增加海洋标题的字体尺寸

发布于 2025-02-08 09:06:31 字数 1251 浏览 2 评论 0原文

我使用以下功能创建了一个散点图：

def plot_all_seaborn(x,y,types,x_axis, y_axis, title):
    tuples = only_floats(x,y,types)
    x,y,types_new = zip(*tuples)
    frame = pd.DataFrame({x_axis:x,y_axis:y,'Model':types_new})
    sns.relplot(data=frame,
                x=x_axis, 
                y=y_axis, 
                hue="Model", 
                style="Model",
                s=500
                ).set(title=title)

对于此图，我想增加表的字体大小。我尝试在set> set方法中添加font_size参数，但这不起作用。我该如何放大标题？

基于此链接（对海出块的字体大小的良好控制），我尝试过：

def plot_all_seaborn(x,y,types,x_axis, y_axis, title):
    tuples = only_floats(x,y,types)
    x,y,types_new = zip(*tuples)
    frame = pd.DataFrame({x_axis:x,y_axis:y,'Model':types_new})
    b = sns.relplot(data=frame,
            x=x_axis, 
            y=y_axis, 
            hue="Model", 
            style="Model",
            s=500
            )
    b.set_title("Title",fontsize=20)
    plt.show()

但是，这返回以下错误：

attributeError：'facetGrid'对象没有属性'set_title'

原文

I have created a scatterplot using the following function:

def plot_all_seaborn(x,y,types,x_axis, y_axis, title):
    tuples = only_floats(x,y,types)
    x,y,types_new = zip(*tuples)
    frame = pd.DataFrame({x_axis:x,y_axis:y,'Model':types_new})
    sns.relplot(data=frame,
                x=x_axis, 
                y=y_axis, 
                hue="Model", 
                style="Model",
                s=500
                ).set(title=title)

For this plot, I want to increase the font size of the table. I have tried adding a font_size parameter in the set method, but this does not work. How can I enlarge the title?

Based on this link (Fine control over the font size in Seaborn plots), I tried:

def plot_all_seaborn(x,y,types,x_axis, y_axis, title):
    tuples = only_floats(x,y,types)
    x,y,types_new = zip(*tuples)
    frame = pd.DataFrame({x_axis:x,y_axis:y,'Model':types_new})
    b = sns.relplot(data=frame,
            x=x_axis, 
            y=y_axis, 
            hue="Model", 
            style="Model",
            s=500
            )
    b.set_title("Title",fontsize=20)
    plt.show()

However, this returns the following error:

AttributeError: 'FacetGrid' object has no attribute 'set_title'

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沫离伤花 2025-02-15 09:06:31

这是可行的，但是您必须将其他参数作为一个指定传递。

def plot_all_seaborn(x,y,types,x_axis, y_axis, title):
    tuples = only_floats(x,y,types)
    x,y,types_new = zip(*tuples)
    frame = pd.DataFrame({x_axis:x,y_axis:y,'Model':types_new})
    b = sns.relplot(data=frame,x=x_axis, y=y_axis)

    b.set_title("Title", {"fontsize"=20})

    plt.show()

This is doable, but you have to pass in the additional parameters as a dict.

def plot_all_seaborn(x,y,types,x_axis, y_axis, title):
    tuples = only_floats(x,y,types)
    x,y,types_new = zip(*tuples)
    frame = pd.DataFrame({x_axis:x,y_axis:y,'Model':types_new})
    b = sns.relplot(data=frame,x=x_axis, y=y_axis)

    b.set_title("Title", {"fontsize"=20})

    plt.show()

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ぺ禁宫浮华殁 2025-02-15 09:06:31

我认为您可以这样取得成果；

def plot_seaborn(x,y,types,x_axis, y_axis, title):
    tuples = only_floats(x,y,types)
    x,y,types_new = zip(*tuples)
    frame = pd.DataFrame({x_axis:x,y_axis:y,'Model':types_new})
    sns.relplot(data=frame,
            x=x_axis, 
            y=y_axis, 
            hue="Model", 
            style="Model",
            s=500
            ).set_title(title, fontsize=23)

I think you can achieve your results like so;

def plot_seaborn(x,y,types,x_axis, y_axis, title):
    tuples = only_floats(x,y,types)
    x,y,types_new = zip(*tuples)
    frame = pd.DataFrame({x_axis:x,y_axis:y,'Model':types_new})
    sns.relplot(data=frame,
            x=x_axis, 
            y=y_axis, 
            hue="Model", 
            style="Model",
            s=500
            ).set_title(title, fontsize=23)

回复收藏 0 原文

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