How to check all the integers in the list are continuous?

Question

I am trying to write a program which will print “YES” if all the numbers in the list are continuous and should return “NO” if the numbers are not continuous. By continuous I mean every number in the list should be greater than one from the previous element.

For example:

It should print “YES” for inputs: [3, 4, 5], [7, 8, 9], [1, 2, 3], [0, 1, 2, 3, 4, 5].. etc
It should print “NO” for inputs: [9, 1, 0], [3, 2, 4], [5, 5], [9, 8, 2, 3, 7].. etc

I used enumerate for this purpose.

Here’s my code:

    inp=[1,2,3,4,5]
    flag=0
    for index,e in enumerate(inp): 
        if index!=len(inp)-1:    
            if inp[index+1]==inp[index]+1:
                flag=1
    if flag==1:
        print ("YES")
    else:
        print ("NO")

The code works fine but i find it redundant.
Is there a better way to do it using enumerate or without using enumerate?

Asked By: Keegs

||

Source

Answer 1

You don’t need enumerate in order to check the elements of your lists are continuous. You can simply achieve it via creating a function using zip and all as:

def check_continuity(my_list):
    return all(a+1==b for a, b in zip(my_list, my_list[1:]))

Same result can be achieved by any with zip as (similar to all but with not and != for comparision):

def check_continuity(my_list):
    return not any(a+1!=b for a, b in zip(my_list, my_list[1:]))

Above functions will return True/False depending upon whether your list is continuous or not.

Sample run:

# Continuous Lists
>>> check_continuity([3, 4, 5])
True
>>> check_continuity([7, 8, 9])
True
>>> check_continuity([1, 2, 3])
True

# Non Continuous Lists
>>> check_continuity([9, 1, 0])
False
>>> check_continuity([3, 2, 4])
False
>>> check_continuity([5, 5])
False

In order to print “YES”/”NO”, you may make a simple if..elsecheck outside the function call as:

>>> "YES" if check_continuity([1, 2, 3]) else "NO"
'YES'

# OR update the return statement in your function to
#    return "NO" if any(a+1!=b for a, b in zip(my_list, my_list[1:])) else "YES"

Answered By: Moinuddin Quadri

Answer 2

I don’t see why you need enumerate at all, given that you never use e. Try using the for...else block:

inp = [1, 2, 3]

for idx in range(len(inp) - 1):
    if inp[idx + 1] - inp[idx] != 1:
        print("NO")
        break
else:
    print("YES")

You can also zip the list with its copy shifted by one (but that, well, makes a copy):

for x, y in zip(inp, inp[1:]):
    if y - x != 1:
        print("NO")
        break
else:
    print("YES")

Or make this into a function:

def increasing_by_one(inp):
    for x, y in zip(inp, inp[1:]):
        if y - x != 1:
            return False
    return True

print('YES' if increasing_by_one(inp) else 'NO')

Answered By: Norrius

Answer 3

a1,a2,a3..an are continuous when (a1+a2+..+an)/n equals (a1+an)/2.
I don’t know python but I am sure you can easily get first,last and count of a list.

update,

sorry, the original answer is a necessary condition, not a sufficient condition. so that’s not right.

Answered By: orzorc

Answer 4

You can easily use this line of code:

l = [1, 2, 3]

True if(list(range(min(l), max(l)+1)) == l) else False

In this sample, the output will be True.

Answered By: farrokhkarimi

How to check all the integers in the list are continuous?

Question:

Answers: