# re

## Regular Expression

```python
import re
regex = r"([a-zA-Z]+) (\d+)"
match = re.search(regex, "I was born on June 24") 
```

## Splitting String to List

Re.split(\[reExpression], \[string]) OR Re.findall(\[reExpression], \[string])

* '\s' #single white space
* '\s+' #split on one or more whitespace
* '\S+' #one or more non whitespace character
* '\w+') #one or more words
* '\W+' #any non word character

## Replace

```
regex = r"[0-9+\. *]"
re.sub(regex, "", '1. Interdimensional \n2. Aliens')
```

'\\\\' as the pattern string, because the regular expression must be \\, and each backslash must be expressed as \ inside a regular Python string literal.

The solution is to use Python’s raw string notation for regular expression patterns; backslashes are not handled in any special way in a string literal prefixed with 'r'