python - 遍历列表并按位置从嵌套字典返回值

Question

假设我有一个字典列表

meta_list = [
    {"id": 9, "name": "Tennis", "type": "SPORT", "url-name": "tennis"}, 
    {"id": 410468520880009, "name": "Live Betting", "type": "OTHER", "url-name": "live-betting"},
    {"id": 423407498380010, "name": "China", "type": "COUNTRY", "url-name": "china"},
    {"id": 613128412140012, "name": "September 7th 2018", "type": "DATE", "url-name": "september-7th-2018"},
    {"id": 643136938410012, "name": "QF", "type": "OTHER", "url-name": "qf"},
    {"id": 899076982160041, "name": "Zhangjiagang ATP Challenger", "type": "COMPETITION", "url-name": "zhangjiagang-atp-challenger"}
]

我想做的是在第 6 个字典中提取“name”的值，

for i in meta_list:
    event_name = i['name'][6]
    print(event_name)

我收到列表索引超出范围的错误。你如何提取名称"name": "张家港ATP挑战者",来自元列表？

Answer 1

你可以做 python 索引:

meta_list[5]['name']

另请注意，python 索引从零开始

说明:

1. 获取meta_list的第六(第五)个元素

2. 获取'name'键

3. 是的，你完成了

您的代码不起作用，因为您通过 meta_list 进行迭代，然后获取 i 的 'name' 键的第 6 个字符词典

最后的输出是:

Zhangjiagang ATP Challenger