Python 数据集模块基于Sqlalchemy,并公开一个函数来返回名为all()的表中的所有记录。all()返回一个可迭代数据集对象。
users = db['user'].all()
for user in db['user']:
print(user['age'])
数据集对象转换为Pandas DataFrame对象的最简单方法是什么?
为了清楚起见,我对Dataset的功能感兴趣,因为它已经将表加载到Dataset对象中。
发布于 2018-06-28 15:02:00
这对我起了作用:
import dataset
import pandas
db = dataset.connect('sqlite:///db.sqlite3')
data = list(db['my_table'].all())
dataframe = pandas.DataFrame(data=data)
发布于 2018-04-23 20:44:37
import pandas as pd
df = pd.DataFrame(data=db['user'])
df
类似的
pd.DataFrame(db['user'])
应该做同样的事
还可以指定列或索引:
https://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.html
发布于 2018-04-24 03:50:40
在对数据集模块投入了大量时间之后,我发现all()可以被迭代成一个列表,然后变成一个熊猫数据。有更好的方法吗?
import dataset
import pandas as pd
# create dataframe
df = pd.DataFrame()
names = ['Bob', 'Jane', 'Alice', 'Ricky']
ages = [31, 30, 31, 30]
df['names'] = names
df['ages'] = ages
print(df)
# create a dict oriented as records from dataframe
user = df.to_dict(orient='records')
# using dataset module instantiate database
db = dataset.connect('sqlite:///mydatabase.db')
# create a reference to a table
table = db['user']
# insert the complete dict into database
table.insert_many(user)
# use Dataset .all() to retrieve all table's rows
from_sql = table.all() # custom ResultIter type (iterable)
# iterate ResultIter type into a list
data = []
for row in from_sql:
data.append(row)
# create dataframe from list and ordereddict keys
df_new = pd.DataFrame(data, columns=from_sql.keys)
# this does not drop the id column, but it should??
df_new.drop(columns=['id'])
print(df_new)
'''
names ages
0 Bob 31
1 Jane 30
2 Alice 31
3 Ricky 30
id names ages
0 1 Bob 31
1 2 Jane 30
2 3 Alice 31
3 4 Ricky 30
'''
https://stackoverflow.com/questions/49989579
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