21
R有没有某种方法可以在没有任何中断的情况下按指定的间隔切割?例如,如果我想要在确切区间[1,10]中的值;默认cut
将此间隔分成更小的间隔。R - 按定义的间隔切割
R有没有某种方法可以在没有任何中断的情况下按指定的间隔切割?例如,如果我想要在确切区间[1,10]中的值;默认cut
将此间隔分成更小的间隔。R - 按定义的间隔切割
要剪切到预定义的时间间隔,可以使用参数breaks
指定断点向量。
定义了一些数据:
x <- sample(0:20, 100, replace=TRUE)
x
现在切割X 0,10和20:
cut(x, breaks=c(0, 10, 20), include.lowest=TRUE)
[1] (10,20] [0,10] [0,10] (10,20] (10,20] (10,20] [0,10] (10,20] (10,20]
[10] (10,20] [0,10] (10,20] (10,20] (10,20] [0,10] (10,20] [0,10] [0,10]
[19] [0,10] (10,20] [0,10] [0,10] [0,10] (10,20] [0,10] (10,20] (10,20]
[28] (10,20] (10,20] [0,10] [0,10] [0,10] [0,10] (10,20] [0,10] [0,10]
[37] [0,10] [0,10] (10,20] (10,20] (10,20] (10,20] [0,10] (10,20] [0,10]
[46] (10,20] [0,10] (10,20] (10,20] [0,10] [0,10] (10,20] (10,20] (10,20]
[55] [0,10] [0,10] (10,20] [0,10] [0,10] [0,10] [0,10] (10,20] (10,20]
[64] (10,20] [0,10] [0,10] (10,20] (10,20] (10,20] (10,20] (10,20] (10,20]
[73] (10,20] [0,10] [0,10] [0,10] (10,20] [0,10] (10,20] [0,10] (10,20]
[82] [0,10] [0,10] (10,20] [0,10] [0,10] [0,10] (10,20] (10,20] [0,10]
[91] [0,10] [0,10] (10,20] (10,20] [0,10] [0,10] [0,10] [0,10] (10,20]
[100] (10,20]
Levels: [0,10] (10,20]
像这样的事情?打破了每个0.2从0到1
> a <- runif(100)
> cut(a, seq(from = 0, to = 1, by = 0.2))
[1] (0,0.2] (0.8,1] (0.8,1] (0.6,0.8] (0.6,0.8] (0,0.2] (0.6,0.8]
[8] (0.2,0.4] (0.8,1] (0.4,0.6] (0.8,1] (0.4,0.6] (0.8,1] (0.6,0.8]
[15] (0.8,1] (0,0.2] (0.8,1] (0.8,1] (0.6,0.8] (0.6,0.8] (0.2,0.4]
[22] (0.4,0.6] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.6,0.8] (0,0.2]
[29] (0,0.2] (0.2,0.4] (0,0.2] (0,0.2] (0,0.2] (0,0.2] (0,0.2]
[36] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.8,1] (0.2,0.4] (0.4,0.6]
[43] (0.4,0.6] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.6,0.8] (0.4,0.6]
[50] (0.6,0.8] (0.6,0.8] (0,0.2] (0.2,0.4] (0.8,1] (0.8,1] (0.8,1]
[57] (0.6,0.8] (0.2,0.4] (0.2,0.4] (0,0.2] (0.8,1] (0.8,1] (0.2,0.4]
[64] (0.8,1] (0.2,0.4] (0.4,0.6] (0.8,1] (0,0.2] (0.4,0.6] (0,0.2]
[71] (0.4,0.6] (0.8,1] (0.6,0.8] (0.4,0.6] (0,0.2] (0.2,0.4] (0.4,0.6]
[78] (0,0.2] (0.2,0.4] (0.8,1] (0,0.2] (0.4,0.6] (0.8,1] (0,0.2]
[85] (0,0.2] (0.2,0.4] (0.2,0.4] (0.4,0.6] (0.8,1] (0.2,0.4] (0,0.2]
[92] (0.6,0.8] (0.2,0.4] (0.2,0.4] (0.8,1] (0.2,0.4] (0.4,0.6] (0,0.2]
[99] (0,0.2] (0,0.2]
Levels: (0,0.2] (0.2,0.4] (0.4,0.6] (0.6,0.8] (0.8,1]
我有关于 '一刀切' 切(X $ MPG,4)给了我 水平的问题:(10.4,16.3](16.3,22.1] (22.1,28](28,33.9) 我们如何提取分界点? 这里的分界点是10.4,16.3,22.1,28,33.9 – darkage 2014-08-08 10:50:33
@darkage这是一个不同的问题,不是真正的相关但是你的问题在'cut'的最后一个例子中得到了回答,如果这样做没有帮助,请提出一个新问题。 – Andrie 2014-08-08 11:24:40