在Python中使用Seaborn为bigrams和trigrams创建副本

Question

我使用Spyder作为Anaconda的一部分，并尝试按事件类型对tweet（文本）进行分类。为此，我使用了包交叉值得分，已经使用TfidVectorizer对我的tweet进行了向量化，然后使用fit\u transform对unigrams、bigrams和trigrams进行了转换，如下所示：

# TF-IDF on unigrams, bigrams and trigrams
tfidf_words = TfidfVectorizer(sublinear_tf=True, min_df=0, norm='l2', encoding='latin-1',
                              ngram_range=(1,1), stop_words='english')

# vectorize for bigrams
tfidf_bigrams = TfidfVectorizer(sublinear_tf=True, min_df=0, norm='l2', encoding='latin-1',
                              ngram_range=(2,2), stop_words='english')

# vecorize for trigrams
tfidf_trigrams = TfidfVectorizer(sublinear_tf=True, min_df=0, norm='l2', encoding='latin-1',
                              ngram_range=(3,3), stop_words='english')

# Transform and fit each of the outputs from TF-IDF (unigrams, bigrams and trigrams)
x_train_words = tfidf_words.fit_transform(x_train_sm.preprocessed).toarray()

# bigrams
x_train_bigrams = tfidf_bigrams.fit_transform(x_train_sm.preprocessed).toarray()

#trigrams
x_train_trigrams = tfidf_trigrams.fit_transform(x_train_sm.preprocessed).toarray()

现在，我使用包cross_val_score执行交叉验证，以计算unigram、bigrams和trigrams的平均精度。一旦完成，我将尝试为达到的精度生成并保存一个箱线图。对于4种不同的型号，此功能已完成：

^{pr2}$

unigrams的输出正是我想要的：

现在，当我运行bigrams和trigrams的代码时（突出显示所有代码并单击“play”），我得到以下结果：

大图：

[

八卦图：

每一个的代码都是相同的，只是它们使用“cv_bigrams”和“cv_trigrams”作为框式图的数据输入。每个代码如下。在

二元代码：

^{3}$

三元代码：

# create blank dataframe with an index equal to number of CV folds * number of models tested
cv_trigrams = pd.DataFrame(index=range(CV * len(models)))

# clear the previous list called 'entries' that was populated with values
entries = []

# calculate the accuracy at each fold and populate the results in the 'entries' list
# populate the dataframe 'cv_trigrams' with the fold and accuracy score at each fold
i = 0
for model in models:
    #model_name = #model.__class__.__name__
    model_name = names[i]
    # model => the model that will be used to fit the data
    # x_train_trigrams => data that is to be fitted by the selected model (trigrams)
    # y_train_sm => y training data after oversampling (event_id)
    # scoring => the type of score you want the function 'cross_val_score' to return
    # cv = number of folds you want to performed with cross-validation
    accuracies = cross_val_score(model, x_train_trigrams, y_train_sm, scoring ='accuracy', cv=CV)
    for fold_idx, accuracy in enumerate(accuracies):
        entries.append((model_name, fold_idx, accuracy))
        cv_trigrams = pd.DataFrame(entries, columns=['model_name_trigrams', 'fold_idx', 'accuracy'])
    i = i + 1

以下是如果我只选择以下代码并运行：

# plot the results of each model as a box plot    
box_bigrams = sns.boxplot(x='model_name_bigrams', y='accuracy', data=cv_bigrams)
box_bigrams = sns.boxplot(x='model_name_bigrams', y='accuracy', data=cv_bigrams)
fig_bigrams = box_bigrams.get_figure()
fig_bigrams.savefig('boxplot_bigrams.png')

同样适用于三角图：

# plot the results of each model as a box plot    
box_trigrams = sns.boxplot(x='model_name_trigrams', y='accuracy', data=cv_trigrams)
box_trigrams = sns.boxplot(x='model_name_trigrams', y='accuracy', data=cv_trigrams)
fig_trigrams = box_trigrams.get_figure()
fig_trigrams.savefig('boxplot_trigrams.png')

输出：

你知道为什么当我一次运行所有代码时（当我把这段代码投入生产时，我需要这样做），而不是突出显示代码段并单独运行时，为什么我会得到相互重叠的重复boxplots？在

Answer 1

回应@ImportanceOfBeingErnest的评论，你的代码太复杂了，你的问题也不够清楚。你想创建3个不同的图形，每种情况一个（单数、二元和三元）？您是否尝试使用一个具有3个轴的图形（matplotlib中称为子图）？你想把三个箱子并排放在一张图上吗？在

对我来说，最简单的方法是创建一个包含3个子图的图形，如下所示：

fig, (ax1, ax2, ax3) = plt.subplots(3,1, figsize=(xx,yy))  # choose appropriate size to fit your needs
sns.boxplot(x='model_name_unigrams', y='accuracy', data=cv_unigrams, ax=ax1)
sns.boxplot(x='model_name_bigrams', y='accuracy', data=cv_bigrams, ax=ax2)
sns.boxplot(x='model_name_trigrams', y='accuracy', data=cv_trigrams, ax=ax3)
fig.savefig('your_figure_name_here.png')

请参阅subplots demo here和有关^{}或^{}的文档。在the documentation for ^{}中，您将看到它是一个“轴级别”函数，这意味着您可以要求它在您选择的任何轴对象上绘图