Python 从数据帧生成传输模型的字典

我有一个运输问题的数据框架

Unnamed: 0 Unnamed: 1    c1   c2   c3   c4   c5  capacity
0        NaN         p1   4    5    6    8   10     500.0
1        NaN         p2   6    4    3    5    8     500.0
2        NaN         p3   9    7    4    2    4     500.0
3     demand        NaN  80  270  250  160  180       NaN

我像这样更改了列名

df.columns = ['Demand', 'Plant', 'c1', 'c2', 'c3', 'c4', 'c5', 'capacity']

 d = {c1:80, c2:270, c3:250, c4:160, c5:180}  # customer demand
 M = {p1:500, p2:500, p3:500}               # factory capacity
 I = [c1,c2,c3,c4,c5]                         # Customers
 J = [p1,p2,p3]                             # Factories
 cost = {(p1,c1):4,    (p1,c2):5,    (p1,c3):6,
 (p1,c4):8,    (p1,c5):10, ......
  } 

我想做一本这样的字典

df.columns = ['Demand', 'Plant', 'c1', 'c2', 'c3', 'c4', 'c5', 'capacity']

 d = {c1:80, c2:270, c3:250, c4:160, c5:180}  # customer demand
 M = {p1:500, p2:500, p3:500}               # factory capacity
 I = [c1,c2,c3,c4,c5]                         # Customers
 J = [p1,p2,p3]                             # Factories
 cost = {(p1,c1):4,    (p1,c2):5,    (p1,c3):6,
 (p1,c4):8,    (p1,c5):10, ......
  } 

对于第一种情况，我使用了以下代码

  M = df.set_index('Plant')['capacity'].to_dict()

它给了我

  {'p1': 500.0, 'p2': 500.0, 'p3': 500.0, nan: nan}  

我不想要任何价值观。请帮助以通用方式查找此总词典（d、M和cost），无需NaN

df1 = df.set_index(["Unnamed: 0", "Unnamed: 1"])
plants = df1.loc[np.NaN]  # remove demand from dataframe

d = dict(df1.loc["demand"].T.squeeze().dropna().iteritems())
M = dict(plants["capacity"].iteritems())
I = list(plants.drop(columns="capacity").columns)
J = list(plants.index)
cost = dict(plants.drop(columns="capacity").stack().iteritems())