR-FRST has more AUC, less green area destruction, more barren to urban areas conversion, and fast tuning process related to SVR-FRST. Finally, we suggest that KLR-FRST and SVR-FRST are, among those evaluated, the most appropriate models for urban growth modelling of the Tabriz mega city of Iran when considering uncertainty.

Appendix

An illustrative example of dependency coefficients of FRST calculations (Jensen 2005)

Suppose the following table as an example with three conditional features (a, b, and c) and a decision field (d). U is a non-empty set of finite objects (the universe) and A is a nonempty finite set of attributes (a, b, and c), with any T ⊆ A.

The conditional features is fuzzified towards two classes (P and Q) as shown in the following table

Suppose X = {0}, F = P_a and y = {1, 2, 3, 4, 5, 6}

Based on the following equation,

$$ {\mu}_{{\underset{\_}{apr}}_R(X)}{(x)}_{x\epsilon U}={\mathit{\sup}}_{F\epsilon U/T}\ \mathit{\min}\left({\mu}_{P_a}(x),{\mathit{\operatorname{inf}}}_{y\epsilon U}\ \mathit{\max}\left\{1-{\mu}_{P_a}(y),{\mu}_X(y)\right\}\right) $$

We calculate the values of the last part of the equation for y = {1, 2, 3, 4, 5, 6}:

$$ y=1\to \mathit{\max}\left\{1-{\mu}_{P_a}(1),{\mu}_X(1)\right\}=\mathit{\max}\left\{1-0.8,1\right\}=\mathit{\max}\left\{0.2,1\right\}=1 $$

$$ y=2\to \mathit{\max}\left\{1-{\mu}_{P_a}(2),{\mu}_X(2)\right\}=\mathit{\max}\left\{1-0.8,0\right\}=\mathit{\max}\left\{0.2,0\right\}=0.2 $$

$$ y=3\to \mathit{\max}\left\{1-{\mu}_{P_a}(3),{\mu}_X(3)\right\}=\mathit{\max}\left\{1-0.6,1\right\}=\mathit{\max}\left\{0.4,1\right\}=1 $$

$$ y=4\to \mathit{\max}\left\{1-{\mu}_{P_a}(4),{\mu}_X(4)\right\}=\mathit{\max}\left\{1-0,0\right\}=\mathit{\max}\left\{1,0\right\}=1 $$

$$ y=5\to \mathit{\max}\left\{1-{\mu}_{P_a}(5),{\mu}_X(5)\right\}=\mathit{\max}\left\{1-0,0\right\}=\mathit{\max}\left\{1,0\right\}=1 $$

$$ y=6\to \mathit{\max}\left\{1-{\mu}_{P_a}(6),{\mu}_X(6)\right\}=\mathit{\max}\left\{1-0,1\right\}=\mathit{\max}\left\{1,1\right\}=1 $$

(So,) ${\mathit{\operatorname{inf}}}_{y\epsilon U}\ \mathit{\max}\left\{1-{\mu}_{P_a}(y),{\mu}_X(y)\right\}=\mathit{\operatorname{inf}}\left\{\mathrm{1,0.2,1},1,1,1\right\}=0.2$

x = {1, 2, 3, 4, 5, 6}:

$$ x=1\to \mathit{\min}\left({\mu}_{P_a}(1),0.2\right)=\mathit{\min}\left(0.8,0.2\right)=0.2 $$

$$ x=2\to \mathit{\min}\left({\mu}_{P_a}(2),0.2\right)=\mathit{\min}\left(0.8,0.2\right)=0.2 $$

$$ x=3\to \mathit{\min}\left({\mu}_{P_a}(3),0.2\right)=\mathit{\min}\left(0.6,0.2\right)=0.2 $$

$$ x=4\to \mathit{\min}\left({\mu}_{P_a}(4),0.2\right)=\mathit{\min}\left(0,0.2\right)=0.2 $$

$$ x=5\to \mathit{\min}\left({\mu}_{P_a}(5),0.2\right)=\mathit{\min}\left(0,0.2\right)=0.2 $$

$$ x=6\to \mathit{\min}\left({\mu}_{P_a}(6),0.2\right)=\mathit{\min}\left(0,0.2\right)=0.2 $$

Suppose X = {0}, F = Q_a and y = {1, 2, 3, 4, 5, 6}

Based on the following equation,

$$ {\mu}_{{\underset{\_}{apr}}_R(X)}{(x)}_{x\epsilon U}={\mathit{\sup}}_{F\epsilon U/T}\ \mathit{\min}\left({\mu}_{Z_a}(x),{\mathit{\operatorname{inf}}}_{y\epsilon U}\ \mathit{\max}\left\{1-{\mu}_{Q_a}(y),{\mu}_X(y)\right\}\right) $$

We calculate the values of the last part of the equation for y = {1, 2, 3, 4, 5, 6}:

$$ y=1\to \mathit{\max}\left\{1-{\mu}_{Q_a}(1),{\mu}_X(1)\right\}=\mathit{\max}\left\{1-0.2,1\right\}=\mathit{\max}\left\{0.8,1\right\}=1 $$

$$ y=2\to \mathit{\max}\left\{1-{\mu}_{Q_a}(2),{\mu}_X(2)\right\}=\mathit{\max}\left\{1-0.2,0\right\}=\mathit{\max}\left\{0.8,0\right\}=0.8 $$

$$ y=3\to \mathit{\max}\left\{1-{\mu}_{Q_a}(3),{\mu}_X(3)\right\}=\mathit{\max}\left\{1-0.4,1\right\}=\mathit{\max}\left\{0.6,1\right\}=1 $$

$$ y=4\to \mathit{\max}\left\{1-{\mu}_{Q_a}(4),{\mu}_X(4)\right\}=\mathit{\max}\left\{1-0.4,0\right\}=\mathit{\max}\left\{0.6,0\right\}=0.6 $$

$$ y=5\to \mathit{\max}\left\{1-{\mu}_{Q_a}(5),{\mu}_X(5)\right\}=\mathit{\max}\left\{1-0.6,0\right\}=\mathit{\max}\left\{0.4,0\right\}=0.4 $$

$$ y=6\to \mathit{\max}\left\{1-{\mu}_{Q_a}(6),{\mu}_X(6)\right\}=\mathit{\max}\left\{1-0.6,1\right\}=\mathit{\max}\left\{0.4,1\right\}=1 $$

$$ {\mathit{\operatorname{inf}}}_{y\epsilon U}\ \mathit{\max}\left\{1-{\mu}_{Q_a}(y),{\mu}_X(y)\right\}=\mathit{\operatorname{inf}}\left\{\mathrm{1,0.8,1},\mathrm{0.6,0.4,1}\right\}=0.4 $$

x = {1, 2, 3, 4, 5, 6}:

$$ x=1\to \mathit{\min}\left({\mu}_{Q_a}(1),0.2\right)=\mathit{\min}\left(0.2,0.4\right)=0.2 $$

$$ x=2\to \mathit{\min}\left({\mu}_{Q_a}(2),0.2\right)=\mathit{\min}\left(0.2,0.4\right)=0.2 $$

$$ x=3\to \mathit{\min}\left({\mu}_{Q_a}(3),0.2\right)=\mathit{\min}\left(0.4,0.4\right)=0.4 $$

$$ x=4\to \mathit{\min}\left({\mu}_{Q_a}(4),0.2\right)=\mathit{\min}\left(0.4,0.4\right)=0.4 $$

$$ x=5\to \mathit{\min}\left({\mu}_{Q_a}(5),0.2\right)=\mathit{\min}\left(0.6,0.4\right)=0.4 $$

$$ x=6\to \mathit{\min}\left({\mu}_{Q_a}(6),0.2\right)=\mathit{\min}\left(0.6,0.4\right)=0.4 $$

Now,${\mu}_{{\underset{\_}{apr}}_R(X)}{(x)}_{x\epsilon U}$ can be calculated for x = {1, 2, 3, 4, 5, 6}:

$$ x=1\to {\mu}_{{\underset{\_}{apr}}_R(X)}{(1)}_{x\epsilon U}=\mathit{\sup}\left\{\mathrm{0.2,0.2}\right\}=0.2 $$

$$ x=2\to {\mu}_{{\underset{\_}{apr}}_R(X)}{(2)}_{x\epsilon U}=\mathit{\sup}\left\{\mathrm{0.2,0.2}\right\}=0.2 $$

$$ x=3\to {\mu}_{{\underset{\_}{apr}}_R(X)}{(3)}_{x\epsilon U}=\mathit{\sup}\left\{\mathrm{0.2,0.4}\right\}=0.4 $$

$$ x=4\to {\mu}_{{\underset{\_}{apr}}_R(X)}{(4)}_{x\epsilon U}=\mathit{\sup}\left\{\mathrm{0.2,0.4}\right\}=0.4 $$

$$ x=5\to {\mu}_{{\underset{\_}{apr}}_R(X)}{(5)}_{x\epsilon U}=\mathit{\sup}\left\{\mathrm{0.2,0.4}\right\}=0.4 $$

$$ x=6\to {\mu}_{{\underset{\_}{apr}}_R(X)}{(6)}_{x\epsilon U}=\mathit{\sup}\left\{\mathrm{0.2,0.4}\right\}=0.4 $$

Same as above calculations, ${\mu}_{{\underset{\_}{apr}}_R(X)}{(x)}_{x\epsilon U}$ can be calculated for X = {1}:

$$ {\mu}_{{\underset{\_}{apr}}_R(X)}{(1)}_{x\epsilon U}=0.2 $$

$$ {\mu}_{{\underset{\_}{apr}}_R(X)}{(2)}_{x\epsilon U}=0.2 $$

$$ {\mu}_{{\underset{\_}{apr}}_R(X)}{(3)}_{x\epsilon U}=0.4 $$

$$ {\mu}_{{\underset{\_}{apr}}_R(X)}{(4)}_{x\epsilon U}=0.4 $$

$$ {\mu}_{{\underset{\_}{apr}}_R(X)}{(5)}_{x\epsilon U}=0.4 $$

$$ {\mu}_{{\underset{\_}{apr}}_R(X)}{(6)}_{x\epsilon U}=0.4 $$

Then, ${\mu}_{POS_a}(x)=\mathit{\sup}{\mu}_{{\underset{\_}{apr}}_R(X)}$ for feature ‘a’ can be calculated:

$$ {\mu}_{POS_a}(1)=\mathit{\sup}{\mu}_{{\underset{\_}{apr}}_R(X)}(1)=\mathit{\sup}\left\{\mathrm{0.2,0.2}\right\}=0.2 $$

$$ {\mu}_{POS_a}(2)=\mathit{\sup}{\mu}_{{\underset{\_}{apr}}_R(X)}(2)=\mathit{\sup}\left\{\mathrm{0.2,0.2}\right\}=0.2 $$

$$ {\mu}_{POS_a}(3)=\mathit{\sup}{\mu}_{{\underset{\_}{apr}}_R(X)}(3)=\mathit{\sup}\left\{\mathrm{0.4,0.4}\right\}=0.4 $$

$$ {\mu}_{POS_a}(4)=\mathit{\sup}{\mu}_{{\underset{\_}{apr}}_R(X)}(4)=\mathit{\sup}\left\{\mathrm{0.4,0.4}\right\}=0.4 $$

$$ {\mu}_{POS_a}(5)=\mathit{\sup}{\mu}_{{\underset{\_}{apr}}_R(X)}(5)=\mathit{\sup}\left\{\mathrm{0.4,0.4}\right\}=0.4 $$

$$ {\mu}_{POS_a}(6)=\mathit{\sup}{\mu}_{{\underset{\_}{apr}}_R(X)}(6)=\mathit{\sup}\left\{\mathrm{0.4,0.4}\right\}=0.4 $$

So, the dependency coefficient of feature ‘a’ is:

$$ {\gamma}_a=\frac{\sum_{x\epsilon U}{\mu}_{POS_{P(Q)}(x)}}{\left|U\right|}=\frac{0.2+0.2+0.4+0.4+0.4+0.4}{6}=\frac{2}{6} $$

Same as above calculations, the dependency coefficients of feature ‘b’ and ‘c’ are:

$$ {\gamma}_b=\frac{2.4}{6},\kern0.75em {\gamma}_c=\frac{1.6}{6} $$

Because of γ_b is the greatest one, so feature ‘b’ is selected in the first level of feature reduction. In the next level, the dependency coefficients of the combinations of ‘b’ with the other features ‘a’ and ‘c’ are calculated:

$$ {\gamma}_{\left\{a,b\right\}}=\frac{3.4}{6},\kern0.75em {\gamma}_{\left\{b,c\right\}}=\frac{3.2}{6} $$

Because of γ_{a, b} is the greatest one, so feature combination {a, b} is selected in the second level of feature reduction

In the third level, the dependency coefficient of the combination {a, b, c} is calculated:

$$ {\gamma}_{\left\{a,b,c\right\}}=\frac{3.4}{6} $$

Because of γ_{{a, b, c}} = γ_{a, b} then {a, b} is the final and minimal set of selected features and feature ‘c’ is removed