Checking number in a list to see if they are beside each other

Checking number in a list to see if they are beside each other - Printable Version

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Checking number in a list to see if they are beside each other - Allaye - Aug-25-2019

List1 =[2,3,4,9,9,5,1]
def checkList(List1):
    for i in range(len(List1 - 1)):
      
if List1[i] == 9 and List1[i+1] == 9:
            return True
          return false

i want to check if the number 9 comes b4 any other number in the list but my problem is the code does not return true or false

RE: Checking number in a list to see if they are beside each other - Yoriz - Aug-25-2019

List1 =[2,3,4,9,9,5,1]

lowercase for variable names.

for i in range(len(List1 - 1)):

-1 should be taken from the result of calling len(List1)
Don't use range to loop over items, loop over them directly.

Indentation is wrong

return false

Should be capital F False

A better way

function check_list that takes in list as it parameter

store the first item of the list in a variable by indexing the list.
loop through the items of the list starting at the second item by slicing the list.
- check if the looped item equals 9 and the previous item equals 9.
  - return True.
- set the previous item to the looped item.
return False

RE: Checking number in a list to see if they are beside each other - Allaye - Aug-25-2019

number = [2, 3, 4, 9, 8, 5, 1]

def checkList(List1):
    for i in range(len(List1) - 1):  #dont understand what you want me to do here 
        if List1[i] == 9 and List1[i + 1] == 9:
            return True
    return False

checkList(number)

i dont get any error when i run the code but i also dont get any output

RE: Checking number in a list to see if they are beside each other - Yoriz - Aug-25-2019

You don't print the result and the list does't have any 9 next to each other so it will return False.

See Never use "for i in range(len(sequence)):"

RE: Checking number in a list to see if they are beside each other - Malt - Aug-26-2019

(Aug-25-2019, 10:13 PM)Allaye Wrote:
number = [2, 3, 4, 9, 8, 5, 1] def checkList(List1): for i in range(len(List1) - 1): #dont understand what you want me to do here if List1[i] == 9 and List1[i + 1] == 9: return True return False checkList(number)
i dont get any error when i run the code but i also dont get any output

You've to print the output instead if returning it, to see the output

RE: Checking number in a list to see if they are beside each other - perfringo - Aug-26-2019

I suggest to follow naming conventions set in PEP8 - Function and Variable Names.

Alternative approach could be using zip, set method .difference and built-in function any. This way in process of writing code one can learn much more of Python Smile

.

zip is for creating pairs of consecutive numbers without need to worry about IndexError:

>>> nums = [1, 9, 9, 2]
>>> list(zip(nums, nums[1:]))
[(1, 9), (9, 9), (9, 2)]

set method .difference can be used for determining whether elements in pair are equal and match desired value:

>>> set((1, 9)).difference((9,))
{1}
>>> set((9, 9)).difference((9,))
set()

In Python empty means False, so to get True if there is empty set one will use 'not'.

any() is built-in function with short circuiting behaviour.

Therefore function could be written (keyword argument is for cases when we need check numbers other than 9):

>>> def check_list(nums, sentinel=9):
...     return any(not set(pair).difference((sentinel,)) for pair in zip(nums, nums[1:]))
...
>>> check_list([1, 2, 3])
False
>>> check_list([1, 9, 9])
True
>>> check_list([1, 2, 2], sentinel=2)
True

One can use built-in help in interactive interpreter to learn about functions and methods used above (help(zip), help(set.difference), help(any)