Как создать Pandas DataFrame из списка списков?

Question

У меня есть список списков.Вот пример этого.

[[], ['1', 'Anthony Taylor', '6', '11.17', '0.68', '0.17', '1.83', '11', '0.17', '1'], ['2', 'Jonathan Moss', '4', '12.75', '0.73', '0.00', '2.00', '8', '0.00', '0'], ['3', 'Martin Atkinson', '3', '11.00', '0.69', '0.67', '1.67', '5', '0.00', '0'], ['4', 'Graham Scott', '3', '10.00', '0.67', '0.00', '0.33', '1', '0.00', '0'], ['5', 'Chris Kavanagh', '3', '10.33', '0.74', '0.00', '1.67', '5', '0.00', '0'], ['6', 'Michael Oliver', '2', '12.00', '0.62', '1.00', '2.00', '4', '0.00', '0'], ['7', 'Paul Tierney', '2', '12.50', '0.66', '0.00', '2.50', '5', '0.00', '0'], ['8', 'Mike Dean', '2', '14.50', '0.73', '0.50', '3.00', '6', '0.00', '0'], ['9', 'Andre Marriner', '2', '8.50', '0.53', '0.00', '2.00', '4', '0.00', '0'], ['10', 'Kevin Friend', '2', '9.50', '0.50', '0.00', '3.00', '6', '0.00', '0'], ['11', 'Craig Pawson', '1', '9.00', '0.45', '0.00', '1.00', '1', '0.00', '0'], ['12', 'Stuart Attwell', '1', '9.00', '0.75', '0.00', '2.00', '2', '0.00', '0'], ['13', 'Lee Probert', '1', '11.00', '0.73', '0.00', '0.00', '0', '0.00', '0'], ['', 'Total / Average', '32', '11.09', '0.66', '0.19', '1.81', '58', '0.03', '1']]

Как создать из этого Pandas DataFrame, который похож на этот формат (каждая строка для отдельного имени)?

['1', 'Anthony Taylor', '6', '11.17', '0.68', '0.17', '1.83', '11', '0.17', '1']
['2', 'Jonathan Moss', '4', '12.75', '0.73', '0.00', '2.00', '8', '0.00', '0']
['3', 'Martin Atkinson', '3', '11.00', '0.69', '0.67', '1.67', '5', '0.00', '0']
['4', 'Graham Scott', '3', '10.00', '0.67', '0.00', '0.33', '1', '0.00', '0']
['5', 'Chris Kavanagh', '3', '10.33', '0.74', '0.00', '1.67', '5', '0.00', '0']
['6', 'Michael Oliver', '2', '12.00', '0.62', '1.00', '2.00', '4', '0.00', '0']
['7', 'Paul Tierney', '2', '12.50', '0.66', '0.00', '2.50', '5', '0.00', '0']
['8', 'Mike Dean', '2', '14.50', '0.73', '0.50', '3.00', '6', '0.00', '0']
['9', 'Andre Marriner', '2', '8.50', '0.53', '0.00', '2.00', '4', '0.00', '0']
['10', 'Kevin Friend', '2', '9.50', '0.50', '0.00', '3.00', '6', '0.00', '0']
['11', 'Craig Pawson', '1', '9.00', '0.45', '0.00', '1.00', '1', '0.00', '0']
['12', 'Stuart Attwell', '1', '9.00', '0.75', '0.00', '2.00', '2', '0.00', '0']
['13', 'Lee Probert', '1', '11.00', '0.73', '0.00', '0.00', '0', '0.00', '0']
['', 'Total / Average', '32', '11.09', '0.66', '0.19', '1.81', '58', '0.03', '1']

Iмне немного помог этот ответ во время моего исследования, но он все еще только помещает все в один столбец, в то время как я бы хотел отдельный столбец для каждого элемента определенного списка.Есть ли способ добиться этого?

Answer 1

Вы можете отфильтровать первый подсписок и перейти к конструктору DataFrame, также возможно задать имена столбцов другим списком:

L = [[], ['1', 'Anthony Taylor', '6', '11.17', '0.68', '0.17', '1.83', '11', '0.17', '1'], ['2', 'Jonathan Moss', '4', '12.75', '0.73', '0.00', '2.00', '8', '0.00', '0'], ['3', 'Martin Atkinson', '3', '11.00', '0.69', '0.67', '1.67', '5', '0.00', '0'], ['4', 'Graham Scott', '3', '10.00', '0.67', '0.00', '0.33', '1', '0.00', '0'], ['5', 'Chris Kavanagh', '3', '10.33', '0.74', '0.00', '1.67', '5', '0.00', '0'], ['6', 'Michael Oliver', '2', '12.00', '0.62', '1.00', '2.00', '4', '0.00', '0'], ['7', 'Paul Tierney', '2', '12.50', '0.66', '0.00', '2.50', '5', '0.00', '0'], ['8', 'Mike Dean', '2', '14.50', '0.73', '0.50', '3.00', '6', '0.00', '0'], ['9', 'Andre Marriner', '2', '8.50', '0.53', '0.00', '2.00', '4', '0.00', '0'], ['10', 'Kevin Friend', '2', '9.50', '0.50', '0.00', '3.00', '6', '0.00', '0'], ['11', 'Craig Pawson', '1', '9.00', '0.45', '0.00', '1.00', '1', '0.00', '0'], ['12', 'Stuart Attwell', '1', '9.00', '0.75', '0.00', '2.00', '2', '0.00', '0'], ['13', 'Lee Probert', '1', '11.00', '0.73', '0.00', '0.00', '0', '0.00', '0'], ['', 'Total / Average', '32', '11.09', '0.66', '0.19', '1.81', '58', '0.03', '1']]

---

cols = ['id','name','a','b','c','d','e','f','g','h']
df = pd.DataFrame(L[1:], columns=cols)
print(df)
    id             name   a      b     c     d     e   f     g  h
0    1   Anthony Taylor   6  11.17  0.68  0.17  1.83  11  0.17  1
1    2    Jonathan Moss   4  12.75  0.73  0.00  2.00   8  0.00  0
2    3  Martin Atkinson   3  11.00  0.69  0.67  1.67   5  0.00  0
3    4     Graham Scott   3  10.00  0.67  0.00  0.33   1  0.00  0
4    5   Chris Kavanagh   3  10.33  0.74  0.00  1.67   5  0.00  0
5    6   Michael Oliver   2  12.00  0.62  1.00  2.00   4  0.00  0
6    7     Paul Tierney   2  12.50  0.66  0.00  2.50   5  0.00  0
7    8        Mike Dean   2  14.50  0.73  0.50  3.00   6  0.00  0
8    9   Andre Marriner   2   8.50  0.53  0.00  2.00   4  0.00  0
9   10     Kevin Friend   2   9.50  0.50  0.00  3.00   6  0.00  0
10  11     Craig Pawson   1   9.00  0.45  0.00  1.00   1  0.00  0
11  12   Stuart Attwell   1   9.00  0.75  0.00  2.00   2  0.00  0
12  13      Lee Probert   1  11.00  0.73  0.00  0.00   0  0.00  0
13      Total / Average  32  11.09  0.66  0.19  1.81  58  0.03  1

Answer 2

Использование pandas.DataFrame.from_records:

import pandas as pd

df = pd.DataFrame.from_records(x[1:]))
print(df)

Выход:

    0                 1   2      3     4     5     6   7     8  9
0    1   Anthony Taylor   6  11.17  0.68  0.17  1.83  11  0.17  1
1    2    Jonathan Moss   4  12.75  0.73  0.00  2.00   8  0.00  0
2    3  Martin Atkinson   3  11.00  0.69  0.67  1.67   5  0.00  0
3    4     Graham Scott   3  10.00  0.67  0.00  0.33   1  0.00  0
4    5   Chris Kavanagh   3  10.33  0.74  0.00  1.67   5  0.00  0
5    6   Michael Oliver   2  12.00  0.62  1.00  2.00   4  0.00  0
6    7     Paul Tierney   2  12.50  0.66  0.00  2.50   5  0.00  0
7    8        Mike Dean   2  14.50  0.73  0.50  3.00   6  0.00  0
8    9   Andre Marriner   2   8.50  0.53  0.00  2.00   4  0.00  0
9   10     Kevin Friend   2   9.50  0.50  0.00  3.00   6  0.00  0
10  11     Craig Pawson   1   9.00  0.45  0.00  1.00   1  0.00  0
11  12   Stuart Attwell   1   9.00  0.75  0.00  2.00   2  0.00  0
12  13      Lee Probert   1  11.00  0.73  0.00  0.00   0  0.00  0
13      Total / Average  32  11.09  0.66  0.19  1.81  58  0.03  1