CODEWITHSUNDEEP
pythonfor-loopwhile-loop
user4139413
user4139413

Reputation:

Order a list from largest to smallest using insertion sort

I'm trying to create a program that orders a random list of 10 numbers from largest to smallest using the insertion sort algorithm. That is, finding the largest number in a list and using .append to add it to a new list. If this is repeated until the end of the list, then the new list will be ordered from largest to smallest. I have already created a program that successfully creates a random list, finds the largest number in the list, and adds it to a new list, the only problem is that I can't find a way to get the program to repeat 10 times. Thanks!

import random
num_list = []
new_list=[]

for num in range(10):
    num_list.append(random.randrange(0,11))

largest=num_list[0]

for repeat in range(len(num_list)):
    for large in num_list:
        if large>largest:
            largest=large
            new_list.append(largest)

print new_list

Please note that the point of this program is to not use any functions that will sort the list for me.

Upvotes: 0

Views: 3596

Answers (1)

VHarisop
VHarisop

Reputation: 2826

You can do it by removing in each step the maximum value from the unsorted list and appending to the new one. It is not very efficient but it's fairly simple.

new_list = []

# init random list
num_list = [random.randrange(0, 11) for _ in range(10)]

# while condition will evaluate to False 
# only when num_list is empty

while num_list:  
    # find index of maximum item
    max_index = num_list.index(max(num_list)) 

    # remove item with pop() and append to sorted list
    new_list.append(num_list.pop(max_index))

Edit: if you want to avoid using the built-in function max(), you can write it on your own using reduce(). 

mx = lambda x: reduce(lambda xs, ys: xs if xs > ys else ys, x)

Then just replace max with mx at the line which finds max_index.

Upvotes: 0

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