我正在尝试使用shelve将一些数据保存/加载到文件中。我有一份字典清单:
inv = [slot0, slot1, slot2, slot3, slot4, slot5, slot6]
7个字典的每一个看起来都是这样的,但数值略有不同:
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/加载词典,我使用了“shelve”模块,下面是保存的代码:
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"]
虽然这段代码工作得很好,但它很长而且效率很低。我尝试使用“for”循环保存数据,如下所示:
for slot in inv:
f["slot_item"] = slot["item"]
f["slot_amount"] = slot["amount"]
…然后像这样加载数据:
for slot in inv:
slot["item"] = f["slot_item"]
slot["amount"] = f["slot_amount"]
但是,当我使用这个方法时,字典在重新启动程序时不会像预期的那样保存/加载更改。如何(如果可能的话)正确地使用“for”循环来有效地保存/加载shelve中的数据
编辑:hud_x
和hud_y
都等于20
我决定改用pickle,因为它更容易处理“for”循环。另外,shelve使用类似字典的语法(pickle没有),所以这可能是问题的一部分
正在保存:
加载:
如果你能设法用一个字典或一个列表来代替
slot1
,slot2
,etc代码就容易多了相关问题 更多 >
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