如何做Clojure计划？

问题描述：

我是学习者，刚刚开始学习关于这个..

I am learner, just now started learning regarding this..

任何人都可以帮助我？

编写一个clojure程序一个函数，most-frequent-word，它有两个参数。第一个参数是一个字符串，第二个参数是一个整数，调用它n。最频繁字返回长度为n的在字符串中最多出现的序列字。例如（最常用字词TCGAAGCTAGACGCTAGTAGCTAGTGTGCA4）返回（CTAGGCTA）

Write a clojure program a function, most-frequent-word, which has two arguments. The first argument is a string, the second argument is an integer, call it n. most-frequent-word returns a sequence word(s) of length n that occurs most in the string. For example (most-frequent-word "TCGAAGCTAGACGCTAGTAGCTAGTGTGCA" 4) returns ("CTAG" "GCTA")

答

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有些提示可供您尝试：

Tips to start:

Some tips to try and get you started:

分区 将字符串转换为字词序列。请记住提供 1 的步骤参数，以便得到所有可能的重叠子序列。

frequencies 计算集合中出现的事物（包括序列）的次数。

max 或 max-key 搜索其输入中的最高值。使用 apply / code> 将集合的内容作为单独的输入插入到它们中。

分区的字符，而不是字符串。您可以使用将它们转换为字符串 clojure.string / join 。

You can use partition to turn the string into a sequence of "words". Remember to provide a step argument of 1, so you get all the possible overlapping subsequences.
frequencies counts how many times things (including sequences) appear in a collection.
max or max-key search for the highest values among their inputs. Use apply to plumb the contents of a collection into them as individual inputs.
partition will output sequences of characters, not strings. You can turn those back into strings with clojure.string/join.

我可以得到更明确的，如果你喜欢，但对于初学者，还有很多的价值，在实验这些在REPL和试图为自己工作。

I can get more explicit if you like, but for a beginner there's also a lot of value in experimenting with these at the REPL and trying to work it out for yourself.

对，这个特定的步骤有点模糊。因为你想要全部有最大频率的字符串，你需要做一些事情，而不仅仅是 max-key 。我这样做是首先找到 max 的频率值，然后过滤掉不同频率的任何键/频率对 。

Right, this particular step was a bit obscure. Since you want all the strings that have maximal frequency you need to do something a bit more than just max-key. The way I did it was to first find the max of the frequency values, then filter out any key/frequency pairs with a different frequency than that.

(defn most-frequent-word [string n]
  (let[freqs (->> string
                  (partition n 1)
                  frequencies)
       biggest-value (apply max (vals freqs))
       maximal-pairs (filter #(= biggest-value (val %)) freqs)]
    (map #(clojure.string/join (key %)) maximal-pairs )))

从性能的角度来看，这并不是很理想，但是似乎比在一次迭代中完成这两个工作更清晰地分离了问题（希望更容易理解）。

This isn't quite ideal from a performance standpoint, but seemed to have a cleaner separation of concerns (and hopefully be easier to understand) than trying to do both jobs in one iteration.

如何做Clojure计划？

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Tips to start:

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