如何以正确的顺序绘制分组条形图

Posted 2023-03-11

【中文标题】如何以正确的顺序绘制分组条形图

【英文标题】：How to plot grouped bars in the correct order

【发布时间】：2021-10-13 12:30:53

【问题描述】：

我正在制作标准化考试熟练程度的分组条形图。这是我的代码：

bush_prof_boy = bush.groupby(['BOY Prof'])['BOY Prof'].count()
bush_prof_pct_boy = bush_prof_boy/bush['BOY Prof'].count() * 100
bush_prof_eoy = bush.groupby(['EOY Prof'])['EOY Prof'].count()
bush_prof_pct_eoy = bush_prof_eoy/bush['EOY Prof'].count() * 100

labels = ['Remedial', 'Below Proficient', 'Proficient', 'Advanced']

fig, ax = plt.subplots()
rects1 = ax.bar(x - width/2, bush_prof_pct_boy, width, label='BOY', 
color='mediumorchid')
rects2 = ax.bar(x + width/2, bush_prof_pct_eoy, width, label='EOY', color='teal')

ax.set_ylabel('% of Students at Proficiency Level', fontsize=18)
ax.set_title('Bushwick Middle Change in Proficiency Levels', fontsize=25)
ax.set_xticks(x)
ax.set_xticklabels(labels, fontsize=25)
ax.legend(fontsize=25)
plt.yticks(fontsize=15)


plt.figure(figsize=(5,15))

plt.show()

“BOY”代表“Beginning of Year”和“EOY”“End of Year”，因此条形图旨在显示在年初和年末进入每个熟练程度的学生的百分比。该图看起来不错，但是当我深入研究数字时，我可以看到 EOY 的标签不正确。这是我的图表：

BOY 的百分比是正确绘制的，但 EOY 的百分比是错误的标签。以下是实际百分比，我确信这是正确的：

BOY %
Advanced            14.0
Below Proficient    38.0
Proficient          34.0
Remedial            14.0

EOY %
Advanced            39.0
Below Proficient    18.0
Proficient          32.0
Remedial            11.0

【参考方案1】：

使用来自Kaggle: ***lyn NY Schools 的数据 单独计算条形组可能会出现问题。

最好在一个数据框中进行计算

，对数据框进行整形，然后进行绘图，因为这样可以确保将条形图绘制在正确的组中。 由于未提供数据，因此从宽格式数字数据开始，然后对数据框进行清理和整形。 使用.cut 将数值转换为分类值 Dataframe用.melt转长格式，然后用.groupby计算'x of Year'内的百分比 用.pivot 重塑，用pandas.DataFrame.plot 绘图

在python 3.8、pandas 1.3.1 和matplotlib 3.4.2 中测试

导入、加载和清理 DataFrame

import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.ticker as mtick
import numpy as np

# data 
data = 'BOY': [11.0, 11.0, 11.0, 11.0, 11.0, 8.0, 11.0, 14.0, 12.0, 13.0, 11.0, 14.0, 10.0, 9.0, 10.0, 10.0, 10.0, 12.0, 12.0, 13.0, 12.0, 11.0, 9.0, 12.0, 16.0, 12.0, 12.0, 12.0, 15.0, 10.0, 10.0, 10.0, 8.0, 11.0, 12.0, 14.0, 10.0, 8.0, 11.0, 12.0, 14.0, 12.0, 13.0, 15.0, 13.0, 8.0, 8.0, 11.0, 10.0, 11.0, 13.0, 11.0, 13.0, 15.0, 10.0, 8.0, 10.0, 9.0, 8.0, 11.0, 13.0, 11.0, 8.0, 11.0, 15.0, 11.0, 12.0, 17.0, 12.0, 11.0, 18.0, 14.0, 15.0, 16.0, 7.0, 11.0, 15.0, 16.0, 13.0, 13.0, 13.0, 0.0, 11.0, 15.0, 14.0, 11.0, 13.0, 16.0, 14.0, 12.0, 8.0, 13.0, 13.0, 14.0, 7.0, 10.0, 16.0, 10.0, 13.0, 10.0, 14.0, 8.0, 16.0, 13.0, 12.0, 14.0, 12.0, 14.0, 16.0, 15.0, 13.0, 13.0, 10.0, 14.0, 8.0, 10.0, 10.0, 11.0, 12.0, 10.0, 12.0, 14.0, 17.0, 13.0, 14.0, 16.0, 15.0, 13.0, 16.0, 9.0, 16.0, 15.0, 11.0, 11.0, 15.0, 14.0, 12.0, 15.0, 11.0, 16.0, 14.0, 14.0, 15.0, 14.0, 14.0, 14.0, 16.0, 15.0, 12.0, 12.0, 14.0, 15.0, 13.0, 14.0, 13.0, 17.0, 14.0, 13.0, 14.0, 13.0, 13.0, 12.0, 10.0, 15.0, 14.0, 12.0, 12.0, 14.0, 12.0, 14.0, 13.0, 15.0, 13.0, 14.0, 14.0, 12.0, 11.0, 15.0, 14.0, 14.0, 10.0], 'EOY': [16.0, 16.0, 16.0, 14.0, 10.0, 14.0, 16.0, 14.0, 15.0, 15.0, 15.0, 11.0, 11.0, 15.0, 10.0, 14.0, 17.0, 14.0, 9.0, 15.0, 14.0, 16.0, 14.0, 13.0, 11.0, 13.0, 12.0, 14.0, 15.0, 13.0, 14.0, 15.0, 12.0, 19.0, 9.0, 13.0, 11.0, 14.0, 17.0, 17.0, 14.0, 13.0, 14.0, 10.0, 16.0, 15.0, 12.0, 11.0, 12.0, 14.0, 15.0, 10.0, 15.0, 14.0, 14.0, 15.0, 18.0, 15.0, 10.0, 10.0, 15.0, 15.0, 13.0, 15.0, 19.0, 13.0, 18.0, 20.0, 21.0, 17.0, 18.0, 17.0, 18.0, 17.0, 12.0, 16.0, 15.0, 18.0, 19.0, 17.0, 20.0, 11.0, 18.0, 19.0, 11.0, 12.0, 17.0, 20.0, 17.0, 15.0, 13.0, 18.0, 14.0, 17.0, 12.0, 12.0, 16.0, 12.0, 14.0, 15.0, 14.0, 10.0, 20.0, 13.0, 18.0, 20.0, 11.0, 20.0, 17.0, 20.0, 13.0, 17.0, 15.0, 18.0, 14.0, 13.0, 13.0, 18.0, 10.0, 13.0, 12.0, 18.0, 20.0, 20.0, 16.0, 18.0, 15.0, 20.0, 22.0, 18.0, 21.0, 18.0, 18.0, 18.0, 17.0, 16.0, 19.0, 16.0, 20.0, 19.0, 19.0, 20.0, 20.0, 14.0, 18.0, 20.0, 20.0, 18.0, 16.0, 21.0, 20.0, 18.0, 15.0, 14.0, 17.0, 19.0, 21.0, 14.0, 18.0, 15.0, 18.0, 21.0, 19.0, 17.0, 16.0, 16.0, 15.0, 20.0, 19.0, 16.0, 21.0, 17.0, 19.0, 15.0, 18.0, 20.0, 18.0, 20.0, 18.0, 16.0, 16.0]
df = pd.DataFrame(data)

# replace numbers with categorical labels; could also create new columns
labels = ['Remedial', 'Below Proficient', 'Proficient', 'Advanced']
bins = [1, 11, 13, 15, np.inf]
df['BOY'] = pd.cut(x=df.BOY, labels=labels, bins=bins, right=True)
df['EOY'] = pd.cut(x=df.EOY, labels=labels, bins=bins, right=True)

# melt the relevant columns into a long form
dfm = df.melt(var_name='Tested', value_name='Proficiency')

# set the categorical label order, which makes the xaxis labels print in the specific order
dfm['Proficiency'] = pd.Categorical(dfm['Proficiency'], labels, ordered=True)

Groupby、百分比计算和绘图形状

# groupby and get the value counts
dfg = dfm.groupby('Tested')['Proficiency'].value_counts().reset_index(level=1, name='Size').rename('level_1': 'Proficiency', axis=1)

# divide by the Tested value counts to get the percent
dfg['percent'] = dfg['Size'].div(dfm.Tested.value_counts()).mul(100).round(1)

# reshape to plot
dfp = dfg.reset_index().pivot(index='Proficiency', columns='Tested', values='percent')

# display(dfp)
Tested             BOY   EOY
Proficiency                 
Remedial          34.8   9.9
Below Proficient  28.7  12.7
Proficient        27.1  25.4
Advanced           8.8  51.9

情节

ax = dfp.plot(kind='bar', figsize=(15, 5), rot=0, color=['orchid', 'teal'])

# formatting
ax.yaxis.set_major_formatter(mtick.PercentFormatter())
ax.set_ylabel('Students at Proficiency Level', fontsize=18)
ax.set_xlabel('')
ax.set_title('Bushwick Middle Change in Proficiency Levels', fontsize=25)
ax.set_xticklabels(ax.get_xticklabels(), fontsize=25)
ax.legend(fontsize=25)
_ = plt.yticks(fontsize=15)

# add bar labels
for p in ax.containers:
    ax.bar_label(p, fmt='%.1f%%', label_type='edge', fontsize=12)
    
# pad the spacing between the number and the edge of the figure
ax.margins(y=0.2)

查看条形标签匹配dfp

【讨论】：

非常感谢！太棒了。在绘图之前更容易操作数据。我是 Python 新手，还在学习。