Python 如何从列表列表创建数据帧？

我有一份清单。这是它的一个样本

[[], ['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']]

我如何用这种格式创建一个类似于这种格式的Pandas数据帧（每一行都有一个单独的名称）

在我的研究过程中，我的答案对我有所帮助，但它仍然只将所有内容放在一个单独的列中，而我希望为特定列表中的每个元素单独设置一个列。

---

有没有办法做到这一点

您可以筛选出第一个子列表并传递给DataFrame构造函数，也可以通过另一个列表设置列名称：

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']]

---

使用：

输出：

    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

---

这对我很有用，谢谢。如果有可能接受两个答案，我会这么做：）这是我今天对熊猫的新认识。非常感谢

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

import pandas as pd

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

    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