﻿ Pandas: Extract only punctuations from the specified column of a given DataFrame - w3resource # Pandas: Extract only punctuations from the specified column of a given DataFrame

## Pandas: String and Regular Expression Exercise-31 with Solution

Write a Pandas program to extract only punctuations from the specified column of a given DataFrame.

Sample Solution:

Python Code :

``````import pandas as pd
import re as re
pd.set_option('display.max_columns', 10)
df = pd.DataFrame({
'company_code': ['c0001.','c000,2','c0003', 'c0003#', 'c0004,'],
'year': ['year 1800','year 1700','year 2300', 'year 1900', 'year 2200']
})
print("Original DataFrame:")
print(df)
def find_punctuations(text):
result = re.findall(r'[!"\\$%&\'()*+,\-.\/:;=#@?\[\\\]^_`{|}~]*', text)
string="".join(result)
return list(string)
df['nonalpha']=df['company_code'].apply(lambda x: find_punctuations(x))
print("\nExtracting punctuation:")
print(df)
``````

Sample Output:

```Original DataFrame:
company_code       year
0       c0001.  year 1800
1       c000,2  year 1700
2        c0003  year 2300
3       c0003#  year 1900
4       c0004,  year 2200

Extracting punctuation:
company_code       year nonalpha
0       c0001.  year 1800      [.]
1       c000,2  year 1700      [,]
2        c0003  year 2300       []
3       c0003#  year 1900      [#]
4       c0004,  year 2200      [,]
```

Python Code Editor:

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## Python: Tips of the Day

Returns True if there are duplicate values in a flat list, False otherwise

Example:

```def tips_duplicates(lst):
return len(lst) != len(set(lst))

x = [2, 4, 6, 8, 4, 2]
y = [1, 3, 5, 7, 9]
print(tips_duplicates(x))
print(tips_duplicates(y))
```

Output:

```True
False
```