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在做一个聚类分析，输入以下语言后： #### 用数据框形式输入数据 V <- data.frame( v1=c(11.26, 24.1, 21.38, 11.12, 12.23, 2.74, 16.74, 19.08, 18.27, 33.89, 13.32, 6.42, 40.27, 11.02, 10.13, 15.6, 10.46, 48.64, 7.03, 18.65, 13.1), v2=c(28.64, 41.99, 48.23, 26.73, 13.64, 4.64, 2.83, 41.62, 21.05, 65.79, 23.78, 4.85, 34.67, 4.67, 9.73, 16.56, 26.73, 48.31, 5.7, 9.1, 11.04), v3=c(14.61, 16.23, 29.10, 18.04, 12.33, 23.16, 19.53, 28.57, 19.94, 7.27, 12.22, 14.90, 22.02, 26.25, 32.47, 17.60, 16.18, 15.94, 32.84, 13.74, 12.25), v4=c(12.16, 19.25, 22.16, 11.12, 13.01, 7.51, 9.29, 20.07, 20.72, 14.59, 15.68, 5.66, 16.02, 23.51, 16.29, 19.88, 9.78, 13.96, 9.00, 12.99, 7.19), v5=c(23.65, 18.94, 21.90, 26.68, 26.47, 29.70, 22.72, 20.14, 18.67, 23.50, 19.14, 18.03, 22.74, 38.63, 29.28, 38.32, 24.02, 20.26, 31.87, 39.74, 28.89), v6=c(7.70, 10.79, 3.59, 3.97, 3.01, -3.77, 15.94, 17.13, 14.14, 34.14, 2.34, 1.67, 32.03, 0.11, 13.14, 18.87, 8.91, 18.19, 1.15, 12.45, -1.19), v7=c(9.35, 9.16, 10.78, 8.12, 9.42, 15.83, 9.87, 8.37, 8.67, 5.25, 6.14, 0.00, 13.40, 10.03, 8.85, 11.20, 7.01, 8.75, 9.83, 14.07, 11.60), v8=c(5.96, 16.55, 7.73, 15.14, 10.46, 1.55, 4.12, 4.20, 0.85, 0.16, 18.72, 0.00, 0.17, 1.01, 1.31, 5.75, 10.19, 0.98, 11.53, 7.36, 13.13), row.names=c("全国", "北京", "天津", "沈阳", "大连", "长春", "哈尔滨", "上海", "南京", "苏州", "杭州", "宁波", "厦门", "济南", "青岛", "武汉", "广州", "深圳", "重庆", "成都", "西安") ） #### 生成距离结构，做系统聚类（离差平方和法） d <- dist(scale(V)) hc=hclust(d, "ward") #### 绘出谱系图和聚类情况（离差平方和法） opar=par(mfrow=c(2,1), mar=c(5.2, 4, 0, 0)) plclust(hc, hang=-1); re2=rect.plclust(hc, k=5, border="red") par(opar) 出现如下结果： > #### 用数据框形式输入数据
> V <- data.frame(
+ v1=c(11.26, 24.1, 21.38, 11.12, 12.23, 2.74, 16.74, 19.08, 18.27, 33.89, 13.32, 6.42, 40.27, 11.02, 10.13, 15.6, 10.46, 48.64, 7.03, 18.65, 13.1),
+ v2=c(28.64, 41.99, 48.23, 26.73, 13.64, 4.64, 2.83, 41.62, 21.05, 65.79, 23.78, 4.85, 34.67, 4.67, 9.73, 16.56, 26.73, 48.31, 5.7, 9.1, 11.04),
+ v3=c(14.61, 16.23, 29.10, 18.04, 12.33, 23.16, 19.53, 28.57, 19.94, 7.27,12.22, 14.90, 22.02, 26.25, 32.47, 17.60, 16.18, 15.94, 32.84, 13.74, 12.25),
+ v4=c(12.16, 19.25, 22.16, 11.12, 13.01, 7.51, 9.29, 20.07, 20.72, 14.59, 15.68, 5.66, 16.02, 23.51, 16.29, 19.88, 9.78, 13.96, 9.00, 12.99, 7.19),
+ v5=c(23.65, 18.94, 21.90, 26.68, 26.47, 29.70, 22.72, 20.14, 18.67, 23.50, 19.14, 18.03, 22.74, 38.63, 29.28, 38.32, 24.02, 20.26, 31.87, 39.74, 28.89),
+ v6=c(7.70, 10.79, 3.59, 3.97, 3.01, -3.77, 15.94, 17.13, 14.14, 34.14, 2.34, 1.67, 32.03, 0.11, 13.14, 18.87, 8.91, 18.19, 1.15, 12.45, -1.19),
+ v7=c(9.35, 9.16, 10.78, 8.12, 9.42, 15.83, 9.87, 8.37, 8.67, 5.25, 6.14, 0.00, 13.40, 10.03, 8.85, 11.20, 7.01, 8.75, 9.83, 14.07, 11.60),
+ v8=c(5.96, 16.55, 7.73, 15.14, 10.46, 1.55, 4.12, 4.20, 0.85, 0.16, 18.72, 0.00, 0.17, 1.01, 1.31, 5.75, 10.19, 0.98, 11.53, 7.36, 13.13), row.names=c("全国", "北京", "天津", "沈阳", "大连", "长春", "哈尔滨", "上海", "南京", "苏州", "杭州", "宁波", "厦门", "济南", "青岛", "武汉", "广州", "深圳", "重庆", "成都", "西安")
+ ）
> #### 生成距离结构，做系统聚类（离差平方和法）
> d <- dist(scale(V))
错误于scale(V) : 找不到对象'V'
> hc=hclust(d, "ward")
The "ward" method has been renamed to "ward.D"; note new "ward.D2"
错误于if (is.na(n) || n > 65536L) stop("size cannot be NA nor exceed 65536") :
需要TRUE/FALSE值的地方不可以用缺少值
> #### 绘出谱系图和聚类情况（离差平方和法）
> opar=par(mfrow=c(2,1), mar=c(5.2, 4, 0, 0))
> plclust(hc, hang=-1); re2=rect.plclust(hc, k=5, border="red")
错误于plot.hclust(x = tree, labels = labels, hang = hang, axes = axes, :
找不到对象'hc'
此外: 警告信息：
'plclust' is deprecated.
Use 'plot' instead.
See help("Deprecated")
> par(opar)
求教，如何修改，以实现聚类分析！万分感谢！ |
发表于 2014-12-31 19:20:06
