Clojure 嵌套 for 循环与索引
Clojure nested for loop with index
我一直在尝试像下面这样按惯用方式循环遍历嵌套向量：
[[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]]
找到值后，我还需要 return 坐标。
例如调用 (find-key-value 3) 应该 return [1 2]
这是我目前所拥有的，但它没有给我我需要的输出 return ([] [] [] [] [] [1 2] [] [] []) 因为我只需要 [1 2]
(defn find-key-value
[array value]
(for [x (range 0 (count array))]
(loop [y 0
ret []]
(cond
(= y (count (nth array x))) [x y]
:else (if (= value (get-in array [x y]))
(recur (+ 1 y) (conj ret [x y]))
(recur (+ 1 y) ret))))))
任何人对我如何修复我的代码以获得我想要的解决方案有任何想法或有更好的方法！
有一个 map-indexed 有时很有用。请参阅 Clojure 备忘单和其他 docs listed here。
==> 能否请您编辑问题以阐明搜索条件？
以下是您可以执行哪些操作来搜索所需答案的概述：
(ns tst.demo.core
(:use demo.core tupelo.core tupelo.test))
(defn coords
[data pred]
(let [result (atom [])]
(doseq [row (range (count data))
col (range (count (first data)))]
(let [elem (get-in data [row col])
keeper? (pred elem)]
(when keeper?
(swap! result conj [row col]))))
(deref result)))
(dotest
(let [data [[11 12 13]
[21 22 23]
[31 32 33]]
ends-in-2? (fn [x] (zero? (mod x 2)))]
(is= (coords data ends-in-2?)
[[0 1]
[1 1]
[2 1]])))
它基于与文档相同的 template project。有很多变体（例如，您可以使用 reduce 而不是原子）。
请查看上面列出的文档。
列表理解可用于查找满足谓词的所有值的坐标：
(defn find-locs [pred coll]
(for [[i vals] (map-indexed vector coll)
[j val] (map-indexed vector vals)
:when (pred val)]
[i j]))
(find-locs #(= 3 %) [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
=> ([1 5])
(find-locs zero? [[0 1 1] [1 1 1] [1 0 1]])
=> ([0 0] [2 1])
提出的问题似乎暗示应忽略输入中的关键字，在这种情况下答案变为：
(defn find-locs-ignore-keyword [pred coll]
(for [[i vals] (map-indexed vector coll)
[j val] (map-indexed vector (remove keyword? vals))
:when (pred val)]
[i j]))
(find-locs-ignore-keyword #(= 3 %) [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
=> ([1 2])
(defn vec-to-map [v] (into {} (into [] (map vec (partition 2 v)))))
(defn vec-vals [v] (vals (vec-to-map v)))
(defn map-vec-index [v el] (.indexOf (vec-vals v) el))
(defn find-val-coord
([arr val] (find-val-coord arr val 0))
([arr val counter]
(let [row (first arr)
idx (map-vec-index row val)]
(cond (< 0 idx) [counter idx]
:else (recur (rest arr) val (inc counter))))))
(find-val-coord arr 3) ;; => [1 2]
我们也可以编写函数来选择值或对应的键
给定坐标时来自数组：
(defn vec-keys [v] (keys (vec-to-map v)))
(defn get-val-coord [arr coord]
(nth (vec-vals (nth arr (first coord))) (second coord)))
(defn get-key-coord [arr coord]
(nth (vec-keys (nth arr (first coord))) (second coord)))
(get-val-coord arr [1 2]) ;; => 3
(get-key-coord arr [1 2]) ;; => :c
我可能稍微过度设计了这个答案，但这里有一个 非递归 和基于 单循环 [=16] 的非惰性方法=] 将适用于 任意和混合级别的嵌套 并且不会因递归而遭受堆栈溢出：
(defn find-key-value [array value]
(loop [remain [[[] array]]]
(if (empty? remain)
nil
(let [[[path x] & remain] remain]
(cond (= x value) path
(sequential? x)
(recur (into remain
(comp (remove keyword?)
(map-indexed (fn [i x] [(conj path i) x])))
x))
:default (recur remain))))))
(find-key-value [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]] 3)
;; => [1 2]
(find-key-value [[:a 1 [[[[[:c]]]] [[[9 [[[3]] :k]] 119]]]] [:a [[[1]]] :b 1]] 3)
;; => [0 1 1 0 0 1 0 0 0]
(find-key-value (last (take 20000 (iterate vector 3))) 3)
;; => [0 0 0 0 0 0 0 0 0 0 0 0 0 ...]
一个更简单的解决方案，假设内部向量是二维数组
键值向量，使用二维数组的展平和 .indexOf.
(defn find-coord [arr val]
(let [m (count (first arr))
idx (.indexOf (flatten arr) val)]
[(quot idx m) (quot (dec (mod idx m)) 2)]))
(find-coord arr 3) ;;=> [1 2]
clojure 核心中有一个函数完全适合任务：keep-indexed。这正是 indexed map + filter:
(defn find-val-idx [v data]
(ffirst (keep-indexed
(fn [i row]
(seq (keep-indexed
(fn [j [_ x]] (when (= v x) [i j]))
(partition 2 row))))
data)))
user> (find-val-idx 3 [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
;;=> [1 2]
user> (find-val-idx 10 [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
;;=> nil
user> (find-val-idx 1 [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
;;=> [0 0]
我一直在尝试像下面这样按惯用方式循环遍历嵌套向量：
[[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]]
找到值后，我还需要 return 坐标。
例如调用 (find-key-value 3) 应该 return [1 2]
这是我目前所拥有的，但它没有给我我需要的输出 return ([] [] [] [] [] [1 2] [] [] []) 因为我只需要 [1 2]
(defn find-key-value
[array value]
(for [x (range 0 (count array))]
(loop [y 0
ret []]
(cond
(= y (count (nth array x))) [x y]
:else (if (= value (get-in array [x y]))
(recur (+ 1 y) (conj ret [x y]))
(recur (+ 1 y) ret))))))
任何人对我如何修复我的代码以获得我想要的解决方案有任何想法或有更好的方法！
有一个 map-indexed 有时很有用。请参阅 Clojure 备忘单和其他 docs listed here。
==> 能否请您编辑问题以阐明搜索条件？
以下是您可以执行哪些操作来搜索所需答案的概述：
(ns tst.demo.core
(:use demo.core tupelo.core tupelo.test))
(defn coords
[data pred]
(let [result (atom [])]
(doseq [row (range (count data))
col (range (count (first data)))]
(let [elem (get-in data [row col])
keeper? (pred elem)]
(when keeper?
(swap! result conj [row col]))))
(deref result)))
(dotest
(let [data [[11 12 13]
[21 22 23]
[31 32 33]]
ends-in-2? (fn [x] (zero? (mod x 2)))]
(is= (coords data ends-in-2?)
[[0 1]
[1 1]
[2 1]])))
它基于与文档相同的 template project。有很多变体（例如，您可以使用 reduce 而不是原子）。
请查看上面列出的文档。
列表理解可用于查找满足谓词的所有值的坐标：
(defn find-locs [pred coll]
(for [[i vals] (map-indexed vector coll)
[j val] (map-indexed vector vals)
:when (pred val)]
[i j]))
(find-locs #(= 3 %) [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
=> ([1 5])
(find-locs zero? [[0 1 1] [1 1 1] [1 0 1]])
=> ([0 0] [2 1])
提出的问题似乎暗示应忽略输入中的关键字，在这种情况下答案变为：
(defn find-locs-ignore-keyword [pred coll]
(for [[i vals] (map-indexed vector coll)
[j val] (map-indexed vector (remove keyword? vals))
:when (pred val)]
[i j]))
(find-locs-ignore-keyword #(= 3 %) [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
=> ([1 2])
(defn vec-to-map [v] (into {} (into [] (map vec (partition 2 v)))))
(defn vec-vals [v] (vals (vec-to-map v)))
(defn map-vec-index [v el] (.indexOf (vec-vals v) el))
(defn find-val-coord
([arr val] (find-val-coord arr val 0))
([arr val counter]
(let [row (first arr)
idx (map-vec-index row val)]
(cond (< 0 idx) [counter idx]
:else (recur (rest arr) val (inc counter))))))
(find-val-coord arr 3) ;; => [1 2]
我们也可以编写函数来选择值或对应的键 给定坐标时来自数组：
(defn vec-keys [v] (keys (vec-to-map v)))
(defn get-val-coord [arr coord]
(nth (vec-vals (nth arr (first coord))) (second coord)))
(defn get-key-coord [arr coord]
(nth (vec-keys (nth arr (first coord))) (second coord)))
(get-val-coord arr [1 2]) ;; => 3
(get-key-coord arr [1 2]) ;; => :c
我可能稍微过度设计了这个答案，但这里有一个 非递归 和基于 单循环 [=16] 的非惰性方法=] 将适用于 任意和混合级别的嵌套 并且不会因递归而遭受堆栈溢出：
(defn find-key-value [array value]
(loop [remain [[[] array]]]
(if (empty? remain)
nil
(let [[[path x] & remain] remain]
(cond (= x value) path
(sequential? x)
(recur (into remain
(comp (remove keyword?)
(map-indexed (fn [i x] [(conj path i) x])))
x))
:default (recur remain))))))
(find-key-value [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]] 3)
;; => [1 2]
(find-key-value [[:a 1 [[[[[:c]]]] [[[9 [[[3]] :k]] 119]]]] [:a [[[1]]] :b 1]] 3)
;; => [0 1 1 0 0 1 0 0 0]
(find-key-value (last (take 20000 (iterate vector 3))) 3)
;; => [0 0 0 0 0 0 0 0 0 0 0 0 0 ...]
一个更简单的解决方案，假设内部向量是二维数组
键值向量，使用二维数组的展平和 .indexOf.
(defn find-coord [arr val]
(let [m (count (first arr))
idx (.indexOf (flatten arr) val)]
[(quot idx m) (quot (dec (mod idx m)) 2)]))
(find-coord arr 3) ;;=> [1 2]
clojure 核心中有一个函数完全适合任务：keep-indexed。这正是 indexed map + filter:
(defn find-val-idx [v data]
(ffirst (keep-indexed
(fn [i row]
(seq (keep-indexed
(fn [j [_ x]] (when (= v x) [i j]))
(partition 2 row))))
data)))
user> (find-val-idx 3 [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
;;=> [1 2]
user> (find-val-idx 10 [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
;;=> nil
user> (find-val-idx 1 [[:a 1 :b 1 :c 1] [:a 1 :b 1 :c 3] [:a 1 :b 1 :c 1]])
;;=> [0 0]