Python | Split String by Number : Shubham Sayon

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December 03, 2022
Python | Split String by Number
by: Shubham Sayon
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✨Summary: To split a string by a number, use the regex split method using the “\d” pattern.

Minimal Example

my_string = "#@1abc3$!*5xyz"

# Method 1
import re

res = re.split('\d+', my_string)
print(res)

# Method 2
import re

res = re.findall('\D+', my_string)
print(res)

# Method 3
from itertools import groupby

li = [''.join(g) for _, g in groupby(my_string, str.isdigit)]
res = [x for x in li if x.isdigit() == False]
print(res)

# Method 4
res = []
for i in my_string:
    if i.isdigit() == True:
        my_string = my_string.replace(i, ",")
print(my_string.split(","))

# Outputs:
# ['#@', 'abc', '$!*', 'xyz']

Problem Formulation

📜Problem: Given a string containing different characters. How will you split the string whenever a number appears?

Method 1: re.split()

The re.split(pattern, string) method matches all occurrences of the pattern in the string and divides the string along the matches resulting in a list of strings between the matches. For example, re.split('a', 'bbabbbab') results in the list of strings ['bb', 'bbb', 'b'].

Code:

import re
my_string = "#@1abc3$!*5xyz"
res = re.split('\d+', my_string)
print(res)

# ['#@', 'abc', '$!*', 'xyz']

Explanation: The \d special character matches any digit between 0 and 9. By using the maximal number of digits as a delimiter, you split along the digit-word boundary. 

Method 2: re.findall()

The re.findall(pattern, string) method scans string from left to right, searching for all non-overlapping matches of the pattern. It returns a list of strings in the matching order when scanning the string from left to right.

Code:

import re
my_string = "#@1abc3$!*5xyz"
res = re.findall('\D+', my_string)
print(res)

# ['#@', 'abc', '$!*', 'xyz']

Explanation: The \D special character matches all characters except any digit between 0 and 9. Thus, you are essentially finding all character groups that appear before the occurrence of a digit.

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Method 3: itertools.groupby()

Code:

from itertools import groupby
my_string = "#@1abc3$!*5xyz"
li = [''.join(g) for _, g in groupby(my_string, str.isdigit)]
res = [x for x in li if x.isdigit() == False]
print(res)

# ['#@', 'abc', '$!*', 'xyz']

Explanation:

  • The itertools.groupby(iterable, key=None) function creates an iterator that returns tuples (key, group-iterator) grouped by each value of key. We use the str.isdigit() function as key function.
  • The str.isdigit() function returns True if the string consists only of numeric characters. Thus, you will have a list created by using numbers as separators. Note that this list will also contain the numbers as items within it.
  • In order to eliminate the numbers, use another list comprehension that checks if an element in the list returned previously is a digit or not with the help of the isdigit method. If it is a digit, the item will be discarded. Otherwise it will be stored in the list.

Method 4: Replace Using a for Loop

Approach: Use a for loop to iterate through the characters of the given string. Check if a character is a digit or not. As soon as a digit is found, replace that character/digit with a delimiter string ( we have used a comma here) with the help of the replace() method. This basically means that you are placing a particular character in the string whenever a number appears. Once all the digits are replaced by the separator string, split the string by passing the separator string as a delimiter to the split method.

Code:

my_string = "#@1abc3$!*5xyz"
res = []
for i in my_string:
    if i.isdigit():
        my_string = my_string.replace(i, ",")
print(my_string.split(","))

# ['#@', 'abc', '$!*', 'xyz']

Conclusion

Phew! We have successfully solved the given problem and managed to do so using four different ways. I hope you found this article helpful and it answered your queries. Please subscribe and stay tuned for more solutions and tutorials.

Happy coding! 🙂

🌐Related Read: How to Split a String Between Numbers and Letters?


December 03, 2022 at 11:35PM
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