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# Python | Extract Numbers in Brackets in String

• Last Updated : 21 Mar, 2023

Sometimes, while working with Python strings, we can have a problem in which we have to perform the task of extracting numbers in strings that are enclosed in brackets. Let’s discuss the certain ways in which this task can be performed.

Method 1: Using regex The way to solve this task is to construct a regex string that can return all the numbers in a string that has brackets around them.

## Python3

 `# Python3 code to demonstrate working of``# Extract Numbers in Brackets in String``# Using regex``import` `re` `# initializing string``test_str ``=` `"gfg is   all geeks"` `# printing original string``print``(``"The original string is : "` `+` `test_str)` `# Extract Numbers in Brackets in String``# Using regex``res ``=` `re.findall(r``"\[\s*\+?(-?\d+)\s*\]"``, test_str)` `# printing result``print``(``"Extracted number list : "` `+` `str``(res))`

Output :

```The original string is : gfg is   all geeks
Extracted number list : ['1', '4']```

Time Complexity: O(N), where N is the length of the given string.
Space Complexity: O(N)

Method 2 : Using startswith(), endswith() and replace() methods

Initially split the string. Iterate over the list after splitting, check for strings in the list that start with “[” and end with”]”.If found remove those square braces and check whether the string after removing braces is numeric or not.

## Python3

 `# Python3 code to demonstrate working of``# Extract Numbers in Brackets in String` `# initializing string``test_str ``=` `"gfg is   all geeks"` `# printing original string``print``(``"The original string is : "` `+` `test_str)` `# Extract Numbers in Brackets in String``x``=``test_str.split()``res``=``[]``for` `i ``in` `x:``    ``if` `i.startswith(``'['``) ``and` `i.endswith(``']'``) :``        ``a``=``i.replace(``'['``,'')``        ``a``=``a.replace(``']'``,'')``        ``if` `a.isdigit():``            ``res.append(a)``# printing result``print``(``"Extracted number list : "` `+` `str``(res))`

Output

```The original string is : gfg is   all geeks
Extracted number list : ['1', '4']```

Time Complexity: O(N), where N is the length of the given string.
Space Complexity: O(N)

Method 3 : Using find() and replace() methods

## Python3

 `# Python3 code to demonstrate working of``# Extract Numbers in Brackets in String` `# initializing string``test_str ``=` `"gfg is   all geeks"` `# printing original string``print``(``"The original string is : "` `+` `test_str)` `# Extract Numbers in Brackets in String``x``=``test_str.split()``res``=``[]``for` `i ``in` `x:``    ``if` `i.find(``'['``)``=``=``0` `and` `i.find(``']'``)``=``=``len``(i)``-``1` `:``        ``a``=``i.replace(``'['``,'')``        ``a``=``a.replace(``']'``,'')``        ``if` `a.isdigit():``            ``res.append(a)``# printing result``print``(``"Extracted number list : "` `+` `str``(res))`

Output

```The original string is : gfg is   all geeks
Extracted number list : ['1', '4']```

Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(N)

Method 4: Using split() and isdigit() methods

Step-by-step approach:

• Initialize a list variable to store the extracted numbers.
• Split the string using the “[” and “]” characters as delimiters. This will create a list of substrings.
• Loop through the substrings and check if each substring contains a number using the isdigit() method.
• If a substring contains a number, append it to the list variable.
• Print the list of extracted numbers.

Below is the implementation of the above approach:

## Python3

 `# Python3 code to demonstrate working of``# Extract Numbers in Brackets in String``# Using split() and isdigit() methods` `# initializing string``test_str ``=` `"gfg is   all geeks"` `# printing original string``print``(``"The original string is : "` `+` `test_str)` `# Extract Numbers in Brackets in String``# Using split() and isdigit() methods``num_list ``=` `[]``for` `substring ``in` `test_str.split(``"["``):``    ``if` `"]"` `in` `substring:``        ``num ``=` `substring.split(``"]"``)[``0``]``        ``if` `num.isdigit():``            ``num_list.append(``int``(num))` `# printing result``print``(``"Extracted number list : "` `+` `str``(num_list))`

Output

```The original string is : gfg is   all geeks
Extracted number list : [1, 4]```

Time complexity: O(n), where n is the length of the input string. This is because we need to loop through each character in the string once.
Auxiliary space: O(m), where m is the number of numbers in brackets in the input string. This is because we need to store each extracted number in a list.

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