CODEWITHSUNDEEP
pythonpython-3.xif-statement
ami96
ami96

Reputation: 9

condition to check if the list include a string except the string that I want

I want to write a condition that check if the list has another string that is not '1' or '2'.
For example: 

list = ['1', '2'] -> is True
list = ['1', '3'] -> is false

I tried something like that but it didn't work:

if list[0] is not '1' or list[0] is not '2' and list[1] is not '1' or list[1] is not '2':
    return False
else:
    return True

Help someone??

Upvotes: 0

Views: 930

Answers (6)

LotusAlice
LotusAlice

Reputation: 135

For starters, welcome! :)

We can try the all() function to check all the elements in the list using if statements (making it a little easier to read):

For example:

list1 = ['1', '3']

if all(elem in ('1', '2') for elem in list1): #if all elements in the list have '1' or '2'
    True     #return true

else:        #if not
    False    #return false

Output:

False

Hope this is what you were looking for.

Upvotes: 0

Sarthak chauhan
Sarthak chauhan

Reputation: 36

Hellooooo!! First thing u can do is to make an iter object.

s=iter("12")

This will create an iter object with "1" and "2".You must be thinking why to use iter rather than a list. The answer lies in the size,memory of iter objects is less than that of list.

import sys
l=["1","2"]
s=iter("12")
sys.getsizeof(l)  #return 72
sys.getsizeof(s)  #return 48

So what i did was that,i got the memory size of l and s. If you want to know more about this function,then go to sys.getsizof() Now coming back to your question,check the code below.

for element in l:
    if element in s:
       print("69 is not just a number!!!")    #You can replace this part by anything u want

Output:

69 is not just a number!!!
69 is not just a number!!!

The benefit of this is that you can use this for much more strings!!!.And the memory will be very less.

You can also do it how Austin did it.

Keep Learning!

Upvotes: 0

Prathamesh
Prathamesh

Reputation: 1057

Try something like this,

l1=['1','2']
l2=['1','3']
def check(l1):
    if '1' in l1 and '2' in l1:
        print('True')
    else:
        print('False')

check(l1)
check(l2)

output:

True
False

Upvotes: -1

joshuaprakasam
joshuaprakasam

Reputation: 134

Don't use list as variable since it is a built-in function.

checking_elements = ['1', '2']

ls = ['1', '3']

for ele in ls:
    if ele not in checking_elements:
        return False

return True

if your ls is going to be only two elements.Then

checking_elements = ['1', '2']

ls = ['1', '3']

return (ls[0] in checking_elements and ls[1] in checking_elements)

Upvotes: 0

Zain Arshad
Zain Arshad

Reputation: 1907

You can use any() for this, simply pythonic:

def check_list(l):
    return not any(x not in ('1', '2') for x in l)

Upvotes: 1

Austin
Austin

Reputation: 26039

Use all in a generator:

def check_value(lst):
    return all(x in ('1', '2') for x in lst)

Usage:

>>> check_value(['1', '2'])
True
>>> check_value(['1', '3'])
False

Upvotes: 2

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