How to create a matrix with values as dictionaries in python

Question

To create a matrix whose values are lists I normally do the following:

T = [[0]*(4) for x in range(4)]
print T

Output:

[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]

I am trying to create a matrix whose values are dictionaries , I try the same thing and I get problem.

T = [{"first":None,"second":None} *(4) for x in range(4)]

TypeError: unsupported operand type(s) for *: 'dict' and 'int'

What is the correct way to do this?

Intention:

At each T[i][j] , I am trying to record the values picked by first player and second player in a game round (i,j).

@StefanPochmann No, my lists version is right. I get the desired output.[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]] — KurinchiMalar, Jan 26 '16 at 04:35
Possible duplicate of [How to define two-dimensional array in python](http://stackoverflow.com/questions/6667201/how-to-define-two-dimensional-array-in-python) — Anantha Krishnan, Jan 26 '16 at 04:39
@AnanthaKrishnan No my question is different. I want a two-dimensional array with values in each T[i][j] as dictionaries. — KurinchiMalar, Jan 26 '16 at 04:52

score 1 · Accepted Answer · edited May 23 '17 at 11:52

If I understand you correctly you should write:

T = [[{"first":None,"second":None} for y in range(4)] for x in range(4)]
print T

`

[[{'second': None, 'first': None}, {'second': None, 'first': None},
{'second': None, 'first': None}, {'second': None, 'first': None}],
 [{'second': None, 'first': None}, {'second': None, 'first': None},
 {'second': None, 'first': None}, {'second': None, 'first': None}], 
[{'second': None, 'first': None}, {'second': None, 'first': None}, 
{'second': None, 'first': None}, {'second': None, 'first': None}], 
[{'second': None, 'first': None}, {'second': None, 'first': None},
 {'second': None, 'first': None}, {'second': None, 'first': None}]]`

Reason you can't do:

T = [[{"first":None,"second":None}]*4 for x in range(4)]

is the same as the reason you can't do:

T = [[[1]]*4 for x in range(4)]

When you write * you are essentially repeating the same object given number of times. So, if you update one of the entries all the entries in that row will get updated as they are all the same object.

This doesn't happen in case of : T = [[0]*4 for x in range(4)]

as 0 is an int and you replace it completely when you assign a new value at any place. You can find better explanation in some other SO answers. 1, 2, 3

For dictionaries + and * are not supported. If this answers your problem then you should do more research before posting a question. Otherwise, please elaborate.

I already tried this. when i try to update.T[0][0]["first"] = 2 T[0][0]["second"] = 3 the entire first list gets updated. which i dont want. I want just the first dic at T[0][0] to get updated. — KurinchiMalar, Jan 26 '16 at 05:25
This is exactly what I wanted! Thankyou... Now I am able to reference T[0][0]["first"] = 2 T[0][0]["second"] = 3 T[0][1]["first"] = 4 and get necessary changes — KurinchiMalar, Jan 26 '16 at 05:40

Tadhg McDonald-Jensen · Answer 2 · 2016-01-26T12:42:02.020

0

if you want a dictionary with 4 unique keys you need to have a generator making the keys as well:

EDIT - removed *(4) from copied code

T = [{i:None for i in range(4)} for x in range(4)]

do take note that the keys of a dictionary do not strictly have to be strings, this makes the keys ints so it would work very similarly to any code meant to support the version using lists

edited Jan 26 '16 at 12:42

answered Jan 26 '16 at 05:00

Tadhg McDonald-Jensen

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actually you might as well do `T = {x:{i:None for i in range(4)} for x in range(4)}` while you are at it... – Tadhg McDonald-Jensen Jan 26 '16 at 05:01
For the answer you suggested I am still getting "TypeError: unsupported operand type(s) for *: 'dict' and 'int'" – KurinchiMalar Jan 26 '16 at 05:38
my bad, copied from your's and forgot to remove that :P now it works – Tadhg McDonald-Jensen Jan 26 '16 at 12:42
I just looked at the question again and based on the _intention_ this will not solve the problem, this will make each row a dict instead of each cell. – Tadhg McDonald-Jensen Jan 26 '16 at 12:52

score 0 · Answer 3 · answered Jan 26 '16 at 05:39

I feel that the other answers do not give you what you exactly want, which is how to hack on things to make them behave the way you want to.

Since the dict class does not have an imul numeric type to deal with multiplication operation, we can simply create our own class which does that for us.

a = { 'first': None , 'second': None }

class MultDict:
    def __mul__(self, other):
        lst = []
        for i in range(other):
            lst.append(a.copy())
        return lst

x = MultDict()

t = [ x * (4) for i in range(4)]

print(t)

t[0][0]['first']=2
print(t)

Output:

[[{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None}],
[{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None}],
[{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None}],
[{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None}]] 


[[{'first': 2, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None}],
[{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None}],
[{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None}],
[{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None},
{'first': None, 'second': None}]]

How to create a matrix with values as dictionaries in python

3 Answers3