Question

我有一个熊猫数据框，其中包含学生的成绩。我想为年级生成词云或数字云。有什么方法可以实现它。我尝试了所有可能的方法，但徒劳无功。基本上我想要的是其中包含数字的词云。 CGPA列中。

这是我尝试过的：

import pandas as pd
from wordcloud import WordCloud
import matplotlib.pyplot as plt
df = pd.read_csv("VTU_marks.csv")
# rounding off
df = df[df['CGPA'].isnull() == False]
df['CGPA'] = df['CGPA'].round(decimals=2)

wordcloud = WordCloud(max_font_size=50,max_words=100,background_color="white").generate(string)
plt.figure()
plt.imshow(wordcloud, interpolation="bilinear")
plt.axis("off")
plt.show()

但是我遇到了错误

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-47-29ec36ebbb1e> in <module>()
----> 1 wordcloud = WordCloud(max_font_size=50, max_words=100, background_color="white").generate(string)
      2 plt.figure()
      3 plt.imshow(wordcloud, interpolation="bilinear")
      4 plt.axis("off")
      5 plt.show()

/usr/local/lib/python3.6/dist-packages/wordcloud/wordcloud.py in generate(self, text)
    603         self
    604         """
--> 605         return self.generate_from_text(text)
    606 
    607     def _check_generated(self):

/usr/local/lib/python3.6/dist-packages/wordcloud/wordcloud.py in generate_from_text(self, text)
    585         """
    586         words = self.process_text(text)
--> 587         self.generate_from_frequencies(words)
    588         return self
    589 

/usr/local/lib/python3.6/dist-packages/wordcloud/wordcloud.py in generate_from_frequencies(self, frequencies, max_font_size)
    381         if len(frequencies) <= 0:
    382             raise ValueError("We need at least 1 word to plot a word cloud, "
--> 383                              "got %d." % len(frequencies))
    384         frequencies = frequencies[:self.max_words]
    385 

ValueError: We need at least 1 word to plot a word cloud, got 0.

您可以找到数据here。我们将非常感谢您提供任何有助于绘制情节的帮助。

Answer 1

设置好数据并根据需要四舍五入后，我们可以计算每个分数的出现频率：

counts = df['CGPA'].value_counts()

我们需要确保这里的索引是字符串，浮点数将引发错误（这是您的示例尝试中的错误）。因此，我们可以将它们转换为以下字符串：

counts.index = counts.index.map(str)

#Below alternative works for pandas versions >= 0.19.0
#counts.index = counts.index.astype(str)

然后我们可以使用.generate_from_frequencies方法来获得您想要的东西：

wordcloud = WordCloud().generate_from_frequencies(counts)
plt.figure()
plt.imshow(wordcloud, interpolation="bilinear")
plt.axis("off")
plt.show()

这给了我以下内容：

完整MWE：

import pandas as pd
from wordcloud import WordCloud
import matplotlib.pyplot as plt

df = pd.read_csv("VTU_marks.csv")
# rounding off
df = df[df['CGPA'].isnull() == False]
df['CGPA'] = df['CGPA'].round(decimals=2)

counts = df['CGPA'].value_counts()

counts.index = counts.index.map(str)
#counts.index = counts.index.astype(str)

wordcloud = WordCloud().generate_from_frequencies(counts)
plt.figure()
plt.imshow(wordcloud, interpolation="bilinear")
plt.axis("off")
plt.show()

生成词云以显示Python中的数字频率

1 个答案: