[英]Make Pandas dataframe from dict of dict that contain index mapped to value
我有一个dict
的dicts
,我试图使成Pandas
DataFrame
。 将该dict
构造为映射到将列索引映射为其值的dict
索引,然后我希望DataFrame
其他所有内容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
我目前正在计划编写一个函数,该函数将从d
每个项目中yield
一个元组,并将其用作创建DataFrame
的可迭代DataFrame
,但是我对是否还有其他人做过类似的事情感兴趣。
只需简单地调用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
好吧,为什么不按常规方式进行处理和移置它:
>>> 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
>>>
经过时间测试其他建议,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传递给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)
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