我有以下数据:

kNN,               kMeans,           expected1.1048387096774193,2.019927536231884,0.01.521505376344086,2.0120481927710845,0.01.271505376344086,2.019927536231884,0.02.020833333333333,2.019927536231884,0.00.7708333333333334,2.019927536231884,0.04.020408163265306,2.0120481927710845,0.00.6210526315789474,2.019927536231884,0.00.7708333333333334,2.019927536231884,0.03.354166666666667,2.019927536231884,0.02.020833333333333,2.019927536231884,1.01.450310559006211,2.0120481927710845,0.00.8719780219780221,2.019927536231884,-1.04.020408163265306,2.019927536231884,-1.03.520618556701031,2.019927536231884,1.01.521505376344086,2.019927536231884,0.0

我想根据kNN和kMeans的值绘制expected。我尝试了散点图，并阅读了关于在R中绘图的内容，但无法决定哪种图表有用。任何帮助都将不胜感激。我尝试了散点图，但它没有给我提供太多信息。

我的目标是探索数据，看看我是否可以训练单层感知器以接近预期值。如果数据是线性可分的（多类别，因此使用一对多方法）。

谢谢。

回答：

这是我能想到的:

dat <-structure(list(kNN = c(1.10483870967742, 1.52150537634409, 1.27150537634409, 2.02083333333333, 0.770833333333333, 4.02040816326531, 0.621052631578947, 0.770833333333333, 3.35416666666667, 2.02083333333333, 1.45031055900621, 0.871978021978022, 4.02040816326531, 3.52061855670103, 1.52150537634409), kMeans = c(2.01992753623188, 2.01204819277108, 2.01992753623188, 2.01992753623188, 2.01992753623188, 2.01204819277108, 2.01992753623188, 2.01992753623188, 2.01992753623188, 2.01992753623188, 2.01204819277108, 2.01992753623188, 2.01992753623188, 2.01992753623188, 2.01992753623188), expected = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, -1, -1, 1, 0)), .Names = c("kNN", "kMeans", "expected"), class = "data.frame", row.names = c(NA, -15L))plot(dat$kNN, dat$kMeans, col=dat$expected+2)