List 将向量元素与相应的列表交错以创建新的向量

我有以下格式的数据：

[["i1" "i2"]
 ['("A" "B" "C" "D") '("red" "blue" "green" "yellow")]]

我想将其转换为新的数据结构：

[["column" "value"]
 ["i1" "A"]
 ["i1" "B"]
 ["i1" "C"]
 ["i1" "D"]
 ["i2" "red"]
 ["i2" "blue"]
 ["i2" "green"]
 ["i2" "yellow"]]

对这个难题的任何帮助都将是巨大的

---

我尝试使用嵌套的“for”语句进行far，但是我无法获得与头向量相同级别的结果向量，尽管我多次尝试转换结果。我还对列的值使用了“interleave”和“repeat”，但这也会在错误的级别创建列表

不是惯用语，但成功了

(defn doit [[is vss]]
  (vec (cons
         ["column" "value"]
         (mapcat (fn [i vs] (mapv (fn [v] [i v]) vs)) is vss))))

((fn [coll]
   (let [ks (first coll) vs (last coll)]
     (cons 
      '("column" "value")
      (partition 2 (concat
                    (interleave (repeat (first ks)) (first vs))
                    (interleave (repeat (last ks)) (last vs)))))))
 [["i1" "i2"]
  ['("A" "B" "C" "D") '("red" "blue" "green" "yellow")]])

---

不是惯用的，但确实做到了

((fn [coll]
   (let [ks (first coll) vs (last coll)]
     (cons 
      '("column" "value")
      (partition 2 (concat
                    (interleave (repeat (first ks)) (first vs))
                    (interleave (repeat (last ks)) (last vs)))))))
 [["i1" "i2"]
  ['("A" "B" "C" "D") '("red" "blue" "green" "yellow")]])

你的例子是：

(conform [["i1" "i2"] ['("A" "B" "C" "D") '("red" "blue" "green" "yellow")]])
=> (["i1" "A"] ["i1" "B"] ["i1" "C"] ["i1" "D"]
    ["i2" "red"] ["i2" "blue"] ["i2" "green"] ["i2" "yellow"])

奖金：

(conform [["i1" "i2"]
          ['("A" "B" "C" "D") '("red" "blue" "green" "yellow")]
          ['("E" "F" "G" "H") '("Mara" "Lara" "Clara" "Foxy")]])

=> (["i1" "A"] ["i1" "B"] ["i1" "C"] ["i1" "D"] 
    ["i2" "red"] ["i2" "blue"] ["i2" "green"] ["i2" "yellow"] 
    ["i1" "E"] ["i1" "F"] ["i1" "G"] ["i1" "H"] 
    ["i2" "Mara"] ["i2" "Lara"] ["i2" "Clara"] ["i2" "Foxy"])

更多奖金：

(conform [["i1" "i2" "i3"]
          ['("A" "B" "C" "D") '("red" "blue" "green" "yellow") ["Ufo"]]
          ['("E" "F" "G" "H") '("Mara" "Lara" "Clara" "Foxy") ["Orange" "Apple"]]])

=> (["i1" "A"] ["i1" "B"] ["i1" "C"] ["i1" "D"] 
    ["i2" "red"] ["i2" "blue"] ["i2" "green"] ["i2" "yellow"] 
    ["i3" "Ufo"] 
    ["i1" "E"] ["i1" "F"] ["i1" "G"] ["i1" "H"] 
    ["i2" "Mara"] ["i2" "Lara"] ["i2" "Clara"] ["i2" "Foxy"] ["i3" "Orange"] ["i3" "Apple"])

根据结果创建地图很容易：

(reduce (fn [acc [k v]]
          (update-in acc [k] (fnil conj []) v)) {} *1 )

=> {"i3" ["Ufo" "Orange" "Apple"], 
    "i2" ["red" "blue" "green" "yellow" "Mara" "Lara" "Clara" "Foxy"], 
    "i1" ["A" "B" "C" "D" "E" "F" "G" "H"]}

你的例子是：

(conform [["i1" "i2"] ['("A" "B" "C" "D") '("red" "blue" "green" "yellow")]])
=> (["i1" "A"] ["i1" "B"] ["i1" "C"] ["i1" "D"]
    ["i2" "red"] ["i2" "blue"] ["i2" "green"] ["i2" "yellow"])

奖金：

(conform [["i1" "i2"]
          ['("A" "B" "C" "D") '("red" "blue" "green" "yellow")]
          ['("E" "F" "G" "H") '("Mara" "Lara" "Clara" "Foxy")]])

=> (["i1" "A"] ["i1" "B"] ["i1" "C"] ["i1" "D"] 
    ["i2" "red"] ["i2" "blue"] ["i2" "green"] ["i2" "yellow"] 
    ["i1" "E"] ["i1" "F"] ["i1" "G"] ["i1" "H"] 
    ["i2" "Mara"] ["i2" "Lara"] ["i2" "Clara"] ["i2" "Foxy"])

更多奖金：

(conform [["i1" "i2" "i3"]
          ['("A" "B" "C" "D") '("red" "blue" "green" "yellow") ["Ufo"]]
          ['("E" "F" "G" "H") '("Mara" "Lara" "Clara" "Foxy") ["Orange" "Apple"]]])

=> (["i1" "A"] ["i1" "B"] ["i1" "C"] ["i1" "D"] 
    ["i2" "red"] ["i2" "blue"] ["i2" "green"] ["i2" "yellow"] 
    ["i3" "Ufo"] 
    ["i1" "E"] ["i1" "F"] ["i1" "G"] ["i1" "H"] 
    ["i2" "Mara"] ["i2" "Lara"] ["i2" "Clara"] ["i2" "Foxy"] ["i3" "Orange"] ["i3" "Apple"])

根据结果创建地图很容易：

(reduce (fn [acc [k v]]
          (update-in acc [k] (fnil conj []) v)) {} *1 )

=> {"i3" ["Ufo" "Orange" "Apple"], 
    "i2" ["red" "blue" "green" "yellow" "Mara" "Lara" "Clara" "Foxy"], 
    "i1" ["A" "B" "C" "D" "E" "F" "G" "H"]}

---

这里有一些有趣的答案。我只想在概念上补充一点，我认为这是一个很好的地方。对于第一个向量中的每个项目，您希望将其与第二个向量中相应列表中的项目配对

(defn convert [[cols vals]]
    (vec (cons ["column" "value"] ;; turn the list into a vector.
            (for [i (range (count cols)) ;; i <- index over columns
                  j (nth vals i)] ;; j <- items from the ith list
              [(nth cols i) j])))) ;; [col val]

user=>(convert data)

---

这里有一些有趣的答案。我只想在概念上补充一点，我认为这是一个很好的地方。对于第一个向量中的每个项目，您希望将其与第二个向量中相应列表中的项目配对

(defn convert [[cols vals]]
    (vec (cons ["column" "value"] ;; turn the list into a vector.
            (for [i (range (count cols)) ;; i <- index over columns
                  j (nth vals i)] ;; j <- items from the ith list
              [(nth cols i) j])))) ;; [col val]

user=>(convert data)

---

如果你先亲自尝试，人们会更感激你，也会更愿意帮助你，并分享你尝试过的东西。你必须引用以下列表：

['（“A”“B”“C”“D”）（“红色”“蓝色”“绿色”“黄色”）]

@shaktiman谢谢你。我添加了这些信息。@谢谢，这是我的一个常见错误。如果您先亲自尝试，人们会更加感激您，并愿意为您提供更多帮助，同时分享您尝试的内容。您必须引用以下列表：

['（“a”“B”“C”“D”）（“红色”“蓝色”“绿色”“黄色”）]

@shaktiman谢谢。我添加了信息。@谢谢，这是我的一个常见错误。已更正。以前的错误版本被完全否决。已更正。上一个错误的版本完全被否决。您的

for

构造是错误的。在

j

中，您正在迭代各个单词的字母。还有，为什么要搞索引？还有，我之所以使用索引，是因为我们不想将

cols

中的每个项目与

vals

中每个列表中的项目配对，我想让

for

完成大部分工作。为什么？因为我看到了四个答案，其中三个几乎完全相同，没有一个使用了

for

。答案没有错，但我们不是来为人们编写代码的，我们是来教书的。对不起，我的错。我发誓我在我的REPL中尝试过，今天早上看到了不同的结果-@galdre我要感谢你使用“for”。我花了很长时间想让它工作，但没能。我认为MAPCAT和CONS是我丢失的部分，我没有考虑的事情。这很有启发性。你的

for

构造是错误的。在

j

中，您正在迭代各个单词的字母。还有，为什么要搞索引？还有，我之所以使用索引，是因为我们不想将

cols

中的每个项目与

vals

中每个列表中的项目配对，我想让

for

完成大部分工作。为什么？因为我看到了四个答案，其中三个几乎完全相同，没有一个使用了

for

。答案没有错，但我们不是来为人们编写代码的，我们是来教书的。对不起，我的错。我发誓我在我的REPL中尝试过，今天早上看到了不同的结果-@galdre我要感谢你使用“for”。我花了很长时间想让它工作，但没能。我认为MAPCAT和CONS是我丢失的部分，我没有考虑的事情。这很有启发性。