在两点之间缩放系列

发布于 2024-10-27 10:30:18 字数 241 浏览 1 评论 0原文

如何缩放一个系列，使该系列中的第一个数字为 0，最后一个数字为 1。我研究了“近似”、“缩放”，但它们没有实现此目标。

# 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)

原文

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)

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明月夜 2024-11-03 10:30:18

scales 包有一个函数可以为您执行此操作：rescale。

library("scales")
rescale(s)

默认情况下，这会将给定的 s 范围缩放到 0 到 1，但其中一个或两个都可以调整。例如，如果您希望它从 0 缩放到 10，

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

或者您希望将 s 的最大值缩放为 1，但为 0（而不是 s 的最小值））缩放到0，你可以使用

rescale(s, from=c(0, max(s)))

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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誰ツ都不明白 2024-11-03 10:30:18

使用基本算术创建一个小函数来执行此操作很简单：

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

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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七秒鱼° 2024-11-03 10:30:18

或者：（

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

未经测试）

它具有将原始居中和缩放因子作为属性附加到输出的功能，因此可以检索它们并在以后用于取消缩放数据（如果需要）。它有一个奇怪的地方，即它总是以（按列）矩阵的形式返回结果，即使它传递的是一个向量；如果您想要一个向量而不是矩阵，可以使用 drop(scale(...))（这通常并不重要，但矩阵格式有时会引起问题）下游......根据我的经验，更常见的是 tibbles/tidyverse，尽管我没有停下来检查这些情况下到底出了什么问题）。

Alternatively:

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

(untested)

This has the feature that it attaches the original centering and scaling factors to the output as attributes, so they can be retrieved and used to un-scale the data later (if desired). It has the oddity that it always returns the result as a (columnwise) matrix, even if it was passed a vector; you can use drop(scale(...)) if you want a vector instead of a matrix (this usually doesn't matter but the matrix format can occasionally cause trouble downstream ... in my experience more often with tibbles/in tidyverse, although I haven't stopped to examine exactly what's going wrong in these cases).

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画中仙 2024-11-03 10:30:18

这应该可以做到：

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

编辑

我对这两种方法的性能很好奇

> 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

从reshape::rescaler.default中提取原始代码

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

会产生类似的结果。

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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烧了回忆取暖 2024-11-03 10:30:18

您还可以使用插入符号包，它将为您提供 preProcess 功能，就像这样简单：

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

更多详细信息请参阅软件包帮助。

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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哽咽笑 2024-11-03 10:30:18

我在 r 中创建了以下函数：

ReScale <- function(x,first,last){(last-first)/(max(x)-min(x))*(x-min(x))+first}

这里，第一个是起点，最后一个是终点。

I created following function in r:

ReScale <- function(x,first,last){(last-first)/(max(x)-min(x))*(x-min(x))+first}

Here, first is start point, last is end point.

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