Make Pandas dataframe from dict of dict that contain index mapped to value

Question

I have a dict of dicts that I'm trying to make into a Pandas DataFrame . The dict is structured to be the index mapped to a dict that maps column indexes to their value, and then I want everything else in the DataFrame to be 0. For example:

d = {0: {0:2, 2:5},
     1: {1:1, 3:2},
     2: {2:5}}

So then then I want the DataFrame to look like

index   c0   c1   c2   c3
    0  2.0  NaN  5.0  NaN
    1  NaN  1.0  NaN  2.0
    2  NaN  NaN  5.0  NaN

I currently am planning on writing a function that will yield a tuple from each item from d and using that as an iterable for creating the DataFrame , but am interested in if anyone else has done anything similar.

Answer 1

Just simple call DataFrame.from_dict

pd.DataFrame.from_dict(d,'index').sort_index(axis=1)
     0    1    2    3
0  2.0  NaN  5.0  NaN
1  NaN  1.0  NaN  2.0
2  NaN  NaN  5.0  NaN

Answer 2

Well, why not doing it in the regular way and transposing it:

>>> pd.DataFrame(d).T
     0    1    2    3
0  2.0  NaN  5.0  NaN
1  NaN  1.0  NaN  2.0
2  NaN  NaN  5.0  NaN
>>> 

Answer 3

After time testing the other suggestions, I found my original method was a lot faster. I am using the following function to make an iterator that I pass into pd.DataFrame

def row_factory(index_data, row_len):
    """
    Make a generator for iterating for index_data

    Parameters:
        index_data (dict): a dict mapping the a value to a dict of index mapped to values. All indexes not in
                           second dict are assumed to be None.
        row_len (int): length of row

    Example:
        index_data = {0: {0:2, 2:1}, 1: {1:1}} would yield [0, 2, None, 1] then [1, None, 1, None]
    """
    for key, data in index_data.items():
        # Initialize row with the key starting, then None for each value
        row = [key] + [None] * (row_len - 1)
        for index, value in data.items():
            # Only replace indexes that have a value
            row[index] = value
        yield row

df = pd.DataFrame(row_factory(d), 5)