Question

我正在尝试创建一个包含以下内容的类：名称空间，文件路径和其中包含的selectionset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“记录”。

到目前为止，我已经知道了：

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)

我的问题是我该如何搜索mainList来查找我想要的记录。即'John'，然后找到名称空间，文件路径和selectionset数据？

很抱歉，如果我的术语在某些方面不正确！

干杯。

Answer 1

虽然它通常用作完善的网站框架，但Django model与您在此处要进行的过滤工作非常接近：

fn

Answer 2

您将需要创建一个函数，该函数将记录列表和名称搜索作为参数。然后它将返回匹配的记录，您将可以从中获取更多信息。

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 3

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 4

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

这应该比循环快，但是如果键不是唯一的，它将覆盖数据。

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类-存储并返回实例

4 个答案: