我要求用户将一个项目添加到列表中,然后保存修改后的列表。但是当我再次运行该程序时,之前添加的元素就消失了。我不明白为什么新元素只是暂时保存,而我的列表会自行重置。有人可以解释并建议我如何保存新项目吗?
import pickle
my_list = ["a", "b", "c", "d", "e"]
def add(item):
my_list.append(item)
with open("my_list.pickle", 'wb') as file:
pickle.dump(my_list, file)
return my_list
while True:
item = input("Add to the list: \n").upper()
if item == "Q":
break
else:
item = add(item)
with open("my_list.pickle", "rb") as file1:
my_items = pickle.load(file1)
print(my_items)
最佳答案
您在填充数据后从文件中读取列表。因此,如果我们分析 main(我删除了 add
函数,并注释了程序),我们得到:
# the standard value of the list
my_list = ["a", "b", "c", "d", "e"]
# adding data to the list
while True:
# we write the new list to the file
item = input("Add to the list: \n").upper()
if item == "Q":
break
else:
item = add(item)
# loading the list we overrided in this program session
with open("my_list.pickle", "rb") as file1:
my_items = pickle.load(file1)
# print the loaded list
print(my_items)
因此,由于您从默认列表开始,并且每次向文件添加元素时都会重写文件,如果您在程序末尾加载列表,您猜怎么着?您将获得刚刚保存的列表。
解决方案是将加载移至程序顶部:
import pickle
<b>import os.path</b>
# the standard value of the list
my_list = ["a", "b", "c", "d", "e"]
# in case the file already exists, we use that list
<b>if os.path.isfile("my_list.pickle"):
with open("my_list.pickle", "rb") as file1:
my_items = pickle.load(file1)</b>
# adding data to the list
while True:
# we write the new list to the file
item = input("Add to the list: \n").upper()
if item == "Q":
break
else:
item = add(item)
# print the final list
print(my_items)
请注意,每次存储新列表的效率相当低。您最好让用户有机会更改列表,并将其存储在程序末尾。
关于python - 使用 pickle 保存修改后的列表时出现问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47500793/