这里有一个简单的方法，你可以替换掉不需要的引号，然后用一个空格分割，使用列表解析得到一个列表：

df2 = [''.join(row).replace('"', '').split(" ") for row in df1]

print(df2)

输出：

[['1', 'P040', '68.13', 'P040_1', '2.55', '8'],
['2', 'P040', '46.82', 'P040_2', '2.53', '8'],
['3', 'P040', '46.82', 'P040_3', '2.51', '8']]

因为list中的每个list只包含一个元素，所以你不需要运行2个for循环，这可以用一行列表解析来解决：

df1 = [
['1 "P040" 68.13 "P040_1" 2.55 8'],
['2 "P040" 46.82 "P040_2" 2.53 8'],
['3 "P040" 46.82 "P040_3" 2.51 8']
]

df2 = [row[0].replace('"', '').split(' ') for row in df1]

print(df2)

>>> [['1', 'P040', '68.13', 'P040_1', '2.55', '8'],
     ['2', 'P040', '46.82', 'P040_2', '2.53', '8'],
     ['3', 'P040', '46.82', 'P040_3', '2.51', '8']]

你可以通过嵌套列表解析来实现它：

df2 = [[item.strip('""') for item in elem.split()] for row in df1 for elem in row]

或具有列表解析的嵌套循环：

df2 = []
for row in df1:
    for elem in row:
        df2.append([item.strip('""') for item in elem.split()])

输出：

['1', 'P040', '68.13', 'P040_1', '2.55', '8']
['2', 'P040', '46.82', 'P040_2', '2.53', '8']
['3', 'P040', '46.82', 'P040_3', '2.51', '8']

df1 = [['1 "P040" 68.13 "P040_1" 2.55 8'],
       ['2 "P040" 46.82 "P040_2" 2.53 8'],
       ['3 "P040" 46.82 "P040_3" 2.51 8']]

df1 = [[v.replace(' ','').split('"') for v in l] for l in df1]
print(df1)

您可以组合使用split()和strip()函数来拆分引号和空格。
此示例将拆分df1中的元素并创建一个新的df2列表：

df2 = []
for row in df1:
    new_row = []
    for elem in row[0].split('"'):
        new_row.extend(elem.strip().split())
    df2.append(new_row)

print(df2)

您可以使用shlex.split（）在'words'上进行拆分，其中单词可能是带引号的字符串，如下所示：

import shlex

for i in range(len(df1)):
    df1[i] = shlex.split(df1[i][0])

这是假设你的每个列表项都是一个包含字符串的列表。这修改了df1，创建了一个新的'df2'，它将是：

import shlex

df2 = []
for row in df1:
    df2.append(shlex.split(df1[i][0]))

输出将为：

[
  ['1', 'P040', '68.13', 'P040_1', '2.55', '8'],
  ['2', 'P040', '46.82', 'P040_2', '2.53', '8'],
  ['3', 'P040', '46.82', 'P040_3', '2.51', '8']
]

使用shlex.split（）的优点是，带引号的字符串可以包含空格，而不会产生普通的'split（）'无法解决的问题。