python - Python3 Pandas Dataframe KeyError 问题
问题描述
当我运行这段代码
crawl_stats = (
crawls['updated']
.groupby(crawls.index.get_level_values('url'))
.agg({
'number of crawls': 'count',
'proportion of updates': 'mean',
'number of updates': 'sum'
})
它显示错误:
KeyError Traceback (most recent call last)
<ipython-input-62-180f1041465d> in <module>
8 crawl_stats = (
9 crawls['updated']
---> 10 .groupby(crawls.index.get_level_values('url'))
11 # .groupby('url')
12 .agg({
/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site-packages/pandas/core/indexes/base.py in _get_level_values(self, level)
3155 """
3156
-> 3157 self._validate_index_level(level)
3158 return self
3159
/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site-packages/pandas/core/indexes/base.py in _validate_index_level(self, level)
1942 elif level != self.name:
1943 raise KeyError('Level %s must be same as name (%s)' %
-> 1944 (level, self.name))
1945
1946 def _get_level_number(self, level):
KeyError: 'Level url must be same as name (None)'
我尝试了这个修改后的代码:
crawl_stats = (
crawls['updated']
# .groupby(crawls.index.get_level_values('url'))
.groupby('url')
.agg({
'number of crawls': 'count',
'proportion of updates': 'mean',
'number of updates': 'sum'
})
它还显示错误:
KeyError Traceback (most recent call last)
<ipython-input-63-8c5f0f6f7c86> in <module>
9 crawls['updated']
10 # .groupby(crawls.index.get_level_values('url'))
---> 11 .groupby('url')
12 .agg({
13 'number of crawls': 'count',
3293 # Add key to exclusions
KeyError: 'url'
我之前已经尝试过在堆栈溢出中做其他指导,但它仍然不起作用。有人可以帮我解决吗?谢谢!
这是我创建 Dataframe 爬网的代码。
def make_crawls_dataframe(crawl_json_records):
"""Creates a Pandas DataFrame from the given list of JSON records.
The DataFrame corresponds to the following relation:
crawls(primary key (url, hour), updated)
Each hour in which a crawl happened for a page (regardless of
whether it found a change) should be represented. `updated` is
a boolean value indicating whether the check for that hour found
a change.
The result is sorted by URL in ascending order and **further**
sorted by hour in ascending order among the rows for each URL.
Args:
crawl_json_records (list): A list of JSON objects such as the
crawl_json variable above.
Returns:
DataFrame: A table whose schema (and sort order) is described
above.
"""
url = []
hour = []
updated = []
# To get the 1000 url, number of checks and positive checks
for i in range(len(crawl_json_records)):
temp_url = [crawl_json_records[i]['url']]
temp_len = crawl_json_records[i]["number of checks"]
temp_checks = crawl_json_records[i]["positive checks"]
# url.append(temp_url*temp_len)
for item0 in temp_url*temp_len:
url.append(item0)
# hour.append(list(range(1,temp_len+1)))
for item1 in list(range(1,temp_len+1)):
hour.append(item1)
temp_updated = [0]*temp_len
for item in temp_checks:
temp_updated[item-1] = 1
# updated.append(temp_updated)
for item2 in temp_updated:
updated.append(item2)
# print('len(url):',len(url))
# 521674
# print('len(hour):',len(hour))
# print('len(updated):',len(updated))
# Above 3 is 521674
#print(type(temp_len))
#print(temp_len)
#print(temp_url*temp_len)
columns = ['url','hour','updated']
data = np.array((url,hour,updated)).T
df = pd.DataFrame(data=data, columns=columns)
df.index += 1
# df.index = df['url']
return df.sort_values(by=['url','hour'], ascending=True)
crawls = make_crawls_dataframe(crawl_json)
crawls.head(50) # crawls shows as the image
解决方案
你需要替换这个:
.groupby(crawls.index.get_level_values('url'))
和:
.groupby('url')
因为您的 DataFrame 中没有索引。
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