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[0][0]|2[0-2][00])\b",text)
    result = re.findall(r"\b(18[0-9]{2}|19[0-8][0-9]|199[0-9]|2[01][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     [1800]
1        c0002  year 1700         []
2        c0003  year 2300         []
3        c0003  year 1900     [1900]
4        c0004  year 2200     [2200]

Python Code Editor:

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