这是作为data.frame
被调用的数据的转储test
:
test <- structure(list(cluster = structure(c(6L, 7L, 17L, 1L, 8L, 11L,
15L, 2L, 4L, 5L, 16L, 12L, 9L, 14L, 13L, 10L, 3L), .Label = c("Cluster_10141",
"Cluster_10883", "Cluster_1419", "Cluster_16641", "Cluster_20143",
"Cluster_20637", "Cluster_20919", "Cluster_21451", "Cluster_26249",
"Cluster_28603", "Cluster_30198", "Cluster_32878", "Cluster_41908",
"Cluster_46928", "Cluster_55982", "Cluster_57942", "Cluster_9642"
), class = "factor"), value = c(0.02, 0.02, 0.147, 0.148, 0.148,
0.148, 0.498, 0.5, 0.5, 0.5, 0.867, 0.868, 0.87, 0.87, 0.871,
0.872, 0.873)), .Names = c("cluster", "value"), row.names = c(NA,
-17L), class = "data.frame")
看起来像:
cluster value
1 Cluster_20637 0.020
2 Cluster_20919 0.020
3 Cluster_9642 0.147
<<snip>>
16 Cluster_28603 0.872
17 Cluster_1419 0.873
生成累积百分比变量
> test$cumperc <- (1:nrow(test))/nrow(test)
> test
cluster value cumperc
1 Cluster_20637 0.020 0.05882353
2 Cluster_20919 0.020 0.11764706
3 Cluster_9642 0.147 0.17647059
<<snip>>
14 Cluster_46928 0.870 0.82352941
15 Cluster_41908 0.871 0.88235294
16 Cluster_28603 0.872 0.94117647
17 Cluster_1419 0.873 1.00000000
然后绘制数据
情节(test$value,test$cumperc,type="l",xlim=c(0,1))
编辑以解决以下评论:
试试这个首先对集群进行分组:
tabvals <- table(test$value)
plot(names(tabvals),(1:length(tabvals))/length(tabvals),xlim=c(0,1),type="l")
这给出了这个情节: