【发布时间】:2017-06-29 11:43:51
【问题描述】:
我正在处理一个小的示例 python 文件。我有一个需要转换为泡菜的 csv 文件。这是我到目前为止的代码。
import csv
import pickle
class primaryDetails:
def __init__(self, name, age, gender, contactDetails):
self.name = name
self.age = age
self.gender = gender
self.contactDetails = contactDetails
def __str__(self):
return "{} {} {} {}".format(self.name, self.age, self.gender, self.contactDetails)
def __iter__(self):
return iter([self.name, self.age, self.gender, self.contactDetails])
class contactDetails:
def __init__(self, cellNum, phNum, Location):
self.cellNum = cellNum
self.phNum = phNum
self.Location = Location
def __str__(self):
return "{} {} {}".format(self.cellNum, self.phNum, self.Location)
def __iter__(self):
return iter([self.cellNum, self.phNum, self.Location])
a_list = []
with open("t_file.csv", "r") as f:
reader = csv.reader(f)
for row in reader:
a = contactDetails(row[3], row[4], row[5])
a_list.append(primaryDetails(row[0], row[1], row[2] , a))
file = open('writepkl.pkl', 'wb')
# pickle.dump(a_list[0], primaryDetails)
pickle.dump(primaryDetails, a_list[0])
file.close()
csv 文件
Bat,45,M,123456789,98764,Gotham
Sup,46,M,290345720,098484,Krypton
Wwomen,30,F,758478574,029383,Themyscira
Flash,27,M,3646348348,839484,Central City
Hulk,50,M,52903852398,298392,Ohio
当我读取文件并将其放入列表时,我无法挑选列表。我还尝试使用a_list[0] 而不是列表来腌制它,它给了我错误 pickle.dump(primaryDetails, a_list[0])
类型错误:文件必须具有“写入”属性。我需要将数据放在一个列表中并腌制,以便我可以将其保存到 db as mentioned here。有人可以帮我弄清楚我做错了什么。
【问题讨论】: