﻿ 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:

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## Python: Tips of the Day

Returns every nth element in a list

Example:

```def tips_every_nth(lst, nth):
return lst[nth - 1::nth]

print(tips_every_nth([1, 2, 3, 4, 5, 6, 7, 8, 9], 3))]
```

Output:

```[3, 6, 9]
```