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In python, how can i get the multiplication of array elements using recursion?

I want to get the multiplication of array elements using recursion.
arr is array, and n is length of the array.
if arr=[1, 2, 3], n=3, answer is 6.
I tried this, but error occurred.
def multiply(arr, n):
if n == 0:
return arr
else:
return arr[n] * \
multyply(arr[n - 1])
please help me.

You should implement it like this
def mul(arr):
if not arr:
return 1
return arr[0] * mul(arr[1:])

Your approach is close. Try change n to count up and the exit condition to check when n exceeds the length of arr -
def multiply(arr, n = 0):
if n >= len(arr):
return 1
else:
return arr[n] * multiply(arr, n + 1)

Grab 'n' numbers from a given list of numbers with minimum difference between them

I put up a similar question a few hours ago, albeit with a few mistakes, and my poor understanding, admittedly
So the question is, from a given list of indefinite numbers, I'm supposed to take an input from the user, say 3, and grab 3 numbers wherein the numbers have the least difference between them.
def findMinDiff(arr):
# Initialize difference as infinite
diff = 10**20
n = len(arr)
# Find the min diff by comparing difference
# of all possible pairs in given array
for i in range(n-1):
for j in range(i+1,n):
if abs(arr[i]-arr[j]) < diff:
diff = abs(arr[i] - arr[j])
# Return min diff
return diff
def findDiffArray(arr):
diff = 10**20
arr_diff = []
n = len(arr)
for i in range(n-1):
arr_diff.append(abs(arr[i]-arr[i+1]))
return arr_diff
def choosingElements(arr, arr_diff):
arr_choose = []
least_index = 0
least = arr_diff[0]
least_index_array = []
flag = 0
flag2 = 0
for z in range(0,3):
for i in range(0,len(arr_diff)-1):
if arr_diff[i] < least:
if flag > 0:
if i == least_index:
continue
least = arr_diff[i]
least_index = i
least_index_array.append(i)
arr_choose.append(arr[i])
flag += 1
arr_choose.append(arr[i+1])
flag += 1
print("least index is", least_index)
return arr_choose
# Driver code
arr = [1, 5, 3, 19, 18, 25]
arr_diff = findDiffArray(arr)
arr_diff2 = arr_diff.copy()
item_number = int(input("Enter the number of gifts"))
arr_choose = choosingElements(arr, arr_diff2)
print("Minimum difference is " + str(findMinDiff(arr)))
print("Difference array")
print(*arr_diff, sep = "\n")
print("Numbers with least difference for specified items are", arr_choose)
This is how much I've tried, and I've thought to find the difference between numbers, and keep picking ones with the least difference between them, and I realised that my approach is probably wrong.
Can anybody kindly help me out? Thanks!

Now, I'm sure the time complexity on this isn't great, and it might be hard to understand, but how about this:
arr = [1, 18, 5, 19, 25, 3]
# calculates length of the overall path
def calc_path_difference(arr, i1, i2, i3):
return abs(arr[i1] - arr[i2]) + abs(arr[i2] - arr[i3])
# returns dictionary with differences to other numbers in arr from each number
def differences_dict(arr):
return {
current: [
abs(number - current) if abs(number - current) != 0 else float("inf")
for number in arr
]
for current in arr
}
differences = differences_dict(arr)
# Just to give some starting point, take the first three elements of arr
current_path = [calc_path_difference(arr, 0, 1, 2), 0, 1, 2]
# Loop 1
for i, num in enumerate(arr):
# Save some time by skippin numbers who's path
# already exceeds the min path we currently have
if not min(differences[num]) < current_path[0]:
continue
# Loop 2
for j, num2 in enumerate(arr):
# So you can't get 2 of the same index
if j == i:
continue
# some code for making indices i and j of differences
# infinite so they can't be the smallest, but not sure if
# this is needed without more tests
# diff_arr_copy = differences[num2].copy()
# diff_arr_copy[i], diff_arr_copy[j] = float("inf"), float("inf")
# Get index of number in arr with smallest difference to num2
min_index = differences[num2].index(min(differences[num2]))
# So you can't get 2 of the same index again
if min_index == i or min_index == j:
continue
# Total of current path
path_total = calc_path_difference(arr, i, j, min_index)
# Change current path if this one is shorter
if path_total < current_path[0]:
current_path = [path_total, i, j, min_index]
Does this work for you? I played around with the order of the elements in the array and it seemed to give the correct output each time but I would have liked to have another example to test it on.

How to iterate a list from the beginning when it reaches the maximum index

Hi i am new in python and i have a problem. I know there is a way to this this with % but i am trying to do it without the modulo if there is a way to do so. I have a list of numbers and i need to every Nth(i have every second number to keep it simple) number to be printed out, but when is reaches the maximux index it will give me an error "list is out of index". I have tried if original_position >= len(s): for j in s[0:] but it doesnt seems to work. This is what i have come up with so far:
def napln(n):
s = []
for i in range(n):
s.append(i)
print(s)
return s
def rozpocitavadlo(s):
rozpocitavadlo = []
for i in s:
original_position = s[i]
changed_position = s[i+2]
original_position = changed_position
if original_position >= len(s):
for j in s[0:]:
original_position = s[j]
changed_position = s[j+2]
original_position = changed_position
rozpocitavadlo.append(s.pop(original_position))
s = napln(6)

Your question i have a list of numbers and i need to every Nth(i have every second number to keep it simple) number to be printed out can be solved with built-in Python functions! No need for extra code.
for i in range(0, len(s), 2):
print(s[i])
The parameters for range are the starting index (0), how many items to loop over (len(s)) and how big the steps should be. This way, Python will print the 0th element and, from there, every second element.
EDIT:
You could try something like this:
s = [0, 1, 2, 3, 4, 5]
out = []
for offset in range(len(s) // (len(s) // 2)):
for i in range(offset, len(s), 2):
out.append(s[i])
offset = offset + 1
print(out)
With this, you are looping through the list multiple times with different starting offsets, so you first get all even numbers, and then all odd numbers. The calculation in the offset loop is there to figure out how many times we need to loop through the list.

If your intent is just to print out one number every two, you can simply run something like:
def rozpocitavadlo(s):
to_print = s[::2]
print(to_print)
Adding a 1 in [1::2] your iterating process will start from the second element skipping the first one, as follow:
def rozpocitavadlo(s):
to_print = s[1::2]
print(to_print)
If you also want to start over and add the skipped element to the final list you can add first the even number and then the odd ones:
def rozpocitavadlo(s):
to_print = s[1::2]
to_print.extend(s[0::2])
print(to_print)
Your result is going to be: [1, 3, 5, 0, 2, 4]
You can deepen here the slice notation that allow you to iterate over a list of a custom number of elements
Obviously as said by #vishal-gahlot, you can parameterize the number of elements to skip by adding a Nth parameter to the function
def rozpocitavadlo(s, Nth: int):
to_print = s[::Nth]
print(to_print)
FINAL: for a more flexible implementation with Nth as a parameter you can iterate over and over the original list until every element is added to the new one:
def rozpocitavadlo(s, nth: int):
to_print = []
for i in reversed(range(nth)):
to_print.extend(s[i::nth])
print(to_print)
UPDATE AFTER COMMENTS: with this you will get the even number, remove them and than reiterate on the remain element until the last remaining:
def rozpocitavadlo(s):
to_print = []
while len(s) > 0:
if len(s) == 1:
to_print.extend(s[:])
break
to_print.extend(s[1::2])
del s[1::2]
print(to_print)
Your output will be: [1, 3, 5, 2, 4, 0]

def rozpocitavadlo(s,Nth): return s[::Nth]
You can use this method to print every nth number of the list

Python code for printing out the second largest number number given a list [duplicate]

I'm learning Python and the simple ways to handle lists is presented as an advantage. Sometimes it is, but look at this:
>>> numbers = [20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7]
>>> numbers.remove(max(numbers))
>>> max(numbers)
74
A very easy, quick way of obtaining the second largest number from a list. Except that the easy list processing helps write a program that runs through the list twice over, to find the largest and then the 2nd largest. It's also destructive - I need two copies of the data if I wanted to keep the original. We need:
>>> numbers = [20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7]
>>> if numbers[0]>numbers[1]):
... m, m2 = numbers[0], numbers[1]
... else:
... m, m2 = numbers[1], numbers[0]
...
>>> for x in numbers[2:]:
... if x>m2:
... if x>m:
... m2, m = m, x
... else:
... m2 = x
...
>>> m2
74
Which runs through the list just once, but isn't terse and clear like the previous solution.
So: is there a way, in cases like this, to have both? The clarity of the first version, but the single run through of the second?

You could use the heapq module:
>>> el = [20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7]
>>> import heapq
>>> heapq.nlargest(2, el)
[90.8, 74]
And go from there...

Since #OscarLopez and I have different opinions on what the second largest means, I'll post the code according to my interpretation and in line with the first algorithm provided by the questioner.
def second_largest(numbers):
count = 0
m1 = m2 = float('-inf')
for x in numbers:
count += 1
if x > m2:
if x >= m1:
m1, m2 = x, m1
else:
m2 = x
return m2 if count >= 2 else None
(Note: Negative infinity is used here instead of None since None has different sorting behavior in Python 2 and 3 – see Python - Find second smallest number; a check for the number of elements in numbers makes sure that negative infinity won't be returned when the actual answer is undefined.)
If the maximum occurs multiple times, it may be the second largest as well. Another thing about this approach is that it works correctly if there are less than two elements; then there is no second largest.
Running the same tests:
second_largest([20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7])
=> 74
second_largest([1,1,1,1,1,2])
=> 1
second_largest([2,2,2,2,2,1])
=> 2
second_largest([10,7,10])
=> 10
second_largest([1,1,1,1,1,1])
=> 1
second_largest([1])
=> None
second_largest([])
=> None
Update
I restructured the conditionals to drastically improve performance; almost by a 100% in my testing on random numbers. The reason for this is that in the original version, the elif was always evaluated in the likely event that the next number is not the largest in the list. In other words, for practically every number in the list, two comparisons were made, whereas one comparison mostly suffices – if the number is not larger than the second largest, it's not larger than the largest either.

You could always use sorted
>>> sorted(numbers)[-2]
74

Try the solution below, it's O(n) and it will store and return the second greatest number in the second variable. UPDATE: I've adjusted the code to work with Python 3, because now arithmetic comparisons against None are invalid.
Notice that if all elements in numbers are equal, or if numbers is empty or if it contains a single element, the variable second will end up with a value of None - this is correct, as in those cases there isn't a "second greatest" element.
Beware: this finds the "second maximum" value, if there's more than one value that is "first maximum", they will all be treated as the same maximum - in my definition, in a list such as this: [10, 7, 10] the correct answer is 7.
def second_largest(numbers):
minimum = float('-inf')
first, second = minimum, minimum
for n in numbers:
if n > first:
first, second = n, first
elif first > n > second:
second = n
return second if second != minimum else None
Here are some tests:
second_largest([20, 67, 3, 2.6, 7, 74, 2.8, 90.8, 52.8, 4, 3, 2, 5, 7])
=> 74
second_largest([1, 1, 1, 1, 1, 2])
=> 1
second_largest([2, 2, 2, 2, 2, 1])
=> 1
second_largest([10, 7, 10])
=> 7
second_largest( [1, 3, 10, 16])
=> 10
second_largest([1, 1, 1, 1, 1, 1])
=> None
second_largest([1])
=> None
second_largest([])
=> None

Why to complicate the scenario? Its very simple and straight forward
Convert list to set - removes duplicates
Convert set to list again - which gives list in ascending order
Here is a code
mlist = [2, 3, 6, 6, 5]
mlist = list(set(mlist))
print mlist[-2]

You can find the 2nd largest by any of the following ways:
Option 1:
numbers = set(numbers)
numbers.remove(max(numbers))
max(numbers)
Option 2:
sorted(set(numbers))[-2]

The quickselect algorithm, O(n) cousin to quicksort, will do what you want. Quickselect has average performance O(n). Worst case performance is O(n^2) just like quicksort but that's rare, and modifications to quickselect reduce the worst case performance to O(n).
The idea of quickselect is to use the same pivot, lower, higher idea of quicksort, but to then ignore the lower part and to further order just the higher part.

This is one of the Simple Way
def find_second_largest(arr):
first, second = 0, 0
for number in arr:
if number > first:
second = first
first = number
elif number > second and number < first:
second = number
return second

If you do not mind using numpy (import numpy as np):
np.partition(numbers, -2)[-2]
gives you the 2nd largest element of the list with a guaranteed worst-case O(n) running time.
The partition(a, kth) methods returns an array where the kth element is the same it would be in a sorted array, all elements before are smaller, and all behind are larger.

there are some good answers here for type([]), in case someone needed the same thing on a type({}) here it is,
def secondLargest(D):
def second_largest(L):
if(len(L)<2):
raise Exception("Second_Of_One")
KFL=None #KeyForLargest
KFS=None #KeyForSecondLargest
n = 0
for k in L:
if(KFL == None or k>=L[KFL]):
KFS = KFL
KFL = n
elif(KFS == None or k>=L[KFS]):
KFS = n
n+=1
return (KFS)
KFL=None #KeyForLargest
KFS=None #KeyForSecondLargest
if(len(D)<2):
raise Exception("Second_Of_One")
if(type(D)!=type({})):
if(type(D)==type([])):
return(second_largest(D))
else:
raise Exception("TypeError")
else:
for k in D:
if(KFL == None or D[k]>=D[KFL]):
KFS = KFL
KFL = k
elif(KFS == None or D[k] >= D[KFS]):
KFS = k
return(KFS)
a = {'one':1 , 'two': 2 , 'thirty':30}
b = [30,1,2]
print(a[secondLargest(a)])
print(b[secondLargest(b)])
Just for fun I tried to make it user friendly xD

>>> l = [19, 1, 2, 3, 4, 20, 20]
>>> sorted(set(l))[-2]
19

O(n): Time Complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount. For example following functions have O(n) time complexity.
// Here c is a positive integer constant
for (int i = 1; i <= n; i += c) {
// some O(1) expressions
}
To find the second largest number i used the below method to find the largest number first and then search the list if thats in there or not
x = [1,2,3]
A = list(map(int, x))
y = max(A)
k1 = list()
for values in range(len(A)):
if y !=A[values]:
k.append(A[values])
z = max(k1)
print z

Objective: To find the second largest number from input.
Input : 5
2 3 6 6 5
Output: 5
*n = int(raw_input())
arr = map(int, raw_input().split())
print sorted(list(set(arr)))[-2]*

def SecondLargest(x):
largest = max(x[0],x[1])
largest2 = min(x[0],x[1])
for item in x:
if item > largest:
largest2 = largest
largest = item
elif largest2 < item and item < largest:
largest2 = item
return largest2
SecondLargest([20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7])

list_nums = [1, 2, 6, 6, 5]
minimum = float('-inf')
max, min = minimum, minimum
for num in list_nums:
if num > max:
max, min = num, max
elif max > num > min:
min = num
print(min if min != minimum else None)
Output
5

Initialize with -inf. This code generalizes for all cases to find the second largest element.
max1= float("-inf")
max2=max1
for item in arr:
if max1<item:
max2,max1=max1,item
elif item>max2 and item!=max1:
max2=item
print(max2)

Using reduce from functools should be a linear-time functional-style alternative:
from functools import reduce
def update_largest_two(largest_two, x):
m1, m2 = largest_two
return (m1, m2) if m2 >= x else (m1, x) if m1 >= x else (x, m1)
def second_largest(numbers):
if len(numbers) < 2:
return None
largest_two = sorted(numbers[:2], reverse=True)
rest = numbers[2:]
m1, m2 = reduce(update_largest_two, rest, largest_two)
return m2
... or in a very concise style:
from functools import reduce
def second_largest(n):
update_largest_two = lambda a, x: a if a[1] >= x else (a[0], x) if a[0] >= x else (x, a[0])
return None if len(n) < 2 else (reduce(update_largest_two, n[2:], sorted(n[:2], reverse=True)))[1]

This can be done in [N + log(N) - 2] time, which is slightly better than the loose upper bound of 2N (which can be thought of O(N) too).
The trick is to use binary recursive calls and "tennis tournament" algorithm. The winner (the largest number) will emerge after all the 'matches' (takes N-1 time), but if we record the 'players' of all the matches, and among them, group all the players that the winner has beaten, the second largest number will be the largest number in this group, i.e. the 'losers' group.
The size of this 'losers' group is log(N), and again, we can revoke the binary recursive calls to find the largest among the losers, which will take [log(N) - 1] time. Actually, we can just linearly scan the losers group to get the answer too, the time budget is the same.
Below is a sample python code:
def largest(L):
global paris
if len(L) == 1:
return L[0]
else:
left = largest(L[:len(L)//2])
right = largest(L[len(L)//2:])
pairs.append((left, right))
return max(left, right)
def second_largest(L):
global pairs
biggest = largest(L)
second_L = [min(item) for item in pairs if biggest in item]
return biggest, largest(second_L)
if __name__ == "__main__":
pairs = []
# test array
L = [2,-2,10,5,4,3,1,2,90,-98,53,45,23,56,432]
if len(L) == 0:
first, second = None, None
elif len(L) == 1:
first, second = L[0], None
else:
first, second = second_largest(L)
print('The largest number is: ' + str(first))
print('The 2nd largest number is: ' + str(second))

You can also try this:
>>> list=[20, 20, 19, 4, 3, 2, 1,100,200,100]
>>> sorted(set(list), key=int, reverse=True)[1]
100

A simple way :
n=int(input())
arr = set(map(int, input().split()))
arr.remove(max(arr))
print (max(arr))

use defalut sort() method to get second largest number in the list.
sort is in built method you do not need to import module for this.
lis = [11,52,63,85,14]
lis.sort()
print(lis[len(lis)-2])

Just to make the accepted answer more general, the following is the extension to get the kth largest value:
def kth_largest(numbers, k):
largest_ladder = [float('-inf')] * k
count = 0
for x in numbers:
count += 1
ladder_pos = 1
for v in largest_ladder:
if x > v:
ladder_pos += 1
else:
break
if ladder_pos > 1:
largest_ladder = largest_ladder[1:ladder_pos] + [x] + largest_ladder[ladder_pos:]
return largest_ladder[0] if count >= k else None

def secondlarget(passinput):
passinputMax = max(passinput) #find the maximum element from the array
newpassinput = [i for i in passinput if i != passinputMax] #Find the second largest element in the array
#print (newpassinput)
if len(newpassinput) > 0:
return max(newpassinput) #return the second largest
return 0
if __name__ == '__main__':
n = int(input().strip()) # lets say 5
passinput = list(map(int, input().rstrip().split())) # 1 2 2 3 3
result = secondlarget(passinput) #2
print (result) #2

if __name__ == '__main__':
n = int(input())
arr = list(map(float, input().split()))
high = max(arr)
secondhigh = min(arr)
for x in arr:
if x < high and x > secondhigh:
secondhigh = x
print(secondhigh)
The above code is when we are setting the elements value in the list
as per user requirements. And below code is as per the question asked
#list
numbers = [20, 67, 3 ,2.6, 7, 74, 2.8, 90.8, 52.8, 4, 3, 2, 5, 7]
#find the highest element in the list
high = max(numbers)
#find the lowest element in the list
secondhigh = min(numbers)
for x in numbers:
'''
find the second highest element in the list,
it works even when there are duplicates highest element in the list.
It runs through the entire list finding the next lowest element
which is less then highest element but greater than lowest element in
the list set initially. And assign that value to secondhigh variable, so
now this variable will have next lowest element in the list. And by end
of loop it will have the second highest element in the list
'''
if (x<high and x>secondhigh):
secondhigh=x
print(secondhigh)

Max out the value by comparing each one to the max_item. In the first if, every time the value of max_item changes it gives its previous value to second_max. To tightly couple the two second if ensures the boundary
def secondmax(self, list):
max_item = list[0]
second_max = list[1]
for item in list:
if item > max_item:
second_max = max_item
max_item = item
if max_item < second_max:
max_item = second_max
return second_max

you have to compare in between new values, that's the trick, think always in the previous (the 2nd largest) should be between the max and the previous max before, that's the one!!!!
def secondLargest(lista):
max_number = 0
prev_number = 0
for i in range(0, len(lista)):
if lista[i] > max_number:
prev_number = max_number
max_number = lista[i]
elif lista[i] > prev_number and lista[i] < max_number:
prev_number = lista[i]
return prev_number

Most of previous answers are correct but here is another way !
Our strategy is to create a loop with two variables first_highest and second_highest. We loop through the numbers and if our current_value is greater than the first_highest then we set second_highest to be the same as first_highest and then the second_highest to be the current number. If our current number is greater than second_highest then we set second_highest to the same as current number
#!/usr/bin/env python3
import sys
def find_second_highest(numbers):
min_integer = -sys.maxsize -1
first_highest= second_highest = min_integer
for current_number in numbers:
if current_number == first_highest and min_integer != second_highest:
first_highest=current_number
elif current_number > first_highest:
second_highest = first_highest
first_highest = current_number
elif current_number > second_highest:
second_highest = current_number
return second_highest
print(find_second_highest([80,90,100]))
print(find_second_highest([80,80]))
print(find_second_highest([2,3,6,6,5]))

Best solution that my friend Dhanush Kumar came up with:
def second_max(loop):
glo_max = loop[0]
sec_max = float("-inf")
for i in loop:
if i > glo_max:
sec_max = glo_max
glo_max=i
elif sec_max < i < glo_max:
sec_max = i
return sec_max
#print(second_max([-1,-3,-4,-5,-7]))
assert second_max([-1,-3,-4,-5,-7])==-3
assert second_max([5,3,5,1,2]) == 3
assert second_max([1,2,3,4,5,7]) ==5
assert second_max([-3,1,2,5,-2,3,4]) == 4
assert second_max([-3,-2,5,-1,0]) == 0
assert second_max([0,0,0,1,0]) == 0

Below code will find the max and the second max numbers without the use of max function. I assume that the input will be numeric and the numbers are separated by single space.
myList = input().split()
myList = list(map(eval,myList))
m1 = myList[0]
m2 = myList[0]
for x in myList:
if x > m1:
m2 = m1
m1 = x
elif x > m2:
m2 = x
print ('Max Number: ',m1)
print ('2nd Max Number: ',m2)

Here I tried to come up with an answer.
2nd(Second) maximum element in a list using single loop and without using any inbuilt function.
def secondLargest(lst):
mx = 0
num = 0
sec = 0
for i in lst:
if i > mx:
sec = mx
mx = i
else:
if i > num and num >= sec:
sec = i
num = i
return sec

Selection Sort (low to high) python

I am trying to write a selection sort algorithm for sorting lists of numbers from lowest to highest.
def sortlh(numList):
if type(numList) != list:
print("Input must be a list of numbers.")
else:
inf = float("inf")
sortList = [0]*len(numList)
count = 0
while count < len(numList):
index = 0
indexLowest = 0
lowest = numList[index]
while index < (len(numList) - 1):
if numList[index + 1] < numList[index]:
lowest = numList[index + 1]
indexLowest = index + 1
index = index + 1
else:
index = index + 1
sortList[count] = lowest
numList[indexLowest] = inf
count = count + 1
return sortList
When I run this code on:
sortlh([9,8,7,6,5,4,3,2,1])
I get (as expected):
[1, 2, 3, 4, 5, 6, 7, 8, 9]
However, when I try another example, I get:
sortlh([1,3,2,4,5,7,6,9,8])
[8, 6, 9, 2, 4, 5, 7, 1, 3]
Does anyone see what is going on here?

Here is how I would suggest rewriting your program.
def sortlh(lst_input):
lst = list(lst_input) # make a copy of lst_input
i = 0
while i < len(lst):
j = i + 1
i_lowest = i
lowest = lst[i_lowest]
while j < len(lst):
if lst[j] < lowest:
i_lowest = j
lowest = lst[i_lowest]
j += 1
lst[i], lst[i_lowest] = lst[i_lowest], lst[i] # swap
i += 1
return lst
test = [9,8,7,6,5,4,3,2,1]
assert sortlh(test) == sorted(test)
test = [1,3,2,4,5,7,6,9,8]
assert sortlh(test) == sorted(test)
We don't test the type of the input. Anything that acts like a list will work, and even an iterator will work.
We don't "mutate" the original input list. We only work on a copy of the data.
When we find the lowest number, we swap it with the first number, and then only look at the remaining numbers. Thus we have less work to do on each loop as we have fewer and fewer unsorted numbers.
EDIT:
If you are a beginner, this part might seem too tricky. If it confuses you or you don't like it, just ignore it for now.
This is a more-advanced way to solve this problem in Python. The inner loop simply finds the lowest number and the index of the lowest number. We can use the Python built-in function min() to do this!
We build a "generator expression" that loops over the list, yielding up tuples. Each tuple is the number and its position. Since we want lower numbers to sort lower, we put the number first in the tuple so that min() can properly compare the tuples. Then min() will find the lowest tuple and we get the value and index.
Also, the outer loop is now a for loop with enumerate rather than a while loop using indexing.
def sortlh(lst_input):
lst = list(lst_input) # make a copy of lst_input
for i, x in enumerate(lst):
lowest, i_lowest = min((n, j) for j, n in enumerate(lst) if j >= i)
lst[i], lst[i_lowest] = lst[i_lowest], lst[i] # swap
return lst
test = [9,8,7,6,5,4,3,2,1]
assert sortlh(test) == sorted(test)
test = [1,3,2,4,5,7,6,9,8]
assert sortlh(test) == sorted(test)