Scale a series between two points

2019-01-03 06:04发布

How do I scale a series such that the first number in the series is 0 and last number is 1. I looked into 'approx', 'scale' but they do not achieve this objective.

# generate series from exponential distr
s = sort(rexp(100))

# scale/interpolate 's' such that it starts at 0 and ends at 1?
# approx(s)
# scale(s)

标签: r scaling
5条回答
来,给爷笑一个
2楼-- · 2019-01-03 06:44

This should do it:

reshape::rescaler.default(s, type = "range")

EDIT

I was curious about the performance of the two methods

> system.time(replicate(100, range01(s)))
   user  system elapsed 
   0.56    0.12    0.69 
> system.time(replicate(100, reshape::rescaler.default(s, type = "range")))
   user  system elapsed 
   0.53    0.18    0.70 

Extracting the raw code from reshape::rescaler.default

range02 <- function(x) {
    (x - min(x, na.rm=TRUE)) / diff(range(x, na.rm=TRUE))
    }

> system.time(replicate(100, range02(s)))
   user  system elapsed 
   0.56    0.12    0.68 

Yields similar result.

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smile是对你的礼貌
3楼-- · 2019-01-03 06:46

Alternatively:

scale(x,center=min(x),scale=diff(range(x)))

(untested)

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何必那么认真
4楼-- · 2019-01-03 06:51

The scales package has a function that will do this for you: rescale.

library("scales")
rescale(s)

By default, this scales the given range of s onto 0 to 1, but either or both of those can be adjusted. For example, if you wanted it scaled from 0 to 10,

rescale(s, to=c(0,10))

or if you wanted the largest value of s scaled to 1, but 0 (instead of the smallest value of s) scaled to 0, you could use

rescale(s, from=c(0, max(s)))
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We Are One
5楼-- · 2019-01-03 06:57

It's straight-forward to create a small function to do this using basic arithmetic:

s = sort(rexp(100))

range01 <- function(x){(x-min(x))/(max(x)-min(x))}

range01(s)

  [1] 0.000000000 0.003338782 0.007572326 0.012192201 0.016055006 0.017161145
  [7] 0.019949532 0.023839810 0.024421602 0.027197168 0.029889484 0.033039408
 [13] 0.033783376 0.038051265 0.045183382 0.049560233 0.056941611 0.057552543
 [19] 0.062674982 0.066001242 0.066420884 0.067689067 0.069247825 0.069432174
 [25] 0.070136067 0.076340460 0.078709590 0.080393512 0.085591881 0.087540132
 [31] 0.090517295 0.091026499 0.091251213 0.099218526 0.103236344 0.105724733
 [37] 0.107495340 0.113332392 0.116103438 0.124050331 0.125596034 0.126599323
 [43] 0.127154661 0.133392300 0.134258532 0.138253452 0.141933433 0.146748798
 [49] 0.147490227 0.149960293 0.153126478 0.154275371 0.167701855 0.170160948
 [55] 0.180313542 0.181834891 0.182554291 0.189188137 0.193807559 0.195903010
 [61] 0.208902645 0.211308713 0.232942314 0.236135220 0.251950116 0.260816843
 [67] 0.284090255 0.284150541 0.288498370 0.295515143 0.299408623 0.301264703
 [73] 0.306817872 0.307853369 0.324882091 0.353241217 0.366800517 0.389474449
 [79] 0.398838576 0.404266315 0.408936260 0.409198619 0.415165553 0.433960390
 [85] 0.440690262 0.458692639 0.464027428 0.474214070 0.517224262 0.538532221
 [91] 0.544911543 0.559945121 0.585390414 0.647030109 0.694095422 0.708385079
 [97] 0.736486707 0.787250428 0.870874773 1.000000000
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我只想做你的唯一
6楼-- · 2019-01-03 07:00

You can also make use of the caret package which will provide you the preProcess function which is just simple like this:

preProcValues <- preProcess(yourData, method = "range")
dataScaled <- predict(preProcValues, yourData)

More details on the package help.

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