Python insert number in list sequence

Question

I am trying to insert a number into a list of a sequence of numbers, for some reason this small program just sits there consuming CPU power... no idea why it's not working:

number = 5
lst = [4,5,6]
if all(x > number for x in lst):
    lst.insert(0,number)
elif all(x < number for x in lst):
    lst.append(number)
else:
    for i,v in enumerate(lst):
        if v>number:
            lst.insert(i-1,number)
print (lst)

expected output:

lst = [4,5,5,6]

You shouldn't change a list while you are iterating over it. Your code is effectively never moving forward because you are constantly inserting a new number and v remains the same each time. It is better to create a new list. — AChampion
– AChampion, Commented May 8, 2017 at 15:47
Just to be sure, you're trying to add a number to a list and keep it sorted? — EndermanAPM
– EndermanAPM, Commented May 8, 2017 at 15:49
You can also have a look at bisect library: docs.python.org/2/library/bisect.html — Don
– Don, Commented May 8, 2017 at 15:49
@Don That would not be a good idea for big lists, as the time complexity would be O(n). If it's already sorted, a binary search can be done in O(log n) time. — Jacob G.
– Jacob G., Commented May 8, 2017 at 15:50
Although it might not be the most efficient way of doing it you could just add the number an sort it. ej: list = [1,3,2] then add whatever you want list.append(2) and sort list.sort() this should output [1,2,2,3] — EndermanAPM
– EndermanAPM, Commented May 8, 2017 at 15:53

Alan Leuthard · Accepted Answer · 2017-05-08 16:11:48Z

2

Your for loop is inserting the number 5 into the middle of the list a theoretically infinite amount of times (or until you run out of whatever limited resource the list consumes, whichever happens first).

1) for i,v in enumerate(lst):
2)    if v>number:
3)       lst.insert(i-1,number)

On the first pass, line 1 starts the loop with v = 4 and i = 0. Line 2 finds v is not greater than number.

On the second pass, line 1 continues the loop with v = 5 and i = 1. Line 2 is also false.

Third pass, line 1: v = 6, i = 2. Line 2 finds a true statement and moves to line 3. Line 3 inserts the object referenced by number into position i - 1, inserting 5 into position 1 of the list.

At this point the list is:

lst = [4, *5*, **5**, 6]

The italicized 5 is the number you added to the list. The bolded 5 is where the current pointer is, i = 2. Notice that the 6 we just checked got moved forward with the insert.

Fourth pass: v = 6, i = 3. Line 2 finds a true statement and moves to line 3. Line 3 inserts the object referenced by number into position i - 1, inserting 5 into position 2 of the list.

At this point the list is:

lst = [4, 5, *5*, **5**, 6]

etc etc etc.

A quick fix:

for i, v in enumerate(lst):
    if v > number:
        lst.insert(i-1, number)
        **break**

You're just checking for and adding a single number, so break out of the loop once you insert it, since you're done.

edited May 8, 2017 at 16:11

answered May 8, 2017 at 16:01

Alan Leuthard

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2 Comments

Stefan Pochmann Over a year ago

Won't run out of stack, list contents aren't stored there.

Alan Leuthard Over a year ago

Had a few doubts about calling it stack. Since I don't know the internal workings of Python objects and their limits just yet, I changed it to be more general.

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