deserialize method
Implementation
@override
void deserialize(Map<String, dynamic>? json) {
if (json == null) {
throw ApiException(
400, 'Failed to deserialize ClassificationResult data model.');
}
if (json.containsKey('ClassName')) {
className = json['ClassName'] as String;
} else {
className = null;
}
if (json.containsKey('ClassProbability')) {
classProbability = json['ClassProbability'] as double;
} else {
classProbability = null;
}
}