Question

我试图通过取出所有类型为None的条目来过滤Python中的列表，但是当我尝试这样做时，出现错误'TypeError：'NoneType'对象不可迭代。我看不到我要去哪里了。

原始错误实际上是与该行一起出现的如果您的“餐馆”不营业[u”类别”]：我在哪里遇到了与先前相同的错误，因此我尝试过滤掉几行以避免发生这种情况，但遇到了同样的问题

这是完整的代码块，尽管错误消息来自过滤器功能

import json

restaurant_ids = set()

# open the businesses file
with codecs.open(businesses_filepath, encoding='utf_8') as f:

    # iterate through each line (json record) in the file
    for business_json in f:
        # convert the json record to a Python dict
        business = json.loads(business_json)

        business[u"categories"] = filter(lambda a: a is not None, business[u"categories"])
        print business[u"categories"]

#         if this business is not a restaurant, skip to the next one
        if u"Restaurants" not in business[u"categories"]:
            continue

        # add the restaurant business id to our restaurant_ids set
        restaurant_ids.add(business[u'business_id'])

# turn restaurant_ids into a frozenset, as we don't need to change it anymore
restaurant_ids = frozenset(restaurant_ids)

# print the number of unique restaurant ids in the dataset
print '{:,}'.format(len(restaurant_ids)), u'restaurants in the dataset.'

我要过滤的文件包含一个JSON，其中包含许多看起来像这样的条目。有些类别中根本没有“ None”，这似乎是问题所在。

{“ business_id”：“ 1SWheh84yJXfytovILXOAQ”，“名称”：“ Arizona Biltmore Golf Club”，“地址”：“ 2818 E Camino Acequia Drive”，“ city”：“ Phoenix”，“ state”：“ AZ” ，“邮政编码”：“ 85016”，“纬度”：33.5221425，“经度”：-112.0184807，“星星”：3.0，“评论计数”：5，“ is_open”：0，“属性”：{“ GoodForKids”：“ False“}，”类别“：”高尔夫，活跃生活“，”小时“：空}

该代码应在if语句之后删除所有非“餐厅”类别，但我似乎甚至无法进入该阶段。

它引发错误

TypeError                                 Traceback (most recent call last)
<ipython-input-84-ac1362de4f26> in <module>()
     11         business = json.loads(business_json)
     12 
---> 13         business[u"categories"] = filter(lambda a: a is not None, business[u"categories"])
     14 #         print business[u"categories"]
     15 

TypeError: 'NoneType' object is not iterable

Answer 1

此：

filter(lambda a: a is not None, business[u"categories"])

尝试从None中删除所有business["categories"]的出现-假定business["categories"]是可迭代的。如果它是None（只是一个None，而不是包含None的列表），则确实会引发以下确切错误：

business = {"categories": None}
filter(lambda a: a is not None, business[u"categories"])

您要在此处测试business[u"categories"] 是否 None，而不是是否包含 None。

使用codecs.open（businesses_filepath，encoding ='utf_8'）作为f：

# iterate through each line (json record) in the file
for business_json in f:
    business = json.loads(business_json)

    categories =   business[u"categories"]
    if categories is None:
        # if it's None it cannot contain "Restaurant" obviously
        continue

    if u"Restaurants" not in categories:
        continue

    # ok it's a restaurant

Answer 2

当获得null（None）值时会产生Nonetype错误，因此首先打印（业务）并在终端中显示它是否打印了任何内容？进一步移动另一行。我认为 json.loads（business_json）不会返回任何内容。您应该尝试 json.load（business_json）或 business_json.json （）

尝试过滤列表时，我得到TypeError：'NoneType'对象不可迭代

2 个答案: