# Python – Sort String by Custom Integer Substrings

Given a list of strings, sort strings by the occurrence of substring from list.

Input : test_list = [“Good at 4”, “Wake at 7”, “Work till 6”, “Sleep at 11”], subord_list = [“11”, “7”, “4”, “6”]
Output : [‘Sleep at 11’, ‘Wake at 7’, ‘Good at 4’, ‘Work till 6’]
Explanation : Strings sorted by substring presence.

Input : test_list = [“Good at 9”, “Wake at 7”, “Work till 6”, “Sleep at 11”], subord_list = [“11”, “7”, “9”, “6”]
Output : [‘Sleep at 11’, ‘Wake at 7’, ‘Good at 9’, ‘Work till 6’]
Explanation : Strings sorted by substring presence.

Method #1 : Using sorted() + zip() + lambda + regex()

The combination of above functions can be used to solve this problem. In this, we perform task of sorting by substring using regex() and sorted(), zip() is used to produce end result.

## Python3

 `# Python3 code to demonstrate working of ``# Sort String by Custom Substrings``# Using sorted() + zip() + lambda + regex()``import` `re` `# initializing list``test_list ``=` `[``"Good at 4"``, ``"Wake at 7"``, ``"Work till 6"``, ``"Sleep at 11"``]` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))` `# initializing substring list ``subord_list ``=` `[``"6"``, ``"7"``, ``"4"``, ``"11"``]`  `# creating inverse mapping with index``temp_dict ``=` `{val: key ``for` `key, val ``in` `enumerate``(subord_list)}` `# custom sorting ``temp_list ``=` `sorted``([[ele, temp_dict[re.search(``"(\d+)\$"``, ele).group()]] \``                ``for` `ele ``in` `test_list], key ``=` `lambda` `x: x[``1``])``# compiling result``res ``=` `[ele ``for` `ele ``in` `list``(``zip``(``*``temp_list))[``0``]]``        ` `# printing result ``print``(``"The sorted list : "` `+` `str``(res))`

Output
```The original list : ['Good at 4', 'Wake at 7', 'Work till 6', 'Sleep at 11']
The sorted list : ['Work till 6', 'Wake at 7', 'Good at 4', 'Sleep at 11']```

Time Complexity: O(nlogn), where n is the elements of list
Auxiliary Space: O(n), where n is the size of list

Method #2 : Using sorted() + zip() + comparator + regex()

This is yet another way in which this task can be performed. In this similar functionality is used as above method, difference is that comparator function is fed to sorted() rather than lambda to sort.

## Python3

 `# Python3 code to demonstrate working of ``# Sort String by Custom Substrings``# Using sorted() + comparator + regex()``import` `re` `# helper function to solve problem``def` `hlper_fnc(ele):``    ``temp ``=` `re.search(``"(\d+)\$"``, ele).group()``    ``return` `temp_dict[temp] ``if` `temp ``in` `temp_dict ``else` `int``(temp)` `# initializing list``test_list ``=` `[``"Good at 4"``, ``"Wake at 7"``, ``"Work till 6"``, ``"Sleep at 11"``]` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))` `# initializing substring list ``subord_list ``=` `[``"6"``, ``"7"``, ``"4"``, ``"11"``]` `# creating inverse mapping with index``temp_dict ``=` `{val: key ``for` `key, val ``in` `enumerate``(test_list)}` `# sorting using comparator``test_list.sort(key ``=` `lambda` `ele: hlper_fnc(ele))``        ` `# printing result ``print``(``"The sorted list : "` `+` `str``(test_list))`

Output
```The original list : ['Good at 4', 'Wake at 7', 'Work till 6', 'Sleep at 11']
The sorted list : ['Good at 4', 'Work till 6', 'Wake at 7', 'Sleep at 11']```

The Time and Space Complexity for all the methods are the same:

Time Complexity: O(n)

Space Complexity: O(n)

Method 3 : using a dictionary+  using the sorted() method  +re module

step by step explanations:

First, a list of strings called “test_list” is initialized with four values.
The original list is printed using the print() function.
A list of substrings called “subord_list” is initialized with the same number of values as the original list.
An empty dictionary called “order_dict” is initialized to store the mappings of substrings to orders.
A for loop is used to iterate through the indices of the “subord_list”. In each iteration, the substring at the current index is added as a key to the “order_dict”, and its corresponding index is added as the value.
The sorted() function is used to sort the “test_list” based on the values of the mappings in the “order_dict”. The key argument of the sorted() function is set to a lambda function that extracts the substring at the end of each string using regular expression and gets its corresponding index from the “order_dict”.
The sorted list is printed using the print() function.

## Python3

 `# Python3 code to demonstrate working of ``# Sort String by Custom Substrings``# Using dictionary and sorted()``import` `re` `# initializing list``test_list ``=` `[``"Good at 4"``, ``"Wake at 7"``, ``"Work till 6"``, ``"Sleep at 11"``]` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))` `# initializing substring list ``subord_list ``=` `[``"6"``, ``"7"``, ``"4"``, ``"11"``]` `# creating dictionary to store mappings of substrings to orders``order_dict ``=` `{}``for` `i ``in` `range``(``len``(subord_list)):``    ``order_dict[subord_list[i]] ``=` `i` `# sorting based on the values of the mappings using sorted()``test_list ``=` `sorted``(test_list, key``=``lambda` `x: order_dict[re.search(``"(\d+)\$"``, x).group()])` `# printing result ``print``(``"The sorted list : "` `+` `str``(test_list))`

Output
```The original list : ['Good at 4', 'Wake at 7', 'Work till 6', 'Sleep at 11']
The sorted list : ['Work till 6', 'Wake at 7', 'Good at 4', 'Sleep at 11']```

The time complexity of this approach is O(nlogn) due to the use of sorted().

Auxiliary space complexity is O(n) due to the use of the order_dict dictionary.

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