Question

我有一个三列的熊猫数据框：'Ethnicity', 'Locus', 'Count'。

当我在jupyter笔记本中渲染图时，该图被合并。我想显示四个不同的情节。

也就是说，我想绘制每个“种族”的轨迹（x轴）与计数（y轴）。我想将其显示为构面，因此在同一视觉中分别显示为四个图。

示例代码

plt.bar(appended_data[appended_data['Ethnicity']==1]['Locus'].values, 
            appended_data[appended_data['Ethnicity']==1]['Count'].values)

而不是上面的代码中的“ 1”，我希望它是一个像“ i”之类的动态变量。

目前，通过粗略解决方案，是将上述代码行放入不同的“ jupyter”单元中，并手动更改[appended_data ['Ethnicity'] == 1]值。

有没有一种方法可以渲染四个图（最好在2 * 2网格中）？

也就是说，

plot1 for Ethnicity = 0, plot2 for Ethnicity = 1
plot1 for Ethnicity = 2, plot2 for Ethnicity = 3

来自以下示例数据：

appended_data= pd.DataFrame({'Ethnicity':[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1,
           1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2,
           2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3,
           3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3],
    'Count': [1, 5, 8, 5, 4, 9, 4, 3, 2, 1, 1, 2, 1, 1, 1, 1, 1, 5, 6, 9, 2, 5,
           2, 1, 1, 3, 2, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 3, 2, 8, 1, 1, 2,
           6, 1, 2, 7, 1, 3, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 8, 2, 8, 4, 5, 1,
           3, 2, 3, 2, 1, 1, 2, 3, 1, 1, 2, 1], 
    'Locus':['13,12', '11,12', '10,10', '12,11', '12,12', '10,12', '10,11',
               '12,10', '11,11', '11,7', '11,13', '11,10', '12,7', '12,13',
               '13,10', '10,8', '9,10', '11,11', '12,11', '10,12', '13,13',
               '12,12', '11,13', '10,13', '12,13', '11,10', '11,12', '12,10',
               '9,12', '10,10', '9,11', '13,10', '7,12', '7,10', '9,10', '10,11',
               '13,8', '11,7', '10,11', '7,12', '10,10', '13,10', '12,7', '11,12',
               '11,11', '11,7', '11,10', '10,12', '9,12', '12,11', '12,8', '8,10',
               '12,12', '12,9', '13,11', '14,10', '7,11', '8,12', '10,7', '10,8',
               '12,12', '13,11', '10,11', '11,10', '11,12', '8,11', '12,10',
               '13,10', '13,12', '12,11', '11,11', '12,7', '10,10', '10,13',
               '11,14', '11,7', '10,12', '12,9']})

Answer 1

import seaborn as sns

appended_data= pd.DataFrame({'Ethnicity':[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1,
           1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2,
           2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3,
           3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3],
    'Count': [1, 5, 8, 5, 4, 9, 4, 3, 2, 1, 1, 2, 1, 1, 1, 1, 1, 5, 6, 9, 2, 5,
           2, 1, 1, 3, 2, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 3, 2, 8, 1, 1, 2,
           6, 1, 2, 7, 1, 3, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 8, 2, 8, 4, 5, 1,
           3, 2, 3, 2, 1, 1, 2, 3, 1, 1, 2, 1], 
    'Locus':['13,12', '11,12', '10,10', '12,11', '12,12', '10,12', '10,11',
               '12,10', '11,11', '11,7', '11,13', '11,10', '12,7', '12,13',
               '13,10', '10,8', '9,10', '11,11', '12,11', '10,12', '13,13',
               '12,12', '11,13', '10,13', '12,13', '11,10', '11,12', '12,10',
               '9,12', '10,10', '9,11', '13,10', '7,12', '7,10', '9,10', '10,11',
               '13,8', '11,7', '10,11', '7,12', '10,10', '13,10', '12,7', '11,12',
               '11,11', '11,7', '11,10', '10,12', '9,12', '12,11', '12,8', '8,10',
               '12,12', '12,9', '13,11', '14,10', '7,11', '8,12', '10,7', '10,8',
               '12,12', '13,11', '10,11', '11,10', '11,12', '8,11', '12,10',
               '13,10', '13,12', '12,11', '11,11', '12,7', '10,10', '10,13',
               '11,14', '11,7', '10,12', '12,9']})

g = sns.FacetGrid(appended_data, col="Ethnicity", col_wrap=2)
g.map(sns.scatterplot, "Locus", "Count")

绘制n个网格状结构

1 个答案: