{"id":20647,"date":"2022-11-10T09:29:42","date_gmt":"2022-11-10T03:59:42","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-iterate-over-a-list-to-create-a-parent-child-dictionary-closed\/"},"modified":"2022-11-10T09:29:42","modified_gmt":"2022-11-10T03:59:42","slug":"solved-iterate-over-a-list-to-create-a-parent-child-dictionary-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-iterate-over-a-list-to-create-a-parent-child-dictionary-closed\/","title":{"rendered":"[Solved] Iterate over a list to create a parent\/child dictionary [closed]"},"content":{"rendered":"<p> [ad_1]<br \/>\n<\/p>\n<div id=\"answer-51569925\" class=\"answer js-answer accepted-answer js-accepted-answer\" data-answerid=\"51569925\" data-parentid=\"51569702\" data-score=\"1\" data-position-on-page=\"1\" data-highest-scored=\"1\" data-question-has-accepted-highest-score=\"1\" itemprop=\"acceptedAnswer\" itemscope itemtype=\"https:\/\/schema.org\/Answer\">\n<div class=\"post-layout\">\n<div class=\"votecell post-layout--left\"><\/div>\n<div class=\"answercell post-layout--right\">\n<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Assuming each name belongs to only one key, and parents are defined before their children, you need to do <em>three<\/em> tasks for each element:<\/p>\n<ol>\n<li>map its name to a yet incomplete child list<\/li>\n<li>map its ID to its name<\/li>\n<li>link it to its parent<\/li>\n<\/ol>\n<p>If parents and children are not ordered, i.e. a child can appear before its parent, you must build 1. and 2. separately from 3.<br \/>\nOtherwise, you can do everything in one pass over your data. Note that instead of indexing into elements, you can destructure the element in the for loop.<\/p>\n<pre><code>key2name = {}\nname2children = {}\nfor key, name, parent in l:\n    name2children[name] = []  # 1.\n    key2name[key] = name      # 2.\n    if parent != 'P':         # root node has no parent\n        name2children[key2name[parent]].append(name) # 3.\n<\/code><\/pre>\n<p>For your example, this produces the structure<\/p>\n<pre><code>{'pharma': ['y', 'x'], 'x': ['z'], 'y': [], 'z': []}\n<\/code><\/pre>\n<p>Note that your data is <em>not<\/em> sorted to match your desired output!<\/p>\n<hr>\n<p>You can pretty print this as desired by walking this tree. The <code>sorted<\/code> call can be dropped if you do not care about ordering.<\/p>\n<pre><code>def printwalk(node, indent=0):\n    print(' '*indent, node)\n    for child in sorted(name2children[node]):\n        printwalk(child, indent+1)\nprintwalk('pharma')\n# pharma\n#  x\n#   z\n#  y\n<\/code><\/pre>\n<hr>\n<p>This is the case if parents and children are not ordered. You must separately initialise the translation (1) and parent-&gt;child (2) containers.<\/p>\n<pre><code>key2name = {}\nname2children = {}\nfor key, name, _ in l:\n    name2children[name] = []  # 1.\n    key2name[key] = name      # 2.\n\nfor key, name, parent in l:\n    if parent != 'P':         # root node has no parent\n        name2children[key2name[parent]].append(name) # 3.\n<\/code><\/pre>\n<\/p><\/div>\n<div class=\"mt24\"><\/div>\n<\/div>\n<p>            <span class=\"d-none\" itemprop=\"commentCount\">2<\/span> <\/p><\/div>\n<\/div>\n<p>[ad_2]<\/p>\n<p>solved Iterate over a list to create a parent\/child dictionary [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"<p>[ad_1] Assuming each name belongs to only one key, and parents are defined before their children, you need to do three tasks for each element: map its name to a yet incomplete child list map its ID to its name link it to its parent If parents and children are not ordered, i.e. a child &#8230; <a title=\"[Solved] Iterate over a list to create a parent\/child dictionary [closed]\" class=\"read-more\" href=\"https:\/\/jassweb.com\/solved\/solved-iterate-over-a-list-to-create-a-parent-child-dictionary-closed\/\" aria-label=\"More on [Solved] Iterate over a list to create a parent\/child dictionary [closed]\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[834,349],"class_list":["post-20647","post","type-post","status-publish","format-standard","hentry","category-solved","tag-dictionary","tag-python"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>[Solved] Iterate over a list to create a parent\/child dictionary [closed] - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-iterate-over-a-list-to-create-a-parent-child-dictionary-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Iterate over a list to create a parent\/child dictionary [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Assuming each name belongs to only one key, and parents are defined before their children, you need to do three tasks for each element: map its name to a yet incomplete child list map its ID to its name link it to its parent If parents and children are not ordered, i.e. a child ... 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