Clojure拉链可移除所有正确的兄弟姐妹
我想为一个zipper编写一个函数,该函数在保持相同位置的同时删除节点的所有正确同级Clojure拉链可移除所有正确的兄弟姐妹,clojure,functional-programming,zipper,Clojure,Functional Programming,Zipper,我想为一个zipper编写一个函数,该函数在保持相同位置的同时删除节点的所有正确同级 (defn remove-all-rights-1 [loc] (if (zip/right loc) (recur (zip/remove (zip/right loc))) loc)) 这里的问题是返回DFS中当前节点之前的位置 因此下面的例子 (-> (clojure.zip/vector-zip [1 [[2] 3]]) (zip/down) (zip/ri
(defn remove-all-rights-1 [loc]
(if (zip/right loc)
(recur (zip/remove (zip/right loc)))
loc))
这里的问题是返回DFS中当前节点之前的位置
因此下面的例子
(-> (clojure.zip/vector-zip [1 [[2] 3]])
(zip/down)
(zip/right)
(zip/down)
(remove-all-rights-1)
(zip/replace :x)
(zip/root))
…给出了[1[:x]]]
而不是[1[:x]]]
,因为zip/remove
跳转到底部叶,而不是仅仅返回左侧
在不改变树中的位置的情况下,如何删除正确的兄弟节点?提前谢谢 推广给出了以下解决方案:
(letfn [(kill-right [loc]
(let [lost (zip/rights loc)
parent (-> loc zip/up zip/node)
node (into (empty parent) (take (- (count parent) (count lost)) parent))]
(-> loc
zip/up
(zip/replace node)
zip/down
zip/rightmost)))]
(-> (clojure.zip/vector-zip [1 [[2] 3]])
zip/down
zip/right
zip/down
kill-right
(zip/replace :x)
zip/root))
(defn remove-all-rights
"Removes all right siblings. Stays at original location."
[loc]
(let [parent-loc (zip/up loc)
|lefts| (inc (count (zip/lefts loc)))]
(->> (zip/make-node loc (zip/node parent-loc) (take |lefts| (zip/children parent-loc)))
(zip/replace parent-loc)
(zip/down)
(zip/rightmost))))
主要思想是构造父节点的副本,其中子节点集合不包含正确的同级节点。泛化给出了以下解决方案:
(defn remove-all-rights
"Removes all right siblings. Stays at original location."
[loc]
(let [parent-loc (zip/up loc)
|lefts| (inc (count (zip/lefts loc)))]
(->> (zip/make-node loc (zip/node parent-loc) (take |lefts| (zip/children parent-loc)))
(zip/replace parent-loc)
(zip/down)
(zip/rightmost))))
主要思想是构造父节点的副本,其中子节点集合不包含正确的兄弟节点。这可以通过Tupelo林库轻松实现:
(dotest
(with-forest (new-forest)
(let [edn-orig [1 [[2] 3]]
root-hid (add-tree (edn->tree edn-orig))
hid (find-hid root-hid [::tf/list ::tf/list])
subtree-edn-orig (-> hid hid->tree tree->edn)
>> (kids-update hid butlast)
subtree-edn-final (-> hid hid->tree tree->edn)
edn-final (-> root-hid hid->tree tree->edn)]
(is= subtree-edn-orig [[2] 3])
(is= subtree-edn-final [[2]])
(is= edn-final [1 [[2]]] ))))
创建的树在第一级和第二级具有:tag
值为:tupelo.forest/list
的节点:
(is= (hid->bush root-hid)
[{:tag :tupelo.forest/list, :tupelo.forest/index nil}
[#:tupelo.forest{:value 1, :index 0}]
[{:tag :tupelo.forest/list, :tupelo.forest/index 1}
[{:tag :tupelo.forest/list, :tupelo.forest/index 0}
[#:tupelo.forest{:value 2, :index 0}]]]] )
HID是指向树节点的指针,因此根HID
指向树的根节点,HID
指向子树[[2]3]
。移除最右边的节点3
后,hid
指向子树[[2]]
对于子节点(kids
),我们使用butlast
函数删除最右边的节点,然后将数据从林/树格式转换回EDN
看到了吗,还有。也有很多。另请参见。这可以通过Tupelo Forest库轻松实现:
(dotest
(with-forest (new-forest)
(let [edn-orig [1 [[2] 3]]
root-hid (add-tree (edn->tree edn-orig))
hid (find-hid root-hid [::tf/list ::tf/list])
subtree-edn-orig (-> hid hid->tree tree->edn)
>> (kids-update hid butlast)
subtree-edn-final (-> hid hid->tree tree->edn)
edn-final (-> root-hid hid->tree tree->edn)]
(is= subtree-edn-orig [[2] 3])
(is= subtree-edn-final [[2]])
(is= edn-final [1 [[2]]] ))))
创建的树在第一级和第二级具有:tag
值为:tupelo.forest/list
的节点:
(is= (hid->bush root-hid)
[{:tag :tupelo.forest/list, :tupelo.forest/index nil}
[#:tupelo.forest{:value 1, :index 0}]
[{:tag :tupelo.forest/list, :tupelo.forest/index 1}
[{:tag :tupelo.forest/list, :tupelo.forest/index 0}
[#:tupelo.forest{:value 2, :index 0}]]]] )
HID是指向树节点的指针,因此根HID
指向树的根节点,HID
指向子树[[2]3]
。移除最右边的节点3
后,hid
指向子树[[2]]
对于子节点(kids
),我们使用butlast
函数删除最右边的节点,然后将数据从林/树格式转换回EDN
看到了吗,还有。也有很多。另请参阅