Python中数据pickle的错误 [duplicate]

我需要将我的训练数据集保存到数据pickle中。我使用了以下代码。

self.featureCounts = collections.defaultdict(lambda :0)self.featureVectors = []self.labelCounts = collections.defaultdict(lambda :0) def SaveOnPickle(self):        f = open('dict.pickle', 'wb')        pickle.dump(self.featureCounts, f)        f.close()

运行时出现了以下错误

Traceback (most recent call last):  File "C:/wamp64/www/M360/TrainClassifier.py", line 76, in <module>    Predic.SaveOnPickle()  File "C:/wamp64/www/M360/TrainClassifier.py", line 50, in SaveOnPickle    pickle.dump(self.featureCounts, f)  File "C:\Python27\lib\pickle.py", line 1376, in dump    Pickler(file, protocol).dump(obj)  File "C:\Python27\lib\pickle.py", line 224, in dump    self.save(obj)  File "C:\Python27\lib\pickle.py", line 331, in save    self.save_reduce(obj=obj, *rv)  File "C:\Python27\lib\pickle.py", line 401, in save_reduce    save(args)  File "C:\Python27\lib\pickle.py", line 286, in save    f(self, obj) # Call unbound method with explicit self  File "C:\Python27\lib\pickle.py", line 568, in save_tuple    save(element)  File "C:\Python27\lib\pickle.py", line 286, in save    f(self, obj) # Call unbound method with explicit self  File "C:\Python27\lib\pickle.py", line 754, in save_global    (obj, module, name))pickle.PicklingError: Can't pickle <function <lambda> at 0x021391B0>: it's not found as __main__.<lambda>

如何解决这个问题?


回答:

您可以使用Pickle

您可以这样保存模型:

import picklef = open('my_classifier.pickle', 'wb')pickle.dump(classifier, f)f.close()

稍后加载时:

import picklef = open('my_classifier.pickle', 'rb')classifier = pickle.load(f)f.close()

Related Posts

L1-L2正则化的不同系数

我想对网络的权重同时应用L1和L2正则化。然而,我找不…

使用scikit-learn的无监督方法将列表分类成不同组别,有没有办法?

我有一系列实例,每个实例都有一份列表,代表它所遵循的不…

f1_score metric in lightgbm

我想使用自定义指标f1_score来训练一个lgb模型…

通过相关系数矩阵进行特征选择

我在测试不同的算法时,如逻辑回归、高斯朴素贝叶斯、随机…

可以将机器学习库用于流式输入和输出吗?

已关闭。此问题需要更加聚焦。目前不接受回答。 想要改进…

在TensorFlow中,queue.dequeue_up_to()方法的用途是什么?

我对这个方法感到非常困惑,特别是当我发现这个令人费解的…

发表回复

您的邮箱地址不会被公开。 必填项已用 * 标注