Python: Sort a given mixed list of integers and strings using lambda
Python Lambda: Exercise-47 with Solution
Write a Python program to sort a given mixed list of integers and strings using lambda. Numbers must be sorted before strings.
Python Code :
def sort_mixed_list(mixed_list): mixed_list.sort(key=lambda e: (isinstance(e, str), e)) return mixed_list mixed_list = [19,'red',12,'green','blue', 10,'white','green',1] print("Original list:") print(mixed_list) print("\nSort the said mixed list of integers and strings:") print(sort_mixed_list(mixed_list))
Original list: [19, 'red', 12, 'green', 'blue', 10, 'white', 'green', 1] Sort the said mixed list of integers and strings: [1, 10, 12, 19, 'blue', 'green', 'green', 'red', 'white']
Visualize Python code execution:
The following tool visualize what the computer is doing step-by-step as it executes the said program:
Python Code Editor:
Have another way to solve this solution? Contribute your code (and comments) through Disqus.Write a Python program to sort a given list of strings(numbers) numerically using lambda.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
Python: Tips of the Day
How do I pass a variable by reference?
Arguments are passed by assignment. The rationale behind this is twofold:
- the parameter passed in is actually a reference to an object (but the reference is passed by value)
- some data types are mutable, but others aren't
- If you pass a mutable object into a method, the method gets a reference to that same object and you can mutate it to your heart's delight, but if you rebind the reference in the method, the outer scope will know nothing about it, and after you're done, the outer reference will still point at the original object.
- If you pass an immutable object to a method, you still can't rebind the outer reference, and you can't even mutate the object.
To make it even more clear, let's have some examples.
List - a mutable type
Let's try to modify the list that was passed to a method:
before, outer_list = ['one', 'two', 'three'] got ['one', 'two', 'three'] changed to ['one', 'two', 'three', 'four'] after, outer_list = ['one', 'two', 'three', 'four']
Since the parameter passed in is a reference to outer_list, not a copy of it, we can use the mutating list methods to change it and have the changes reflected in the outer scope.
Now let's see what happens when we try to change the reference that was passed in as a parameter:
def try_to_change_list_reference(the_list): print('got', the_list) the_list = ['and', 'we', 'can', 'not', 'lie'] print('set to', the_list) outer_list = ['we', 'like', 'proper', 'English'] print('before, outer_list =', outer_list) try_to_change_list_reference(outer_list) print('after, outer_list =', outer_list)
before, outer_list = ['we', 'like', 'proper', 'English'] got ['we', 'like', 'proper', 'English'] set to ['and', 'we', 'can', 'not', 'lie'] after, outer_list = ['we', 'like', 'proper', 'English']
Since the the_list parameter was passed by value, assigning a new list to it had no effect that the code outside the method could see. The the_list was a copy of the outer_list reference, and we had the_list point to a new list, but there was no way to change where outer_list pointed.
String - an immutable type
It's immutable, so there's nothing we can do to change the contents of the string
Now, let's try to change the reference
def try_to_change_string_reference(the_string): print('got', the_string) the_string = 'In a kingdom by the sea' print('set to', the_string) outer_string = 'It was many and many a year ago' print('before, outer_string =', outer_string) try_to_change_string_reference(outer_string) print('after, outer_string =', outer_string)
before, outer_string = It was many and many a year ago got It was many and many a year ago set to In a kingdom by the sea after, outer_string = It was many and many a year ago
Again, since the the_string parameter was passed by value, assigning a new string to it had no effect that the code outside the method could see. The the_string was a copy of the outer_string reference, and we had the_string point to a new string, but there was no way to change where outer_string pointed.
I hope this clears things up a little.
EDIT: It's been noted that this doesn't answer the question that @David originally asked, "Is there something I can do to pass the variable by actual reference?". Let's work on that.
How do we get around this?
As @Andrea's answer shows, you could return the new value. This doesn't change the way things are passed in, but does let you get the information you want back out:
def return_a_whole_new_string(the_string): new_string = something_to_do_with_the_old_string(the_string) return new_string # then you could call it like my_string = return_a_whole_new_string(my_string)
If you really wanted to avoid using a return value, you could create a class to hold your value and pass it into the function or use an existing class, like a list:
def use_a_wrapper_to_simulate_pass_by_reference(stuff_to_change): new_string = something_to_do_with_the_old_string(stuff_to_change) stuff_to_change = new_string # then you could call it like wrapper = [my_string] use_a_wrapper_to_simulate_pass_by_reference(wrapper) do_something_with(wrapper)
Although this seems a little cumbersome.
- New Content published on w3resource:
- HTML-CSS Practical: Exercises, Practice, Solution
- Java Regular Expression: Exercises, Practice, Solution
- Scala Programming Exercises, Practice, Solution
- Python Itertools exercises
- Python Numpy exercises
- Python GeoPy Package exercises
- Python Pandas exercises
- Python nltk exercises
- Python BeautifulSoup exercises
- Form Template
- Composer - PHP Package Manager
- PHPUnit - PHP Testing
- Laravel - PHP Framework