Converting A Dictionary Into A Square Matrix

I am wanting to learn how to convert a dictionary into a square matrix. From what I have read, I may need to convert this into a numpy array and then reshape it. I do not want to u

Solution 1:

If you can get your data in the format

{name1: {name1:data, name2:data, name3:data, ...}, 
 name2: {name1:data, name2:data, name3:data, ...},
 ...
}

then you can just hand it to a pandas DataFrame and it will make it for you. The data at position row = name1 and col = name2 will be the value of name1 vs name2. Here is the code that will do it:

from collections import defaultdict
import pandas

result = defaultdict(dict)
for key,value in dict1.items():
     names = key.split()
     name1 = names[0]
     name2 = names[2]    
     result[name1][name2] = value

df = pandas.DataFrame(result).transpose()
print(df)

This gives the following output:

                              Ann                                  Bob                                        Jen                                      Sarah
Ann    {'shepherd': 13, 'collie': 2, 'poodle': 4}  {'shepherd': 6, 'collie': 2, 'poodle': 5}  {'shepherd': 2, 'collie': 8, 'poodle': 2}  {'shepherd': 0, 'collie': 2, 'poodle': 4}
Bob     {'shepherd': 3, 'collie': 2, 'poodle': 1}  {'shepherd': 3, 'collie': 2, 'poodle': 2}  {'shepherd': 3, 'collie': 2, 'poodle': 2}  {'shepherd': 1, 'collie': 5, 'poodle': 8}
Jen     {'shepherd': 3, 'collie': 7, 'poodle': 2}  {'shepherd': 4, 'collie': 8, 'poodle': 1}  {'shepherd': 9, 'collie': 7, 'poodle': 2}  {'shepherd': 7, 'collie': 9, 'poodle': 2}
Sarah   {'shepherd': 4, 'collie': 6, 'poodle': 3}  {'shepherd': 3, 'collie': 2, 'poodle': 4}  {'shepherd': 1, 'collie': 5, 'poodle': 8}  {'shepherd': 3, 'collie': 2, 'poodle': 2}

A simple conversion to a numpy array would look like:

numpy_array = df.as_matrix()
print(numpy_array)

[[{'shepherd': 13, 'collie': 2, 'poodle': 4}
  {'shepherd': 6, 'collie': 2, 'poodle': 5}
  {'shepherd': 2, 'collie': 8, 'poodle': 2}
  {'shepherd': 0, 'collie': 2, 'poodle': 4}]
 [{'shepherd': 3, 'collie': 2, 'poodle': 1}
  {'shepherd': 3, 'collie': 2, 'poodle': 2}
  {'shepherd': 3, 'collie': 2, 'poodle': 2}
  {'shepherd': 1, 'collie': 5, 'poodle': 8}]
 [{'shepherd': 3, 'collie': 7, 'poodle': 2}
  {'shepherd': 4, 'collie': 8, 'poodle': 1}
  {'shepherd': 9, 'collie': 7, 'poodle': 2}
  {'shepherd': 7, 'collie': 9, 'poodle': 2}]
 [{'shepherd': 4, 'collie': 6, 'poodle': 3}
  {'shepherd': 3, 'collie': 2, 'poodle': 4}
  {'shepherd': 1, 'collie': 5, 'poodle': 8}
  {'shepherd': 3, 'collie': 2, 'poodle': 2}]]

Solution 2:

You may achieve this via generating a nested dictionary like:

{'Bob':   {'Bob': .., 'Sarah': .., 'Ann': .., 'Jen':..}
 'Sarah': {.. .. ..},
 'Ann':   {.. .. ..},
 'Jen':   {.. .. ..},
 }

Below is the sample code:

>>> my_dict = {'Bob VS Sarah': {'shepherd': 1,'collie': 5,'poodle': 8},
... 'Bob VS Ann': {'shepherd': 3,'collie': 2,'poodle': 1},
... 'Bob VS Jen': {'shepherd': 3,'collie': 2,'poodle': 2},
... 'Sarah VS Bob': {'shepherd': 3,'collie': 2,'poodle': 4},
... 'Sarah VS Ann': {'shepherd': 4,'collie': 6,'poodle': 3},
... 'Sarah VS Jen': {'shepherd': 1,'collie': 5,'poodle': 8},
... 'Jen VS Bob': {'shepherd': 4,'collie': 8,'poodle': 1},
... 'Jen VS Sarah': {'shepherd': 7,'collie': 9,'poodle': 2},
... 'Jen VS Ann': {'shepherd': 3,'collie': 7,'poodle': 2},
... 'Ann VS Bob': {'shepherd': 6,'collie': 2,'poodle': 5},
... 'Ann VS Sarah': {'shepherd': 0,'collie': 2,'poodle': 4},
... 'Ann VS Jen': {'shepherd': 2,'collie': 8,'poodle': 2},
... 'Bob VS Bob': {'shepherd': 3,'collie': 2,'poodle': 2},
... 'Sarah VS Sarah': {'shepherd': 3,'collie': 2,'poodle': 2},
... 'Ann VS Ann': {'shepherd': 13,'collie': 2,'poodle': 4},
... 'Jen VS Jen': {'shepherd': 9,'collie': 7,'poodle': 2}}
>>> new_dict = {}
>>> for key, value in my_dict.iteritems():
...     first_name, second_name = map(lambda x: x.strip(), key.split('VS'))
... if first_name notin new_dict:
...         new_dict[first_name] = {}
...     new_dict[first_name][second_name] = value
... >>> new_dict
{'Sarah': {'Sarah': {'shepherd': 3, 'collie': 2, 'poodle': 2}, 
            'Ann': {'shepherd': 4, 'collie': 6, 'poodle': 3}, 
            'Jen': {'shepherd': 1, 'collie': 5, 'poodle': 8}, 
            'Bob': {'shepherd': 3, 'collie': 2, 'poodle': 4}
          }, 
 'Bob':   {'Sarah': {'shepherd': 1, 'collie': 5, 'poodle': 8}, 
           'Bob': {'shepherd': 3, 'collie': 2, 'poodle': 2}, 
           'Jen': {'shepherd': 3, 'collie': 2, 'poodle': 2}, 
           'Ann': {'shepherd': 3, 'collie': 2, 'poodle': 1}}, 
 'Jen':   {'Sarah': {'shepherd': 7, 'collie': 9, 'poodle': 2}, 
           'Bob': {'shepherd': 4, 'collie': 8, 'poodle': 1}, 
           'Jen': {'shepherd': 9, 'collie': 7, 'poodle': 2}, 
           'Ann': {'shepherd': 3, 'collie': 7, 'poodle': 2}
          }, 
 'Ann': {'Sarah': {'shepherd': 0, 'collie': 2, 'poodle': 4}, 
                   'Bob': {'shepherd': 6, 'collie': 2, 'poodle': 5}, 
                   'Jen': {'shepherd': 2, 'collie': 8, 'poodle': 2},                   
                   'Ann': {'shepherd': 13, 'collie': 2, 'poodle': 4}
        }
}