A numerical example

This example shall demonstrate the high efficiency of the RANK method. It is shown in Table 5.2, where the results are also summarized. The procedure starts by calculating the resemblance matrix of species, R. This implies

Table 5.2 Data set for illustrating the RANK algorithm.

relevé

1

2

3

4

rank no.

expl. variance, %

species

1

2

2

1

4.

0.0

species

2

2

1

1

2.

17.0

species

3

1

1

1.

78.5

species

4

2

1

3.

4.5

standardization of the vectors (Formula 5.10) and the computation of the cross product (Formula 5.11). We get:

As can be expected from the data, species 1 and 2 as well as 3 and 4 are highly correlated (r = 0.85 and 0.91 respectively). The dispersions (variances) explained by the individual attributes are found in the respective rows or columns of the correlation matrix R (see Formula 5.12). The correlation coefficient simplifies matters as Shh, the diagonal values, are always equal to 1. The variances the attributes account for are:

ss1 = - [(l.O)2 + (0.85)2 + (-0.91)2 + (—0.64)2] = 2.95

552 = - [(0.85)2 + (1.0)2 + (-0.71)2 + (-0.43)2] = 2.41

553 = - [(—0.91)2 + (—0.71)2 + (1.0)2 + (0.91)2] = 3.14

554 = - [(—0.64)2 + (-0.43)2 + (0.91)2 + (1.0)2] = 2.41

Species number 3 has the highest explanatory power and will get rank no. 1. It is important to note that the other species achieve high values as well. Taking species number 1 instead of number 3, for example, would reduce the efficiency only moderately. This situation is typical for high-dimensional vegetation data.

The correlation matrix is now reduced by the fraction of variance explained by species 3 according to Formula 5.12:

r[ 1 = 1.0 - (-0.91 * -0.91) = 0.17 r'12 = 0.85 - (-0.71 * -0.91) = 0.20 r'13 = -0.91 - (1.0 * -0.91) = 0.0 r'u = -0.64 - (-0.91 * 0.91) = 0.19

In the new, reduced matrix R' the rows and columns related to species 3 are now all zero:

0.20 0.0 0.19 \ 0.50 0.0 0.21 0.0 0.0 0.0 0.21 0.0 0.17 /

The procedure according to Formula 5.12 is now applied to matrix R'. As can be seen in Table 5.2 the variance explained decreases rapidly, confirming the efficiency of the method. It even turns out that species number 1 no longer contributes to the total variance, indicating that the dimension of the total resemblance matrix is equal to 3 only.

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