用Python转置表的一部分

我需要帮助在python中构建一个脚本来转换csv的一部分，如下例所示：

[[0A,0B,0C,0D,0E,0F]
[01,02,03,04,05,06]
[07,08,09,10,11,12]
[13,14,15,16,17,18]
[19,20,21,22,23,24]]

it:

[[0A, 0B, 0C]
[01,02,03]
[01,02,04]
[01,02,05]
[01,02,06]
[07,08,09]
[07.08, 10]
[07,08,11]
 .
 .
 .
[19,20,24]]

有人用python对csv/xls进行过这种转换吗？

您可以使用列表切片、嵌套列表综合和itertools.chain:的组合

M = [['0A','0B','0C','0D','0E','0F'],
     ['01','02','03','04','05','06'],
     ['07','08','09','10','11','12'],
     ['13','14','15','16','17','18'],
     ['19','20','21','22','23','24']]
from itertools import chain
result = [M[0][:3]] + list(chain(*[[x[:2] + [y] for y in x[2:]] for x in M[1:]]))

之后，result为

[['0A', '0B', '0C'],
 ['01', '02', '03'],
 ['01', '02', '04'],
 ['01', '02', '05'],
 ...
 ['19', '20', '23'],
 ['19', '20', '24']]

[row[:3] for row in my_table]

将只给你前3列

或者，我想你可以把它转换成列，得到前3个，然后把它转换回

transposed_table = zip(*my_table)
my_cols = transposed_table[:3]
transposed_back_to_rows = zip(*my_cols)

或者如果你真的想把它放在一行上

zip(*list(zip(*my_table))[:3])

如果你能使用numpy库，你可以使用切片来轻松实现这一点-

示例-

import numpy as np
n = np.array([['0A','0B','0C','0D','0E','0F'],
['01','02','03','04','05','06'],
['07','08','09','10','11','12'],
['13','14','15','16','17','18'],
['19','20','21','22','23','24']])
print('n - ' + str(n))
n1 = np.append(n[:,:3],n[:,3:], axis = 0)
print('n1 - ' + str(n1))

输出-

n - [['0A' '0B' '0C' '0D' '0E' '0F']
 ['01' '02' '03' '04' '05' '06']
 ['07' '08' '09' '10' '11' '12']
 ['13' '14' '15' '16' '17' '18']
 ['19' '20' '21' '22' '23' '24']]
n1 - [['0A' '0B' '0C']
 ['01' '02' '03']
 ['07' '08' '09']
 ['13' '14' '15']
 ['19' '20' '21']
 ['0D' '0E' '0F']
 ['04' '05' '06']
 ['10' '11' '12']
 ['16' '17' '18']
 ['22' '23' '24']]

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