Pandas_构建dataframe的多种方法

1. 使用字典构建
   import pandas as pd

data = {
‘Column1’: [1, 2, 3],
‘Column2’: [‘A’, ‘B’, ‘C’]
}
df = pd.DataFrame(data)

2. 使用列表的列表构建
   data = [
   [1, ‘A’],
   [2, ‘B’],
   [3, ‘C’]
   ]
   df = pd.DataFrame(data, columns=[‘Column1’, ‘Column2’])

3. 使用NumPy数组构建
   import numpy as np

data = np.array([
[1, ‘A’],
[2, ‘B’],
[3, ‘C’]
])
df = pd.DataFrame(data, columns=[‘Column1’, ‘Column2’])

4. 使用Series对象构建
   s1 = pd.Series([1, 2, 3], name=‘Column1’)
   s2 = pd.Series([‘A’, ‘B’, ‘C’], name=‘Column2’)
   df = pd.DataFrame([s1, s2])

5. 使用从文件读取构建
   • 从CSV文件读取
   df = pd.read_csv(‘path_to_file.csv’)

• 从Excel文件读取
df = pd.read_excel(‘path_to_file.xlsx’)

• 从SQL数据库读取
import sqlite3
conn = sqlite3.connect(‘path_to_db.db’)
df = pd.read_sql_query(‘SELECT * FROM table_name’, conn)

6. 使用pd.concat或pd.merge合并多个DataFrame
   df1 = pd.DataFrame({‘A’: [1, 2, 3]})
   df2 = pd.DataFrame({‘B’: [4, 5, 6]})
   df = pd.concat([df1, df2], axis=1)

7. 使用pd.DataFrame.from_records
   records = [
   (1, ‘A’),
   (2, ‘B’),
   (3, ‘C’)
   ]
   df = pd.DataFrame.from_records(records, columns=[‘Column1’, ‘Column2’])

8. 使用pd.DataFrame.from_dict构建
   • 将字典的值作为列
   data = {
   ‘Column1’: [1, 2, 3],
   ‘Column2’: [‘A’, ‘B’, ‘C’]
   }
   df = pd.DataFrame.from_dict(data)

• 将字典的键作为列
data = {
‘Row1’: [1, ‘A’],
‘Row2’: [2, ‘B’],
‘Row3’: [3, ‘C’]
}
df = pd.DataFrame.from_dict(data, orient=‘index’)

9. 使用pd.DataFrame.from_items
   items = [
   (‘Column1’, [1, 2, 3]),
   (‘Column2’, [‘A’, ‘B’, ‘C’])
   ]
   df = pd.DataFrame.from_items(items)

10. 使用pd.DataFrame的构造函数
    df = pd.DataFrame(index=[0, 1, 2], columns=[‘Column1’, ‘Column2’])
    df.loc[0] = [1, ‘A’]
    df.loc[1] = [2, ‘B’]
    df.loc[2] = [3, ‘C’]