将Pandas数据框中字符串出现的次数附加到另一列

Question

I'd like to create an extra column on this dataframe:

Index                  Value
0                22,88,22,24
1                      24,24
2                      22,24
3    11,22,24,12,24,24,22,24
4                         22

So that the number of times a value occurs is stored in a new column:

Index                  Value     22 Count
0                22,88,22,24            2
1                      24,24            1
2                      22,24            1
3    11,22,24,12,24,24,22,24            2
4                         22            1

I'd like to repeat this process for a number of different values within the value column.

My minimal Python knowledge is telling me something like:

df['22 Count'] = df['Value'].count('22')

I've tried this and a few other versions but I must be missing something.

Answer 1

If want count only one value use str.count :

df['22 Count'] = df['Value'].str.count('22')
print (df)
                         Value  22 Count
Index                                   
0                  22,88,22,24         2
1                        24,24         0
2                        22,24         1
3      11,22,24,12,24,24,22,24         2
4                           22         1

For all columns count need:

from collections import Counter

df1 = df['Value'].apply(lambda x: pd.Series(Counter(x.split(','))), 1).fillna(0).astype(int)

Or:

df1 = pd.DataFrame([Counter(x.split(',')) for x in df['Value']]).fillna(0).astype(int)

Or:

from sklearn.feature_extraction.text import CountVectorizer

countvec = CountVectorizer()
counts = countvec.fit_transform(df['Value'].str.replace(',', ' '))
df1 = pd.DataFrame(counts.toarray(), columns=countvec.get_feature_names())

---

print (df1)
   11  12  22  24  88
0   0   0   2   1   1
1   0   0   0   2   0
2   0   0   1   1   0
3   1   1   2   4   0
4   0   0   1   0   0

Last if need add to original:

df = df.join(df1.add_suffix(' Count'))
print (df)
                         Value  11 Count  12 Count  22 Count  24 Count  \
Index                                                                    
0                  22,88,22,24         0         0         2         1   
1                        24,24         0         0         0         2   
2                        22,24         0         0         1         1   
3      11,22,24,12,24,24,22,24         1         1         2         4   
4                           22         0         0         1         0   

       88 Count  
Index            
0             1  
1             0  
2             0  
3             0  
4             0  

Answer 2

Isolated count

You are close. But your syntax attempts to treat a series as if it were a list. Instead, you can use the count method after conversion to list :

from operator import methodcaller

df['22_Count'] = df['Value'].str.split(',').apply(methodcaller('count', '22'))

print(df)

   Index                    Value  22_Count
0      0              22,88,22,24         2
1      1                    24,24         0
2      2                    22,24         1
3      3  11,22,24,12,24,24,22,24         2
4      4                       22         1

Multiple counts

Use the methods shown by @jezrael .