Python class - 存储和引用实例

Question

我正在尝试创建一个 class，其中包含：名称空间、文件路径和选择集数据（列表）数据。

我有一个 UI 可以让我为输入的每个字符添加一个新的 'record'。

到目前为止我得到了这个：

mainlist = []

class chRecord:
    def __init__(self, namespace, filepath, selSets =[]):
        self.namespace = namespace
        self.filepath = filepath
        self.selSets = selSets


aName = "John:"
aAge = "C:/temp/"
aSelSets = "Animation:stuff_Junk 0-10, 
animation:bob_jeff 10-30"

characterRecord = chRecord(aName,aAge,aSelSets)

mainlist.append(characterRecord)

aName = "John2:"
aAge = "C:/temp/"
aSelSets = "Animation:stuff_Junk2 0-10, 
animation:bob_jeff2 10-30"

characterRecord = chRecord(aName,aAge,aSelSets)

mainlist.append(characterRecord)

我的问题是如何搜索 mainList 以找到我要查找的记录。 IE。 'John'然后找到命名空间、文件路径和选择集数据？

抱歉，如果我的术语在某些方面有误！

干杯。

Answer 1

您需要创建一个函数，将记录列表作为参数并按名称进行搜索。然后它将 return 匹配的记录，您将能够从中获得附加信息。

示例：

mainlist = []


def find_record_by_name(records, name):
    for record in records:

        if record.namespace == name:
            return record


class chRecord:
    def __init__(self, namespace, filepath, selSets=[]):
        self.namespace = namespace
        self.filepath = filepath
        self.selSets = selSets


aName = "John:"
aAge = "C:/temp/"
aSelSets = "Animation:stuff_Junk 0-10, animation:bob_jeff 10-30"

characterRecord = chRecord(aName, aAge, aSelSets)

mainlist.append(characterRecord)

aName = "John2:"
aAge = "C:/temp/"
aSelSets = "Animation:stuff_Junk2 0-10,animation:bob_jeff2 10-30"

characterRecord = chRecord(aName, aAge, aSelSets)

mainlist.append(characterRecord)

selected_record = find_record_by_name(mainlist, "John:")

if selected_record:
    print(selected_record.namespace)
    print(selected_record.filepath)
    print(selected_record.selSets)

Answer 2

浏览列表并匹配 namespace 属性。

for record in mainlist:
    if record.namespace == 'John':
        # the record is found, now you can access the attributes with `record.filepath`
        print(record)

Answer 3

虽然它通常用作成熟的网站框架，但 Django model 与您在此处尝试执行的有关过滤的操作非常接近：

from django.db import models


class CHRecord(models.Model):
    namespace = models.CharField(max_length=32)
    filepath = models.FileField()
    # Not sure what this represents in your question, but if you use
    # PostgreSQL as your backend, might want an Array
    sel_sets = models.CharField(max_length=512)

# Saves a row to a database (local SQLite by default)
CHRecord.objects.create(namespace="John:", filepath="C:/temp/", sel_sets="xyz")

# Get matching rows translating to a SQL query
CHRecord.objects.filter(namespace__contains="John")

Answer 4

如果您需要强制使用唯一值，您可以使用字典。只需将键设置为唯一值或应该唯一的值。

这应该运行比循环更快，但如果键不是唯一的，则会覆盖数据。

mainlist = {}

class chRecord:
    def __init__(self, namespace, filepath, selSets =[]):
        self.namespace = namespace
        self.filepath = filepath
        self.selSets = selSets


aName = "John:"
aAge = "C:/temp/"
aSelSets = "Animation:stuff_Junk 0-10, animation:bob_jeff 10-30"

mainlist[aName] = chRecord(aName,aAge,aSelSets)
print(mainlist[aName].selSets)

aName = "John2:"
aAge = "C:/temp/"
aSelSets = "Animation:stuff_Junk2 0-10, animation:bob_jeff2 10-30"

mainlist[aName] = chRecord(aName,aAge,aSelSets)

print(mainlist.get('John2:').namespace)
print(mainlist.get('John2:').filepath)
print(mainlist.get('John2:').selSets)

Python class - 存储和引用实例

Python class - storing and referring back to the instances

python

class

instance