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如何在python中解决未来警告 - >%(min_groups,self.n_splits),警告)?

当我在我的程序中运行mean_acc()方法时,有%(min_groups,self.n_splits)),警告)错误...

def mean_acc():
    models = [
        RandomForestClassifier(n_estimators=200, max_depth=3, random_state=0),
        LinearSVC(),
        MultinomialNB(),
        LogisticRegression(random_state=0)]
    CV = 6
    cv_df = pd.DataFrame(index=range(CV * len(models)))
    entries = []
    for model in models:
        model_name = model.__class__.__name__
        accuracies = cross_val_score(model, features, labels, scoring='accuracy', cv=CV)
        for fold_idx, accuracy in enumerate(accuracies):
            entries.append((model_name, fold_idx, accuracy))
    cv_df = pd.DataFrame(entries, columns=['model_name', 'fold_idx', 'accuracy'])

    print(cv_df.groupby('model_name').accuracy.mean())
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这些是我使用mean_acc()方法运行程序时显示的错误.我可以知道如何在下面解决这些错误吗?请帮助我看看上面导致这些错误的代码,谢谢!

 % (min_groups, self.n_splits)), Warning)
C:\Users\L31307\PycharmProjects\FYP\venv\lib\site-packages\sklearn\model_selection\_split.py:626: Warning: The least populated class in y has only 1 members, which is too few. The minimum number of members in any …
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python scikit-learn linearmodels future-warning

9
推荐指数
1
解决办法
2万
查看次数

如何将文本文件作为字符串读取?

以下是我尝试在名为的方法中读取文本文件中的文本的代码 check_keyword()

def check_keyword():
    with open(unknown.txt, "r") as text_file:
        unknown = text_file.readlines()

    return unknown
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这就是我调用该方法的方式:

dataanalysis.category_analysis.check_keyword()
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文本文件中的文本:

Hello this is a new text file 
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上述方法没有输出:((

python text-files

6
推荐指数
2
解决办法
2万
查看次数

如何解决python中属性错误“ float”对象没有属性“ split”?

当我运行下面的这些代码时,它给我错误,说有属性错误“ float”对象在python中没有属性“ split”。

我想知道为什么会出现此错误,请帮助我查看下面的代码,谢谢:((

pd.options.display.max_colwidth = 10000
df = pd.read_csv(output, sep='|')


def text_processing(df):
    """""=== Lower case ==="""
    '''First step is to transform comments into lower case'''
    df['content'] = df['content'].apply(lambda x: " ".join(x.lower() for x in x.split() if x not in stop_words))

    '''=== Removal of stop words ==='''
    df['content'] = df['content'].apply(lambda x: " ".join(x for x in x.split() if x not in stop_words))

    '''=== Removal of Punctuation ==='''
    df['content'] = df['content'].str.replace('[^\w\s]', '')

    '''=== Removal of Numeric ==='''
    df['content'] = df['content'].str.replace('[0-9]', '') …
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python string series pandas

2
推荐指数
2
解决办法
9234
查看次数