Question

我试图使用搁置来保存/加载文件中的某些数据。我有一个词典列表：

inv = [slot0, slot1, slot2, slot3, slot4, slot5, slot6]

7个词典中的每一个都是这样的，但值略有不同：

slot0 = {"item_pos": [hud_x + 592, hud_y + 4], "text_pos": [hud_x + 612, hud_y + 25], "item": None, "amount": 0}

为了保存/加载字典，我使用＆＃34;搁置＆＃34;模块，这是保存的代码：

with shelve.open((os.path.join(saves_path, "inventory", "inventory")), "c") as f:

        f["slot0_item"] = slot0["item"]
        f["slot0_amount"] = slot0["amount"]
        f["slot1_item"] = slot1["item"]
        f["slot1_amount"] = slot1["amount"]
        f["slot2_item"] = slot2["item"]
        f["slot2_amount"] = slot2["amount"]
        f["slot3_item"] = slot3["item"]
        f["slot3_amount"] = slot3["amount"]
        f["slot4_item"] = slot4["item"]
        f["slot4_amount"] = slot4["amount"]
        f["slot5_item"] = slot5["item"]
        f["slot5_amount"] = slot5["amount"]
        f["slot6_item"] = slot6["item"]
        f["slot6_amount"] = slot6["amount"]

以下是加载代码：

with shelve.open((os.path.join(saves_path, "inventory", "inventory")), "c") as f:

        slot0["item"] = f["slot0_item"]
        slot0["amount"] = f["slot0_amount"]
        slot1["item"] = f["slot1_item"]
        slot1["amount"] = f["slot1_amount"]
        slot2["item"] = f["slot2_item"]
        slot2["amount"] = f["slot2_amount"]
        slot3["item"] = f["slot3_item"]
        slot3["amount"] = f["slot3_amount"]
        slot4["item"] = f["slot4_item"]
        slot4["amount"] = f["slot4_amount"]
        slot5["item"] = f["slot5_item"]
        slot5["amount"] = f["slot5_amount"]
        slot6["item"] = f["slot6_item"]
        slot6["amount"] = f["slot6_amount"]

虽然这段代码运行良好，但它很长且效率很低。我尝试使用＆＃34; for＆＃34;循环以保存这样的数据：

for slot in inv:
            f["slot_item"] = slot["item"]
            f["slot_amount"] = slot["amount"]

...并加载如下数据：

for slot in inv:
            slot["item"] = f["slot_item"]
            slot["amount"] = f["slot_amount"]

然而，当我使用这种方法时，字典不会在重新启动程序时保存/加载他们的更改，就像他们应该的那样。我怎样才能（如果可能的话）正确使用＆＃34; for＆＃34;循环以有效地保存/加载搁置的数据？

编辑：hud_x和hud_y都等于20

Answer 1

如果您可以设法使用dictinary或列表而不是slot1，slot2等代码可以更容易。

import json
from pathlib import Path
import random

# repliacting some data 
def generate_dict():
   return {"item_pos": [rnd() + 592, rnd() + 4], 
           "text_pos": [rnd() + 612, rnd() + 25], 
           "item": None, 
           "amount": 0}

def rnd():
   return random.randint(0, 1000)    

# change of data structure
dict_list = [generate_dict() for id in range(7)]

# easily seriliasable into json
json_str = json.dumps(dict_list, sort_keys=True, indent=4)

# you can save json as file 
Path('store.json').write_text(json_str)

# read back
assert json.loads(Path('store.json').read_text()) == dict_list

Answer 2

我决定使用泡菜，因为它更容易使用＆＃34;对于＆＃34;循环。此外，搁置使用类似字典的语法（pickle不会），因此这可能是问题的一部分。

保存：

with open((os.path.join(saves_path, "inventory", "inventory.txt")), "wb") as f:
        for slot in inv:
            pickle.dump(slot["item"], f)
            pickle.dump(slot["amount"], f)

装载：

with open((os.path.join(saves_path, "inventory", "inventory.txt")), "rb") as f:
        for slot in self.inventory.inv:
            slot["item"] = pickle.load(f)
            slot["amount"] = pickle.load(f)

如何使用＆＃34; for＆＃34;来保存/加载数据。环？

2 个答案: