枚举org-mode中的所有标签

Question

如何生成组织模式文件中所有标签(例如:tag:)的枚举列表？假设我有一个表单列表：

* Head1        :foo:bar:
** Subhead1    :foo:
* Head2
** Subhead2    :foo:bar:

我想生成这个文件中所有标签的列表，以及每个标签被使用的次数。像这样说，

:foo:    3
:bar:    2  

Answer 1

这是一个简短的版本。

(defun get-tag-counts ()
  (let ((all-tags '()))
    (org-map-entries
     (lambda ()
       (let ((tag-string (car (last (org-heading-components)))))
     (when tag-string   
       (setq all-tags
         (append all-tags (split-string tag-string ":" t)))))))


    ;; now get counts
    (loop for tag in (-uniq all-tags) 
      collect (cons tag (cl-count tag all-tags :test 'string=)))))

Answer 2

这是一种方法。

(setq my-hash (make-hash-table :test 'equal))

(org-map-entries
 (lambda ()
   (let ((tag-string (car (last (org-heading-components))))
     (current-count))
     (when tag-string   
       (dolist (tag (split-string tag-string ":" t))

      (setq current-count (gethash tag my-hash))
     (if current-count;
         (puthash tag (+ 1 current-count)  my-hash)
       (puthash tag 1 my-hash))
     )
       )
     )
   )
 )

;; https://github.com/Wilfred/ht.el
(require 'ht)
(ht-map
 (lambda (key value)
   (list key value))
  my-hash)

Answer 3

我无法使用John Kitchin发布的代码，因为它需要一个交互功能。来自IRC freenode/#emacs的bpalmer非常友好地帮助了我。请找到一个工作示例，它显示了相应树下面的所有标签。

; use this in order to be able to use loop on its own
(require 'cl)

;; count tags (see John's answer)
(defun get-tag-counts ()
  (let ((all-tags '()))
    (org-map-entries
     (lambda ()
       (let ((tag-string (car (last (org-heading-components)))))
     (when tag-string  
       (setq all-tags
         (append all-tags (split-string tag-string ":" t)))))))

    ;; now get counts
    (loop for tag in (seq-uniq all-tags)
      collect (cons tag (cl-count tag all-tags :test 'string=)))))

;; wrap get-tag-counts in an interactive function
(defun create-tag-counts-buffer () (interactive)
    (let
        ((tags (get-tag-counts)) (b (get-buffer-create "*Org Tag Count*")))
        (dolist (tag tags) (insert (car tag)) (insert "\n")) (display-buffer b)))

请注意，我不能发表评论约翰的答案，因为我的声誉太低。