Using a While Loop to Iterate a List and Stop at a given value

Question

Write a function called stop_at_four that iterates through a list of numbers. Using a while loop, append each number to a new list until the number 4 appears. The function should return the new list.

My answer

def stop_at_four():  
    list_=[3,6,4,1,3]
    accum_lst=[]
    accum_var=0
    
    while list_[accum_var] !=4:
        accum_lst.append(list_[accum_var])
    accum_var+=1
    return

P.S. any criticism on how to improve my code efficiency would be greatly appreciated.

You should give the list as a parameter and return the new list. Also the increment of accum_var have to be inside the while loop and you should check whether if you reached the end of the list to prevent a list index out of range:

def stop_at_four(my_list):
    accum_lst=[]
    accum_var=0

    while (accum_var < len(my_list)) and (my_list[accum_var] != 4):
        accum_lst.append(my_list[accum_var])
        accum_var+=1
    return accum_lst

print(stop_at_four([3,6,4,1,3]))

Well... this code will run in all possible conditions`

def stop_at_four(lst):
    n = 0
    new_lst = []
    if len(lst) == 0:
        return new_lst
    else:
        while lst[n] != 4 and n < len(lst):
            new_lst.append(lst[n])
            n += 1
        return new_lst
lst = [3, 5, 7, 9.5, 7, 1, 4]
print(stop_at_four(lst))`

.

As mentioned by Arunesh Singh, the correct code block should be

def stop_at_four():  
    list_=[3,6,4,1,3]
    accum_lst=[]
    accum_var=0

    while list_[accum_var] !=4:
        accum_lst.append(list_[accum_var])
        accum_var+=1
    return accum_lst

Incrementation should be part of while loop.

It works for me!!! Just give a try...

def stop_at_four():
    list_=[3,6,4,1,3]
    accum_lst=[]
    accum_var=0

    while list_[accum_var] != 4 :
        accum_lst.append(list_[accum_var])
        accum_var += 1
    return accum_lst

print(stop_at_four())

This works better!

def stop_at_four(lst):
    n = 0
    new_lst = []
    while lst[n] != 4:
        new_lst.append(lst[n])
        n += 1
    return new_lst
lst = [1,2,4,6,12]
print(stop_at_four(lst))

def stop_at_four(lis):
    new_list = []
    start = 0
    while start < len(lis) and lis[start] != 4:
        new_list.append(lis[start])
        start += 1
    return new_list    
    
list1 = [1, 6, 2, 3, 9]     
print(stop_at_four(list1))

Approached this another way but not sure if there is a restriction on using the in operator:

def stop_at_four(lst):
    if 4 not in lst:
        return lst
    idx = 0
    accum_list = []
    while lst[idx] != 4:
        accum_list.append(lst[idx])
        idx += 1
    return accum_list

Try this instead:

def stop_at_four(lst):
    sublist = []
    y = (num for num in lst)  
    num = next(y, 4)  
    while num != 4:
        sublist.append(num)
        num = next(y,4)  
return sublist

This works well also :

def stop_at_four(lst):
i = 0 
new_lst = []
while i < len(lst):
    if lst[i] != 4 :
        new_lst.append(lst[i])
    else :
        break
    i = i + 1
return new_lst