将列表拆分为子列表：

Question

我搜刮了一个囚犯搜索网站，有一个囚犯每次被捕都有多项指控的例子。我计划将它们映射到 django 中的一对多架构数据库中。但是，每当我抓取时，犯人多次指控的情况结果如下所示：

[['MURDER   /   Disposed   /   ,000.00', 'MANUF., POSS OF OTHER SUB. IN SCH. I,IIIII OR PWID - 1ST OFFENSE   /   Disposed   /   ,000.00']...]

因此，不是一对多，而是一对一。这两项费用在 35,000 之后用逗号分隔。有没有人知道这是否可能，因为在某些情况下每个囚犯被指控 10 次以上？

请注意，每当我引入数据时，我都会这样拆分新行：

charges = []
charges_info = driver.find_elements_by_class_name("clear-cell-border")
for p in range(len(charges_info)):
    charges.append(charges_info[p].text.strip().split('\n'))

谢谢！

Answer 1

由于它是列表中的列表，您需要将其解压两次

非list-comprehension

charges_info = [['MURDER   /   Disposed   /   ,000.00', 'MANUF., POSS OF OTHER SUB. IN SCH. I,IIIII OR PWID - 1ST OFFENSE   /   Disposed   /   ,000.00']]

new_list = []
for charges in charges_info:
    for i in charges:
        new_list.append(i)

for item in new_list:
    print(item)

>>> MURDER   /   Disposed   /   ,000.00
>>> MANUF., POSS OF OTHER SUB. IN SCH. I,IIIII OR PWID - 1ST OFFENSE   /   Disposed   /   ,000.00

List-comprehension

charges_info = [['MURDER   /   Disposed   /   ,000.00', 'MANUF., POSS OF OTHER SUB. IN SCH. I,IIIII OR PWID - 1ST OFFENSE   /   Disposed   /   ,000.00']]

new_list = [i for charges in charges_info for i in charges]

for item in new_list:
    print(item)

>>> MURDER   /   Disposed   /   ,000.00
>>> MANUF., POSS OF OTHER SUB. IN SCH. I,IIIII OR PWID - 1ST OFFENSE   /   Disposed   /   ,000.00

将列表拆分为子列表：

Splitting a list into sublist:

python

selenium

list-comprehension

django-models

one-to-many