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首页> 外文期刊>Medical Imaging, IEEE Transactions on >Massive-Training Artificial Neural Network Coupled With Laplacian-Eigenfunction-Based Dimensionality Reduction for Computer-Aided Detection of Polyps in CT Colonography
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Massive-Training Artificial Neural Network Coupled With Laplacian-Eigenfunction-Based Dimensionality Reduction for Computer-Aided Detection of Polyps in CT Colonography

机译:大规模训练人工神经网络结合基于Laplacian特征函数的降维技术在CT结肠造影术中对息肉进行计算机辅助检测

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摘要

A major challenge in the current computer-aided detection (CAD) of polyps in CT colonography (CTC) is to reduce the number of false-positive (FP) detections while maintaining a high sensitivity level. A pattern-recognition technique based on the use of an artificial neural network (ANN) as a filter, which is called a massive-training ANN (MTANN), has been developed recently for this purpose. The MTANN is trained with a massive number of subvolumes extracted from input volumes together with the teaching volumes containing the distribution for the “likelihood of being a polyp;” hence the term “massive training.” Because of the large number of subvolumes and the high dimensionality of voxels in each input subvolume, the training of an MTANN is time-consuming. In order to solve this time issue and make an MTANN work more efficiently, we propose here a dimension reduction method for an MTANN by using Laplacian eigenfunctions (LAPs), denoted as LAP-MTANN. Instead of input voxels, the LAP-MTANN uses the dependence structures of input voxels to compute the selected LAPs of the input voxels from each input subvolume and thus reduces the dimensions of the input vector to the MTANN. Our database consisted of 246 CTC datasets obtained from 123 patients, each of whom was scanned in both supine and prone positions. Seventeen patients had 29 polyps, 15 of which were 5–9 mm and 14 were 10–25 mm in size. We divided our database into a training set and a test set. The training set included 10 polyps in 10 patients and 20 negative patients. The test set had 93 patients including 19 polyps in seven patients and 86 negative patients. To investigate the basic properties of a LAP-MTANN, we trained the LAP-MTANN with actual polyps and a single source of FPs, which were rectal tubes. We applied the trained LAP-MTANN to simulated polyps and rectal tubes. The results showed that the performance of LAP-MTANNs with 20 LAPs was advantageous over th-n-nat of the original MTANN with 171 inputs. To test the feasibility of the LAP-MTANN, we compared the LAP-MTANN with the original MTANN in the distinction between actual polyps and various types of FPs. The original MTANN yielded a 95% (18/19) by-polyp sensitivity at an FP rate of 3.6 (338/93) per patient, whereas the LAP-MTANN achieved a comparable performance, i.e., an FP rate of 3.9 (367/93) per patient at the same sensitivity level. With the use of the dimension reduction architecture, the time required for training was reduced from 38 h to 4 h. The classification performance in terms of the area under the receiver-operating-characteristic curve of the LAP-MTANN (0.84) was slightly higher than that of the original MTANN (0.82) with no statistically significant difference (p-value$=0.48$).
机译:当前CT结肠造影(CTC)中息肉的计算机辅助检测(CAD)的主要挑战是在保持高灵敏度水平的同时减少假阳性(FP)检测的次数。为此目的,最近已经开发了一种基于人工神经网络(ANN)作为过滤器的模式识别技术,称为大规模训练ANN(MTANN)。 MTANN接受了从输入卷中提取的大量子卷以及包含“息肉可能性”分布的教学卷的训练。因此称为“大规模培训”。由于大量子体积和每个输入子体积中体素的高维度,MTANN的训练非常耗时。为了解决这个时间问题,并使MTANN更有效地工作,我们在此提出一种使用拉普拉斯特征函数(LAP)的MTANN降维方法,称为LAP-MTANN。 LAP-MTANN代替输入体素,使用输入体素的依存结构来计算每个输入子体积中输入体素的选定LAP,从而减小了MTANN的输入向量的维数。我们的数据库包含从123例患者中获得的246个CTC数据集,每个患者均在仰卧位和俯卧位进行了扫描。 17名患者有29例息肉,其中15例5-9毫米,14例10-25毫米。我们将数据库分为训练集和测试集。训练组包括10例息肉和20例阴性患者的10个息肉。该测试集有93例患者,其中7例患者中有19例息肉和86例阴性患者。为了研究LAP-MTANN的基本特性,我们用实际息肉和单一来源的FP(即直肠管)训练了LAP-MTANN。我们将训练有素的LAP-MTANN应用于模拟息肉和直肠管。结果表明,具有20个LAP的LAP-MTANN的性能优于具有171个输入的原始MTANN的th-n-nat。为了测试LAP-MTANN的可行性,我们将LAP-MTANN与原始MTANN进行了比较,以区分实际息肉和各种类型的FP。原始的MTANN以每位患者FP的3.6(338/93)FP率产生95%(18/19)的息肉敏感性,而LAP-MTANN表现可比,即FP率为3.9(367/93) 93)在相同的敏感度水平下每位患者。通过使用降维架构,培训所需的时间从38小时减少到4小时。就LAP-MTANN的接收器操作特性曲线下的面积(0.84)而言,分类性能略高于原始MTANN的分类性能(0.82),无统计学差异(p值$ = 0.48 $) 。

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