pandas.DataFrame.from_dict

classmethod DataFrame.from_dict(data, orient='columns', dtype=None, columns=None) [source]

Construct DataFrame from dict of array-like or dicts.

Creates DataFrame object from dictionary by columns or by index allowing dtype specification.

Parameters:

data : dict Of the form {field : array-like} or {field : dict}. orient : {‘columns’, ‘index’}, default ‘columns’ The “orientation” of the data. If the keys of the passed dict should be the columns of the resulting DataFrame, pass ‘columns’ (default). Otherwise if the keys should be rows, pass ‘index’. dtype : dtype, default None Data type to force, otherwise infer. columns : list, default None Column labels to use when orient='index'. Raises a ValueError if used with orient='columns'. New in version 0.23.0.

Returns:

DataFrame

See also

DataFrame.from_records

DataFrame from ndarray (structured dtype), list of tuples, dict, or DataFrame.

DataFrame

DataFrame object creation using constructor.

Примеры

По умолчанию ключи словаря становятся столбцами DataFrame:

>>> data = {'col_1': [3, 2, 1, 0], 'col_2': ['a', 'b', 'c', 'd']}
>>> pd.DataFrame.from_dict(data)
   col_1 col_2
0      3     a
1      2     b
2      1     c
3      0     d

Укажите orient='index' для создания DataFrame, используя ключи словаря как строки:

>>> data = {'row_1': [3, 2, 1, 0], 'row_2': ['a', 'b', 'c', 'd']}
>>> pd.DataFrame.from_dict(data, orient='index')
       0  1  2  3
row_1  3  2  1  0
row_2  a  b  c  d

При использовании ориентации ‘index’, имена столбцов можно указать вручную:

>>> pd.DataFrame.from_dict(data, orient='index',
...                        columns=['A', 'B', 'C', 'D'])
       A  B  C  D
row_1  3  2  1  0
row_2  a  b  c  d