如何在Pandas中创建一个在列和列表之间计数的列？

Question

我想创建一个列df['score']，该列返回单元格和列表之间共有的值计数。

输入：

correct_list = ['cats','dogs']
  answer       
0 cats, dogs, pigs
1 cats, dogs        
2 dogs, pigs        
3 cats              
4 pigs     

def animal_count(dataframe):
    count = 0
    for term in df['answer']:
        if term in symptom_list:
            df['score'] = count + 1

animal_count(df)         

目标输出：

correct_list = ['cats','dogs']

  answer            score
0 cats, dogs, pigs  2
1 cats, dogs        2
2 dogs, pigs        1
3 cats              1
4 pigs              0

有什么想法吗？谢谢！

Answer 1

使用Series.str.count的另一个解决方案：

Series.str.count

[out]

df['score'] = df['answer'].str.count('|'.join(correct_list))

---

更新

正如@PrinceFrancis指出的，如果不应该将

             answer  score
0  cats, dogs, pigs      2
1        cats, dogs      2
2        dogs, pigs      1
3              cats      1
4              pigs      0

计为catsdogs，则可以更改正则表达式模式以适合：

2

[out]

df = pd.DataFrame({'answer': ['cats, dogs, pigs', 'cats, dogs', 'dogs, pigs', 'cats', 'pigs', 'catsdogs']})

pat = '|'.join([fr'\b{x}\b' for x in correct_list])
df['score'] = df['answer'].str.count(pat)

Answer 2

您可以执行以下操作

             answer  score
0  cats, dogs, pigs      2
1        cats, dogs      2
2        dogs, pigs      1
3              cats      1
4              pigs      0
5          catsdogs      0

它将给您以下结果

correct_list = ['cats','dogs']
df['score'] = df['answer'].str.split(',')
df['score'] = df['score'].apply(lambda x: sum(el in x for el in correct_list))
df

Answer 3

    answer  score
0   cats,dogs,pigs  2
1   cats,dogs   2
2   dogs,pigs   1
3   cats    1
4   pigs    0

Answer 4

import pandas as pd
correct_list = ['cats', 'dogs']
answer = ['cats,dogs,pigs','cats,dogs','dogs,pigs','cats','pigs']
answer = [ans.split(',') for ans in answer]
score = [0] * len(answer)
df = pd.DataFrame({'answer':answer,'score':score})
print(df,'\n')
df.score = df.answer.apply(lambda cell: len(set(cell) & set(correct_list)))
print(df)

"""
               answer  score
0  [cats, dogs, pigs]      0
1        [cats, dogs]      0
2        [dogs, pigs]      0
3              [cats]      0
4              [pigs]      0 

               answer  score
0  [cats, dogs, pigs]      2
1        [cats, dogs]      2
2        [dogs, pigs]      1
3              [cats]      1
4              [pigs]      0
"""

Answer 5

我建议：

def my_func(x):
    return sum([1 for y in x.split(',') if y.strip() in correct_list])

df['score'] = df['answer'].apply(my_func)

结果：

correct_list = ['cats','dogs']
df = pd.DataFrame(['cats, dogs, pigs', 'cats, dogs', 'dogs, pigs', 'cats', 'pigs'], columns=['answer'])

df['score'] = df.answer.str.split(', ').apply(lambda x: sum([1 for a in x if a in correct_list]))

Answer 6

您可以测量两个

             answer  score
0  cats, dogs, pigs      2
1        cats, dogs      2
2        dogs, pigs      1
3              cats      1
4              pigs      0

之间的相交set：

list