python字典dict和DataFrame、Series之间的相互转换

1.包含字典的嵌套字典，转DataFrame

import pandas as pd

pop = {'Nevada': {2001: 2.4, 2002: 2.9},
       'Ohio': {2000: 1.5, 2001: 1.7, 2002: 3.6}}
frame = pd.DataFrame(pop)

frame
Out[3]: 
      Nevada  Ohio
2001     2.4   1.7
2002     2.9   3.6
2000     NaN   1.5

类似于numpy的转置：

frame.T
Out[4]: 
        2001  2002  2000
Nevada   2.4   2.9   NaN
Ohio     1.7   3.6   1.5

2. 利用包含等长度列表或NumPy数组的 字典dict 来形成DataFrame

注意：等长度 的列表或数组

data = {'state': ['Ohio', 'Ohio', 'Ohio', 'Nevada', 'Nevada', 'Nevada'],
        'year': [2000, 2001, 2002, 2001, 2002, 2003],
        'pop': [1.5, 1.7, 3.6, 2.4, 2.9, 3.2]}
frame = pd.DataFrame(data)

frame
Out[8]: 
    state  year  pop
0    Ohio  2000  1.5
1    Ohio  2001  1.7
2    Ohio  2002  3.6
3  Nevada  2001  2.4
4  Nevada  2002  2.9
5  Nevada  2003  3.2

指定列的顺序

pd.DataFrame(data, columns=['year', 'state', 'pop'])
Out[9]: 
   year   state  pop
0  2000    Ohio  1.5
1  2001    Ohio  1.7
2  2002    Ohio  3.6
3  2001  Nevada  2.4
4  2002  Nevada  2.9
5  2003  Nevada  3.2

3. pd.Series 转为 dict

C = [6367, 18]
pd.Series(C).value_counts().to_dict()  # C 为list

输出结果：

{0: 6367, -1: 1103, 1: 18}

4.快速给dataframe重命名 columns = dict(zip(list1, list2))

df_10min = df_10min.rename(columns=dict(zip(df_10min.columns, wind_profile_clustering.height)))