freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/entropy.md

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---
id: 599d15309e88c813a40baf58
title: 熵
challengeType: 5
videoUrl: ''
---
# --description--
任务:
计算给定输入字符串的香农熵H.
给定谨慎的随机变量$ X $,它是$ N $“符号”(总字符)的字符串,由$ n $个不同的字符组成对于二进制n = 2位/符号中X的香农熵是
$ H*2X= - \\ sum* {i = 1} ^ n \\ frac {count_i} {N} \\ log_2 \\ left\\ frac {count_i} {N} \\ right$
其中$ count_i $是字符$ n_i $的计数。
# --hints--
`entropy`是一种功能。
```js
assert(typeof entropy === 'function');
```
`entropy("0")`应该返回`0`
```js
assert.equal(entropy('0'), 0);
```
`entropy("01")`应该返回`1`
```js
assert.equal(entropy('01'), 1);
```
`entropy("0123")`应该返回`2`
```js
assert.equal(entropy('0123'), 2);
```
`entropy("01234567")`应该返回`3`
```js
assert.equal(entropy('01234567'), 3);
```
`entropy("0123456789abcdef")`应返回`4`
```js
assert.equal(entropy('0123456789abcdef'), 4);
```
`entropy("1223334444")`应返回`1.8464393446710154`
```js
assert.equal(entropy('1223334444'), 1.8464393446710154);
```
# --solutions--