Question

我有一个dict的一个dicts，我正尝试将其变成一个Pandas DataFrame。 dict的结构是映射到dict的索引，该索引将列索引映射到它们的值，然后我希望DataFrame中的其他所有内容都为0。例如：

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

所以然后我希望DataFrame看起来像

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

我目前正计划编写一个函数，该函数将从yield的每个项目中d个元组，并将其用作创建DataFrame的可迭代项，但是我对是否有人感兴趣其他都做了类似的事情。

Answer 1

只需简单调用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，为什么不按常规方式进行处理和转置呢？

>>> 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

在对其他建议进行时间测试之后，我发现我原来的方法要快得多。我正在使用以下函数来制作迭代器，并将其传递给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)

使用包含映射到值的索引的dict来使Pandas数据框

3 个答案: