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The Application of Fuzzy Clustering Analysis in the Comprehensive Evaluation of Military Physical Quality
DOI: 10.12677/SA.2021.105092, PDF, HTML, XML, 下载: 58  浏览: 765  科研立项经费支持

Abstract: In order to scientifically evaluate the physical fitness of soldiers in grass-roots company, the five-kilometer run with bare hands in endurance, the 100-meter run with speed and the pull-up of strength (single pole one) are taken as the evaluation indexes of military physical fitness. According to the fuzzy clustering analysis method in fuzzy mathematics, the best classification method is determined by using statistics. The results show that the grass-roots company soldiers can be divided into five categories according to their physical qualities. The results provide a theoretical basis for the comprehensive evaluation of the physical qualities of grass-roots company soldiers.

1. 引言

2. 模糊聚类分析方法

2.1. 数据标准化 [1] [5]

$\left[\begin{array}{cccc}{x}_{11}& {x}_{\text{12}}& \cdots & {x}_{1m}\\ {x}_{21}& {x}_{22}& \cdots & {x}_{2m}\\ ⋮& ⋮& \ddots & ⋮\\ {x}_{n1}& {x}_{n2}& \cdots & {x}_{nm}\end{array}\right]$.

1) 平移–标准差变换

${{x}^{\prime }}_{ik}=\frac{{x}_{ik}-{\stackrel{¯}{x}}_{k}}{{s}_{k}}\text{}\left(i=1,2,\cdots ,n;\text{\hspace{0.17em}}k=1,2,\cdots ,m\right)$,(1)

2) 平移–极差变换

${{x}^{″}}_{ik}=\frac{{{x}^{\prime }}_{ik}-\underset{1\le i\le n}{\mathrm{min}}\left\{{{x}^{\prime }}_{ik}\right\}}{\underset{1\le i\le n}{\mathrm{max}}\left\{{{x}^{\prime }}_{ik}\right\}-\underset{1\le i\le n}{\mathrm{min}}\left\{{{x}^{\prime }}_{ik}\right\}}\text{}\left(k=1,2,\cdots ,m\right)$,(2)

2.2. 建立模糊相似矩阵 [1] [5]

${r}_{ij}=\frac{\underset{k=1}{\overset{m}{\sum }}{x}_{ik}\cdot {x}_{jk}}{\sqrt{\underset{k=1}{\overset{m}{\sum }}{x}_{ik}^{2}}\cdot \sqrt{\underset{k=1}{\overset{m}{\sum }}{x}_{jk}^{2}}},\text{}i,j=1,2,\cdots ,n$. (3)

2.3. 聚类 [1] [5]

2.4. 最佳阀值 $\lambda$ 的确定 [1] [5]

1) 按照实际需要，根据动态聚类情况，调整 $\lambda$ 的值以得到适当的分类，而不需要事先估计样本应分成几类。当然，也可由经验丰富的专家结合具体专业知识来确定阀值 $\lambda$，从而得出在 $\lambda$ 水平上的等价分类。

2) 用F统计量确定 $\lambda$ 最佳值

${\stackrel{¯}{x}}_{k}^{\left(j\right)}=\frac{1}{{n}_{j}}\underset{i=1}{\overset{{n}_{j}}{\sum }}{x}_{ik}^{\left(j\right)}\text{}\left(k=1,2,\cdots ,m\right)$, (4)

$F=\frac{\underset{j=1}{\overset{r}{\sum }}{n}_{j}{‖{\stackrel{¯}{x}}^{\left(j\right)}-\stackrel{¯}{x}‖}^{2}/\left(r-1\right)}{\underset{j=1}{\overset{r}{\sum }}\underset{i=1}{\overset{{n}_{j}}{\sum }}{‖{x}_{i}{}^{\left(j\right)}-{\stackrel{¯}{x}}^{\left(j\right)}‖}^{2}/\left(n-r\right)}$,(5)

3. 实例分析

3.1. 数据标准化

Table 1. Original data table

$\stackrel{\to }{x}=\left(21.4542,13.6130,12.0500\right)$,

$\stackrel{\to }{s}=\left(0.3903,0.5394,1.6576\right)$.

$\underset{1\le i\le 60}{\mathrm{min}}\left\{{{x}^{\prime }}_{ik}\right\}=\left(-2.9059,-2.0636,-2.4434\right)$,

$\underset{1\le i\le 60}{\mathrm{max}}\left\{{{x}^{\prime }}_{ik}\right\}=\left(1.8341,2.9795,3.5896\right)$,

$\underset{1\le i\le 60}{\mathrm{max}}\left\{{{x}^{\prime }}_{ik}\right\}-\underset{1\le i\le 60}{\mathrm{min}}\left\{{{x}^{\prime }}_{ik}\right\}=\left(4.7399,5.0431,6.0330\right)$, $k=1,2,3$.

A = [0.3125 0.4000 0.6811

0.1949 0.4000 0.5405

0.4228 0.7000 0.7568

0.2022 0.5000 0.4324

0.31990.2000 0.9189

0.2463 0.4000 0.3027

0.2279 0.4000 0.5730

0.2390 0.4000 0.9892

0.4118 0.3000 0.7568

0.5331 0.4000 0.3514

1.0000 0.1000 0.9189

0.6397 0.7000 0.5405

0.3713 0.4000 0.5405

0.5846 0.3000 0.3784

0.1103 0.7000 0.3189

0.0000 0.6000 0.3027

0.6250 0.4000 0.8649

0.4779 0.0000 0.5730

0.5735 0.4000 0.6811

0.3309 0.4000 0.5730

0.2206 0.4000 0.5730

0.2279 0.3000 0.5568

0.5882 0.2000 0.4486

0.4044 0.4000 0.5568

0.2757 0.5000 0.8541

0.2904 0.4000 0.8541

0.6397 0.4000 0.5730

0.3382 0.4000 0.7568

0.5147 0.4000 0.7027

0.7353 0.3000 0.6919

0.2206 0.2000 0.9892

0.4779 0.4000 0.3135

0.2206 0.4000 0.7892

0.4779 0.0000 0.2757

0.2721 0.4000 0.7568

0.5184 0.4000 0.3514

0.2941 0.4000 0.3405

0.5515 0.4000 0.0000

0.6324 0.4000 0.6757

0.2610 0.4000 0.6811

0.4191 0.4000 0.6811

0.0515 0.4000 0.6973

0.6618 0.0000 0.3784

0.5147 0.4000 0.6811

0.1728 0.5000 0.6811

0.2721 0.4000 0.6811

0.5809 0.4000 0.1027

0.7353 0.4000 0.6811

0.1875 0.4000 0.7622

0.4816 0.8000 0.7622

0.1471 0.5000 0.6811

0.1912 0.4000 0.5405

0.5184 0.5000 0.5405

0.3971 0.6000 0.7297

0.6213 0.4000 1.0000

0.7721 0.4000 0.8649

0.6618 0.4000 0.5730

0.3713 0.4000 0.5730

0.4779 0.4000 0.7568

0.3309 1.0000 0.6811]。

3.2. 标定(建立模糊相似矩阵)

$\sqrt{\underset{k=1}{\overset{3}{\sum }}{x}_{ik}^{2}}=\sqrt{{x}_{i1}^{2}+{x}_{i2}^{2}+{x}_{i3}^{2}}$,

$\underset{k=1}{\overset{3}{\sum }}{x}_{ik}{x}_{jk}={x}_{i1}{x}_{j1}+{x}_{i2}{x}_{j2}+{x}_{i3}{x}_{j3}$,

${r}_{ij}=\frac{\underset{k=1}{\overset{3}{\sum }}{x}_{ik}{x}_{jk}}{\sqrt{\underset{k=1}{\overset{3}{\sum }}{x}_{ik}^{2}}\sqrt{\underset{k=1}{\overset{3}{\sum }}{x}_{jk}^{2}}}=\frac{{x}_{i1}{x}_{j1}+{x}_{i2}{x}_{j2}+{x}_{i3}{x}_{j3}}{\sqrt{{x}_{i1}^{2}+{x}_{i2}^{2}+{x}_{i3}^{2}}\sqrt{{x}_{j1}^{2}+{x}_{j2}^{2}+{x}_{j3}^{2}}}$, $i,j=1,2,\cdots ,60$,

$i\ne j$ 时，由于模糊相似矩阵具有对称性的性质，我们只需计算一个上三角矩阵。 ${x}_{i}$${x}_{j}$ 的相似程度计算举例说明如下：

$i=2,j=4$

$\sqrt{\underset{k=1}{\overset{3}{\sum }}{x}_{2k}^{2}}=\sqrt{{x}_{21}^{2}+{x}_{22}^{2}+{x}_{23}^{2}}=\sqrt{{\left(0.5045\right)}^{2}+{\left(0.1949\right)}^{2}+{\left(0.4000\right)}^{2}}=0.7001$,

$\sqrt{\underset{k=1}{\overset{3}{\sum }}{x}_{4k}^{2}}=\sqrt{{x}_{41}^{2}+{x}_{42}^{2}+{x}_{43}^{2}}=\sqrt{{\left(0.4324\right)}^{2}+{\left(0.2022\right)}^{2}+{\left(0.5000\right)}^{2}}=0.6913$,

$\begin{array}{c}\underset{k=1}{\overset{3}{\sum }}{x}_{2k}{x}_{4k}={x}_{21}{x}_{41}+{x}_{22}{x}_{42}+{x}_{23}{x}_{43}\\ =\left(0.5045×0.4324\right)+\left(0.1949×0.2002\right)+\left(0.4000×0.5000\right)\\ =0.4731\end{array}$,

${r}_{24}=\frac{\underset{k=1}{\overset{3}{\sum }}{x}_{2k}{x}_{4k}}{\sqrt{\underset{k=1}{\overset{3}{\sum }}{x}_{2k}^{2}}\sqrt{\underset{k=1}{\overset{3}{\sum }}{x}_{4k}^{2}}}=\frac{0.4731}{0.7001×0.6913}=0.9775$,

3.3. 聚类

${R}^{2}=R\circ R$, ${R}^{4}={R}^{2}\circ {R}^{2}$, ${R}^{8}={R}^{4}\circ {R}^{4}$,

${R}^{8}={R}^{4}\circ {R}^{4}={R}^{4}$，则得到传递闭包 $t\left(R\right)={R}^{4}$ (模糊等价矩阵 ${R}^{*}$ )。

$\stackrel{\to }{l}=\left(0.94,0.97,0.98,0.99,1.00\right)$,

$\lambda =0.94$，分为1类： $\left\{{x}_{1},{x}_{2},\cdots ,{x}_{60}\right\}$

$\lambda =0.97$，分为2类： $\left\{{x}_{1},{x}_{2},\cdots ,{x}_{i},\cdots ,{x}_{60}\right\}\text{\hspace{0.17em}}\left(i\ne \text{38,47}\right)$$\left\{{x}_{38},{x}_{47}\right\}$

$\lambda =0.98$，分为5类： $\left\{{x}_{1},{x}_{2},\cdots ,{x}_{i},\cdots ,{x}_{60}\right\}\text{\hspace{0.17em}}\left(i\ne 11,15,16,18,34,\text{38,43,47}\right)$$\left\{{x}_{38},{x}_{47}\right\}$$\left\{{x}_{11},{x}_{18}\right\}$$\left\{{x}_{15},{x}_{16}\right\}$$\left\{{x}_{34},{x}_{43}\right\}$

$\lambda =0.99$，分为10类： $\left\{{x}_{5},{x}_{31}\right\}$$\left\{{x}_{11}\right\}$$\left\{{x}_{42}\right\}$$\left\{{x}_{15}\right\}$$\left\{{x}_{16}\right\}$$\left\{{x}_{34},{x}_{43}\right\}$$\left\{{x}_{47}\right\}$$\left\{{x}_{38}\right\}$$\left\{{x}_{18}\right\}$$\left\{{x}_{1},{x}_{2},\cdots ,{x}_{i},\cdots ,{x}_{60}\right\}\text{\hspace{0.17em}}\left(i\ne 5,11,15,16,18,31,34,\text{38,42,43,47}\right)$

$\lambda =1.00$，分为60类： $\left\{{x}_{1}\right\}$$\left\{{x}_{2}\right\}$$\cdots$$\left\{{x}_{i}\right\}$$\cdots$$\left\{{x}_{60}\right\}$

3.4. 最佳阈值的确定

$\lambda =0.94$$r=1$$F=0$

$\lambda =0.97$$r=2$$F=0.0139$

$\lambda =0.98$$r=5$$F=0.1798$

$\lambda =0.99$$r=10$$F=0.1091$

$\lambda =1.00$$r=60$$F=0$

4. 结论

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