如何从嵌套在两个列表中的字典中获取值？

Question

我仍然熟悉python和使用API​​。我从API中提取了以下数据：

    [[{'term': 'Boys', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300247598'},
  {'term': 'Fish', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300266085'},
  {'term': 'Boats', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300178749'}],
 [{'term': 'Bears', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300266516'}],
 None,
  [{'term': 'Interiors',
   'AAT_URL': 'http://vocab.getty.edu/page/aat/300391239'},
  {'term': 'Jewelry', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300209286'},
  {'term': 'Couples', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300379217'},
  {'term': 'Men', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300025928'},
  {'term': 'Women', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300025943'},
  {'term': 'Weights and Measures',
   'AAT_URL': 'http://vocab.getty.edu/page/aat/300386648'}],
...]

我想像这样在最终列表中提取'term'键（但保留子列表）中的值：

[['Boys','Fish','Boats'],['Bears'],['Interiors','Jewlery','Couples','Men','Women','Weights and Measures']...]

每当我尝试遍历这样的列表时：

for x in tags:
    for y in x:
        print (y['term'])

我得到以下结果：

Boys
Fish
Boats
Bears

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-48-f2ff632ad612> in <module>
     11 
     12 for x in tags:
---> 13     for y in x:
     14         print (y['term'])

TypeError: 'NoneType' object is not iterable

我该如何解决？预先谢谢你c：

Answer 1

在遍历x之前先将其与None进行比较：

for x in tags:
    if x is not None:
        for y in x:
            print (y['term'])

Answer 2

要将数据提取到具有相同结构的列表中，请检查None，仅在有数据的情况下才添加内容：

data = [[{'term': 'Boys', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300247598'},
         {'term': 'Fish', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300266085'},
         {'term': 'Boats', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300178749'}],
        [{'term': 'Bears', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300266516'}], 
        None,
        [{'term': 'Interiors', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300391239'},
         {'term': 'Jewelry', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300209286'},
         {'term': 'Couples', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300379217'},
         {'term': 'Men', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300025928'},
         {'term': 'Women', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300025943'},
         {'term': 'Weights and Measures', 'AAT_URL': 'http://vocab.getty.edu/page/aat/300386648'}]]

result = []
for inner_list in data:
    if inner_list:  # None and [] are falsy, ignore them

        # got something, add a new empty list to add things into
        result.append([])
        for d in inner_list:
            # add all dicts term to the last item of result
            result[-1].append(d["term"])  

print(result)

将输出：

[['Boys', 'Fish', 'Boats'], ['Bears'], 
 ['Interiors', 'Jewelry', 'Couples', 'Men', 'Women', 'Weights and Measures']]

Answer 3

首先过滤掉无，然后使用地图功能

    data = [
        [
            {"term": "Boys", "AAT_URL": "http://vocab.getty.edu/page/aat/300247598"},
            {"term": "Fish", "AAT_URL": "http://vocab.getty.edu/page/aat/300266085"},
            {"term": "Boats", "AAT_URL": "http://vocab.getty.edu/page/aat/300178749"},
        ],
        [{"term": "Bears", "AAT_URL": "http://vocab.getty.edu/page/aat/300266516"}],
        None,
        [
            {"term": "Interiors", "AAT_URL": "http://vocab.getty.edu/page/aat/300391239"},
            {"term": "Jewelry", "AAT_URL": "http://vocab.getty.edu/page/aat/300209286"},
            {"term": "Couples", "AAT_URL": "http://vocab.getty.edu/page/aat/300379217"},
            {"term": "Men", "AAT_URL": "http://vocab.getty.edu/page/aat/300025928"},
            {"term": "Women", "AAT_URL": "http://vocab.getty.edu/page/aat/300025943"},
            {"term": "Weights and Measures", "AAT_URL": "http://vocab.getty.edu/page/aat/300386648"},
        ],
    ]


    result = map(lambda x: list(map(lambda i: i["term"], x)), filter(lambda x: x is not None, data))
    print(list(result))

Answer 4

如果您看到从API提取的数据中还存在None元素。因此，您必须检查自己是否在迭代时没有遇到。

for x in tags:
    if(x is not None):
        for y in x:
            print(y['term'])

这将解决问题