在python中解包大列表答案

【问题标题】：Unpacking big list in python在python中解包大列表
【发布时间】：2021-05-29 02:30:02
【问题描述】：

我必须拆开这个大清单并将每一项都放入数据库中。除了这些数据之外，一切都运行良好。

我试过了：

([styles], [seasons], [day], [id]) = search_

但我收到这样的错误：

exception=ValueError('too many values to unpack (expected 4)')>

我使用这样的查询：

insert= "INSERT INTO cantina (styles,seasons,day,id) VALUES (?,?,?,?)"
db.execute(insert,(styles,seasons,day,id))

我知道我犯了一个错误，但是这里每行有 4 个字段，似乎是正确的

search_ = [[["Redwine"], ['04'], ['01'], ['81535']],
[["Redwine"], ['04'], ['02'], ['81536']],
[["Redwine"], ['04'], ['03'], ['81537']],
[["Redwine"], ['04'], ['04'], ['81538']],
[["Redwine"], ['04'], ['05'], ['81539']],
[["Redwine"], ['04'], ['06'], ['81540']],
[["Redwine"], ['04'], ['07'], ['81541']],
[["Redwine"], ['04'], ['08'], ['81542']],
[["Redwine"], ['04'], ['09'], ['81543']],
[["Redwine"], ['04'], ['10'], ['81544']],
[["Redwine"], ['04'], ['11'], ['81545']],
[["Redwine"], ['04'], ['12'], ['81546']],
[["Redwine"], ['02'], ['01'], ['97179']],
[["Redwine"], ['02'], ['02'], ['97180']],
[["Redwine"], ['02'], ['03'], ['97181']],
[["Redwine"], ['02'], ['04'], ['97182']],
[["Redwine"], ['02'], ['05'], ['97183']],
[["Redwine"], ['02'], ['06'], ['97184']],
[["Redwine"], ['02'], ['07'], ['97185']],
[["Redwine"], ['02'], ['08'], ['97186']],
[["Redwine"], ['02'], ['09'], ['97187']],
[["Redwine"], ['02'], ['10'], ['97188']],
[["Redwine"], ['02'], ['11'], ['97189']],
[["Redwine"], ['02'], ['12'], ['97190']]]

([styles], [seasons], [day], [id]) = search_

insert= "INSERT INTO cantina (styles,seasons,day,id) VALUES (?,?,?,?)"
db.execute(insert,(styles,seasons,day,id))

编辑：

嗨，Barmar，因为该列表来自我加入的另一个函数四个列表：

def cantina ( cantina_id)
  for ... in ...:
     styles = 
     season  = 
     day = 
     id = 
     result.append([[styles], [season], [day], [id]]])
  return result

而'append'只接受一个参数。

我必须根据“cantina_id”处理一些数据；将其添加到四个列表并保存到 db

【问题讨论】：

为什么每个值都在另一个列表中，即为什么是["Redwine"] 而不是"Redwine"？如果你不这样做，你可以使用executemany()。
嗨，Barmar，谢谢，请参阅编辑
没有理由像在append 调用中那样将每个值放在自己的列表中。只是不要这样做并使用executemany。
你应该使用result.append([styles, season, day, id])

标签： python list sqlite unpack

【解决方案1】：

您需要遍历search_ 并分别处理每一行。

insert= "INSERT INTO cantina (styles,seasons,day,id) VALUES (?,?,?,?)"
for row in search_:
    ([styles], [seasons], [day], [id]) = row
    cur.execute(insert,(styles,seasons,day,id))
db.commit()

如果您更改 cantina 函数使其不会创建所有这些单元素列表，您可以使用 executemany()

def cantina ( cantina_id)
  result = []
  for ... in ...:
     styles = 
     season  = 
     day = 
     id = 
     result.append([styles, season, day, id])
  return result

search_ = cantina(some_id)
insert= "INSERT INTO cantina (styles,seasons,day,id) VALUES (?,?,?,?)"
cur.executemany(insert, search_)
db.commit()

【讨论】：

谢谢巴马尔，，，