开发者

how to substract numbers from levels

I would like to cut a vector of values ranging 0-70 to x number of categories, and would like the upper limit of each category. So far, I have tried this using cut() and am trying to extract the limits from levels. I have a list of levels, from which I would like to extract the second number from each level. How can I extract the values between space and ] (which is the number I'm interested in)?

I have:

> levels(bins)
 [1] "(-0.07,6.94]" "(6.94,14]"    "(14,21]"      "(21,28]"      "(28,35]"     
 [6] "(35,42]"      "(42,49]"      "(49,56]"      "(56,63.1]"    "(63.1,70.1]" 

and would like to get:

[1] 6.94 14 开发者_运维百科21 28 35 42 49 56 63.1 70.1

Or is there a better way of calculating the upper bounds of categories?


This could be one solution

k <- sub("^.*\\,","", levels(bins))
as.numeric(substr(k,1,nchar(k)-1))

gives

 [1]  6.94 14.00 21.00 28.00 35.00 42.00 49.00 56.00 63.10 70.10


If you want exact values of breaks then you should compute them yourself, cause cut round limits for interval:

x <- seq(0,1,by=.023)
levels(cut(x, 4))
# [1] "(-0.000989,0.247]" "(0.247,0.494]"     "(0.494,0.742]"     "(0.742,0.99]"     
levels(cut(x, 4, dig.lab=10))
# [1] "(-0.000989,0.2467555]" "(0.2467555,0.4945]"    "(0.4945,0.7422445]"   
# [4] "(0.7422445,0.989989]" 

You could look on code to cut.default how breaks are compute:

if (length(breaks) == 1L) {
    if (is.na(breaks) | breaks < 2L) 
        stop("invalid number of intervals")
    nb <- as.integer(breaks + 1)
    dx <- diff(rx <- range(x, na.rm = TRUE))
    if (dx == 0) 
        dx <- abs(rx[1L])
    breaks <- seq.int(rx[1L] - dx/1000, rx[2L] + dx/1000, 
        length.out = nb)
}

So easy way is to grab this code and put into a function:

compute_breaks <- function(x, breaks) 
    if (length(breaks) == 1L) {
        if (is.na(breaks) | breaks < 2L) 
            stop("invalid number of intervals")
        nb <- as.integer(breaks + 1)
        dx <- diff(rx <- range(x, na.rm = TRUE))
        if (dx == 0) 
            dx <- abs(rx[1L])
        breaks <- seq.int(rx[1L] - dx/1000, rx[2L] + dx/1000, 
            length.out = nb)
            breaks
    }

Result is

compute_breaks(x,4)
# [1] -0.000989  0.246755  0.494500  0.742244  0.989989
0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜