python - 如何构建 pandas 数据框中项目的频率计数表？

Question

假设我在 csv 文件 example.csv 中有以下数据:

Word    Score
Dog     1
Bird    2
Cat     3
Dog     2
Dog     3
Dog     1
Bird    3
Cat     1
Bird    1
Cat     3

我想计算每个分数的每个单词的频率。预期输出如下:

        1   2   3
Dog     2   1   1
Bird    0   1   1
Cat     1   0   2

我执行此操作的代码如下:

将 pandas 导入为 pd

x1 = pd.read_csv(r'path\to\example.csv')

def getUniqueWords(allWords) :
    uniqueWords = [] 
    for i in allWords:
        if not i in uniqueWords:
            uniqueWords.append(i)
    return uniqueWords

unique_words = getUniqueWords(x1['Word'])
unique_scores = getUniqueWords(x1['Score'])

scores_matrix = [[0 for x in range(len(unique_words))] for x in range(len(unique_scores)+1)]   
# The '+1' is because Python indexing starts from 0; so if a score of 0 is present in the data, the 0 index will be used for that. 

for i in range(len(unique_words)):
    temp = x1[x1['Word']==unique_words[i]]
    for j, word in temp.iterrows():
        scores_matrix[i][j] += 1  # Supposed to store the count for word i with score j

但这会出现以下错误:

IndexError                                Traceback (most recent call last)
<ipython-input-123-141ab9cd7847> in <module>()
     19     temp = x1[x1['Word']==unique_words[i]]
     20     for j, word in temp.iterrows():
---> 21         scores_matrix[i][j] += 1

IndexError: list index out of range

此外，即使我可以修复此错误，scores_matrix 也不会显示标题(Dog、Bird、Cat 作为行索引，1、2、3 作为列索引)。我希望能够访问每个分数的每个单词的计数 - 达到这样的效果:

scores_matrix['Dog'][1]
>>> 2

scores_matrix['Cat'][2]
>>> 0

那么，我该如何解决/修复这两个问题？

Answer 1

使用groupby与 sort=False 和 value_counts或size与 unstack :

df1 = df.groupby('Word', sort=False)['Score'].value_counts().unstack(fill_value=0)

<小时/>

df1 = df.groupby(['Word','Score'], sort=False).size().unstack(fill_value=0)

print (df1)
Score  1  2  3
Word          
Dog    2  1  1
Bird   1  1  1
Cat    1  0  2

如果顺序不重要，请使用 crosstab :

df1 = pd.crosstab(df['Word'], df['Score'])
print (df1)
Score  1  2  3
Word          
Bird   1  1  1
Cat    1  0  2
Dog    2  1  1

最后通过带有 DataFrame.loc 的标签进行选择:

print (df.loc['Cat', 2])
0