python-3.x – 如何从复杂的字典中提取嵌套数据并从中构建新的字
发布时间:2020-12-20 12:03:03 所属栏目:Python 来源:网络整理
导读:如何从复杂的字典D中提取嵌套数据并从中构建新的字典NS_D? IN: D= {'10': [{'tea': ['indian','green']}],'13': [{'dancing': ['tango','step']}],'12': [{'walk': ['running','walking']}]}OUT: NS_D = {'tea': ['indian','green'],'dancing': ['tango','s
如何从复杂的字典D中提取嵌套数据并从中构建新的字典NS_D?
IN: D= {'10': [{'tea': ['indian','green']}],'13': [{'dancing': ['tango','step']}],'12': [{'walk': ['running','walking']}]} OUT: NS_D = {'tea': ['indian','green'],'dancing': ['tango','step'],'walk': ['running','walking']} 解决方法
您可以执行以下操作
如您所知,可以使用键,值或键以及值列表项来调用词典(所有信息都是here) In [78]: NS_D = {} In [79]: for keys in D.values(): #first loop: parsing through the values ...: for key in keys: #second loop: parsing through the keys ...: for za,az in key.items(): #third loop: printing keys and values ...: NS_D[za] = az #you append the keys and values in the dictionary In [80]: print(NS_D) {'tea': ['indian','walking']} 有点不那么pythonic但它的工作原理. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |