如何将多个序列中的每个元素交织成向量

Question

我想从下面返回的序列的每个元素创建一个向量(vector.我知道(vector得到我想要的东西是错误的.它只是一个占位符.

我正在努力学习如何从三个序列中取出第一个元素并得到["abc-de-fghi"smith barney] [def-hi-jklm"ox todd],然后是第二个元素,第三个,依此类推.

谢谢您的帮助.

(defn test-key-exclusion
    "This function takes csv-data1 (the includees) and tests to see
     if each includee is in csv-data2 (the includeds). This function
     also gathers enough other data, so that excludees (those not found),
     can be identified."

    [csv-data1 pkey-idx1 csv-data2 pkey-idx2 lnam-idx fnam-idx]

    (let [full-row-ret (map #(ret-non-match-rows csv-data2 pkey-idx2 pkey-idx1 %1) csv-data1)
          filtered-row-ret (filter (complement nil?) full-row-ret)]
        filtered-row-ret
        (vector     (map #(nth %1 pkey-idx1 nil)  filtered-row-ret)
                    (map #(nth %1 lnam-idx nil)   filtered-row-ret)
                    (map #(nth %1 fnam-idx nil)   filtered-row-ret))))

编辑:解决方案

我结束了以下操作,从多个序列的每个元素创建一个向量.

(defn test-key-exclusion
    "This function takes csv-data1 (the includees) and tests to see
     if each includee is in csv-data2 (the includeds). This function
     also gathers enough other data, so that excludees (those not found),
     can be identified."

    [csv-data1 pkey-idx1 csv-data2 pkey-idx2 lnam-idx fnam-idx]

    (map (fn [row]
            (vector (nth row pkey-idx1 nil)
                    (nth row lnam-idx nil)
                    (nth row fnam-idx nil)))

         (filter (complement nil?)
            (map #(ret-non-match-rows csv-data2 pkey-idx2 pkey-idx1 %1) csv-data1))))

Answer 1

我认为你需要的只是map.它可能需要多个集合:

user> (map vector [1 2 3] [:a :b :c] ["a" "b" "c"])
([1 :a "a"] [2 :b "b"] [3 :c "c"])