使用包含映射到值的索引的dict來使Pandas數據框

Question

我有一個dict的dicts ，我試圖使成Pandas DataFrame 。 將該dict構造為映射到將列索引映射為其值的dict索引，然后我希望DataFrame其他所有內容DataFrame 0。例如：

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

所以那么我希望DataFrame看起來像

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

我目前正在計划編寫一個函數，該函數將從d每個項目中yield一個元組，並將其用作創建DataFrame的可迭代DataFrame ，但是我對是否還有其他人做過類似的事情感興趣。

Answer 1

只需簡單地調用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

好吧，為什么不按常規方式進行處理和移置它：

>>> 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

經過時間測試其他建議，我發現我原來的方法要快得多。 我正在使用以下函數來制作傳遞給pd.DataFrame的迭代器

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)