﻿ Pandas: Extract year between 1800 to 2200 from the specified column of a given DataFrame - w3resource # Pandas: Extract year between 1800 to 2200 from the specified column of a given DataFrame

## Pandas: String and Regular Expression Exercise-29 with Solution

Write a Pandas program to extract year between 1800 to 2200 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','c0002','c0003', 'c0003', 'c0004'],
'year': ['year 1800','year 1700','year 2300', 'year 1900', 'year 2200']
})
print("Original DataFrame:")
print(df)
def find_year(text):
#line=re.findall(r"\b(18|2[0-2])\b",text)
result = re.findall(r"\b(18[0-9]{2}|19[0-8][0-9]|199[0-9]|2[0-9]{2}|2200)\b",text)
return result
df['year_range']=df['year'].apply(lambda x: find_year(x))
print("\Extracting year between 1800 to 2200:")
print(df)
``````

Sample Output:

```Original DataFrame:
company_code       year
0        c0001  year 1800
1        c0002  year 1700
2        c0003  year 2300
3        c0003  year 1900
4        c0004  year 2200
\Extracting year between 1800 to 2200:
company_code       year year_range
0        c0001  year 1800     
1        c0002  year 1700         []
2        c0003  year 2300         []
3        c0003  year 1900     
4        c0004  year 2200     
```

Python Code Editor:

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.

﻿

## Python: Tips of the Day

Debug With the Print() Function:

```>>> for i in range(5):
...     print(i, end=', ' if i < 4 else '\n')
...
0, 1, 2, 3, 4
>>> for i in range(5):
...     print(f'{i} & {i*i}', end=', ' if i < 4 else '\n')
...
0 & 0, 1 & 1, 2 & 4, 3 & 9, 4 & 16
```