Question

我正在尝试将以下字典转换为数据框：

city_data = {
    'San Francisco': {'x': [1, 2, 3], 'y': [4, 1, 2]},
    'Montreal': {'x': [1, 2, 3], 'y': [2, 4, 5]},
    'New York City': {'x': [1, 2, 3], 'y': [2, 2, 7]},
    'Cincinnati': {'x': [1, 2, 3], 'y': [1, 0, 2]},
    'Toronto': {'x': [1, 2, 3], 'y': [4, 7, 3]},
    'Ottawa': {'x': [1, 2, 3], 'y': [2, 3, 3]}
}

数据框看起来像这样：

city            |  x  |  y
San Francisco   |  1  |  4
San Francisco   |  2  |  1
San Francisco   |  3  |  2
...

使用我在这里找到的解决方案Unfold a nested dictionary with lists into a pandas DataFrame，我尝试过：

data = city_data

def unroll(data):
    if isinstance(data, dict):
        for key, value in data.items():
            # Recursively unroll the next level and prepend the key to each row.
            for row in unroll(value):
                yield [key] + row
    if isinstance(data, list):
        # This is the bottom of the structure (defines exactly one row).
        yield data

df = pd.DataFrame(list(unroll(nested_dict)))
df.rename(columns=lambda i: 'col{}'.format(i+1))

但是，我最终得到以下结果：

Answer 1

几乎是骗子，但输入使它变得有些棘手。使用@Wen中的unnesting。

df = pd.DataFrame.from_dict(city_data, orient='index')
unnesting(df, ['x', 'y'])

               x  y 
Cincinnati     1  1 
Cincinnati     2  0 
Cincinnati     3  2 
Montreal       1  2 
Montreal       2  4 
Montreal       3  5 
New York City  1  2 
New York City  2  2 
New York City  3  7 
Ottawa         1  2 
Ottawa         2  3 
Ottawa         3  3 
San Francisco  1  4 
San Francisco  2  1 
San Francisco  3  2 
Toronto        1  4 
Toronto        2  7 
Toronto        3  3

链接的答案中的文氏函数：

def unnesting(df, explode):
    idx = df.index.repeat(df[explode[0]].str.len())
    df1 = pd.concat([
        pd.DataFrame({x: np.concatenate(df[x].values)}) for x in explode], axis=1)
    df1.index = idx

    return df1.join(df.drop(explode, 1), how='left')

Answer 2

借助@roganjosh的提示，我能够回答我的问题。这是我最终使用的解决方案：

city_data = {
        'San Francisco': {'x': [1, 2, 3], 'y': [4, 1, 2]},
        'Montreal': {'x': [1, 2, 3], 'y': [2, 4, 5]},
        'New York City': {'x': [1, 2, 3], 'y': [2, 2, 7]},
        'Cincinnati': {'x': [1, 2, 3], 'y': [1, 0, 2]},
        'Toronto': {'x': [1, 2, 3], 'y': [4, 7, 3]},
        'Ottawa': {'x': [1, 2, 3], 'y': [2, 3, 3]}
    }

## Prepare my data
data = []
for city in city_data:
    data.append({'x': city_data[city]['x'], 'y': city_data[city]['y'],
             'city': city})


### use function from linked SO question

def explode(df, lst_cols, fill_value='', preserve_index=False):
    # make sure `lst_cols` is list-alike
    if (lst_cols is not None
        and len(lst_cols) > 0
        and not isinstance(lst_cols, (list, tuple, np.ndarray, pd.Series))):
        lst_cols = [lst_cols]
    # all columns except `lst_cols`
    idx_cols = df.columns.difference(lst_cols)
    # calculate lengths of lists
    lens = df[lst_cols[0]].str.len()
    # preserve original index values    
    idx = np.repeat(df.index.values, lens)
    # create "exploded" DF
    res = (pd.DataFrame({
                col:np.repeat(df[col].values, lens)
                for col in idx_cols},
                index=idx)
             .assign(**{col:np.concatenate(df.loc[lens>0, col].values)
                            for col in lst_cols}))
    # append those rows that have empty lists
    if (lens == 0).any():
        # at least one list in cells is empty
        res = (res.append(df.loc[lens==0, idx_cols], sort=False)
                  .fillna(fill_value))
    # revert the original index order
    res = res.sort_index()
    # reset index if requested
    if not preserve_index:        
        res = res.reset_index(drop=True)
    return res

df = pd.DataFrame(data)
df = explode(df, ['x','y'], fill_value='')

输出：

    city            x   y
0   San Francisco   1   4
1   San Francisco   2   1
2   San Francisco   3   2
3   Montreal        1   2
4   Montreal        2   4

虽然这是一个非常冗长的代码，但如果有人可以分享一种更简洁的解决方法，我将感到非常高兴。

Answer 3

尝试：

df = pd.DataFrame(list(unroll(city_data)))
new_df = (df.set_index([0,1])
           .groupby(level=0)
           .apply(lambda x: x.reset_index(level=0,drop=True).T)
        )

new_df.reset_index(level=1, drop=True)

输出：

1              x  y
0                  
Cincinnati     1  1
Cincinnati     2  0
Cincinnati     3  2
Montreal       1  2
Montreal       2  4
Montreal       3  5
New York City  1  2
New York City  2  2
New York City  3  7
Ottawa         1  2
Ottawa         2  3
Ottawa         3  3
San Francisco  1  4
San Francisco  2  1
San Francisco  3  2
Toronto        1  4
Toronto        2  7
Toronto        3  3

Answer 4

此解决方案特定于您的输入数据形式。

out = pd.concat((pd.DataFrame(data) for data in city_data.values()), 
                 keys=city_data.keys(), names = ["city", ""], sort=False)
out.reset_index(level=0, inplace=True)
out.reset_index(drop=True, inplace=True)

输出

             city  x  y
0   San Francisco  1  4
1   San Francisco  2  1
2   San Francisco  3  2
3        Montreal  1  2
4        Montreal  2  4
5        Montreal  3  5
6   New York City  1  2
7   New York City  2  2
8   New York City  3  7
9      Cincinnati  1  1
10     Cincinnati  2  0
11     Cincinnati  3  2
12        Toronto  1  4
13        Toronto  2  7
14        Toronto  3  3
15         Ottawa  1  2
16         Ottawa  2  3
17         Ottawa  3  3

如何将列表作为值的嵌套字典转换为Pandas DataFrame

4 个答案: