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comparing 2 dimensional list - glennford49 - Mar-23-2020

i have this code:

    num1=[[1,2,3],[1,2,4],[1,2,5]]
    num2=[[1,2,3]]
    for i in num1:
        if (num2 == i):
    	    print("same")
    	else:
    	   print("unique")




whats wrong with my code?
num1 has 3 elements, num2 has single element with same value element of num1,
it should print "same" on output since num2 has a the same value element on num1

RE: comparing 2 dimensional list - Larz60+ - Mar-23-2020

first you didn't define i
you don't understand indexing. See: https://docs.python.org/3/tutorial/datastructures.html

RE: comparing 2 dimensional list - glennford49 - Mar-24-2020

Im sorry im a noob in coding, i just cant figure it out , so that it will print "unique" in the output

RE: comparing 2 dimensional list - Larz60+ - Mar-24-2020

you can use:
if [1,2,3] in num1:
    print("it is")
else:
    print("Not found")


RE: comparing 2 dimensional list - perfringo - Mar-24-2020

On row #5 You compare item (list) in first matrice (list of lists) with whole second matrice. They can’t be equal. You could use indexing to refer to item in second matrice (num2[0]) in comparison.

RE: comparing 2 dimensional list - glennford49 - Mar-24-2020

Thanks for your response,maybe comparing those 2 list is of no solution, in my code ,num2 data on it is changing, thats why in my mind i have to iterate num1 list of list , and compare num2 if there's the same value on num1

RE: comparing 2 dimensional list - perfringo - Mar-24-2020

I probably don't understand the problem, but anyway:

>>> num1=[[1,2,3],[1,2,4],[1,2,5]]
>>> num2=[[1,2,3]]
>>> for i, item in enumerate(num1, start=1):
...     if item == num2[0]:
...         print(f'item no {i} is equal')
...     else:
...         print(f'item no {i} is not equal')
... 
item no 1 is equal
item no 2 is not equal
item no 3 is not equal


RE: comparing 2 dimensional list - glennford49 - Mar-24-2020

(Mar-24-2020, 12:13 PM)perfringo Wrote: I probably don't understand the problem, but anyway:

>>> num1=[[1,2,3],[1,2,4],[1,2,5]]
>>> num2=[[1,2,3]]
>>> for i, item in enumerate(num1, start=1):
...     if item == num2[0]:
...         print(f'item no {i} is equal')
...     else:
...         print(f'item no {i} is not equal')
... 
item no 1 is equal
item no 2 is not equal
item no 3 is not equal
thanks, this saves my day!

RE: comparing 2 dimensional list - DeaD_EyE - Mar-24-2020

num2 should be a list and not a list in a list.

Your original code with the change:
num1 = [[1,2,3], [1,2,4], [1,2,5]]
num2 = [1,2,3]
for i in num1:
    if (num2 == i):
        print("same")
    else:
        print("unique")
Output:
same
unique
unique


RE: comparing 2 dimensional list - glennford49 - Mar-24-2020

(Mar-24-2020, 01:29 PM)DeaD_EyE Wrote: num2 should be a list and not a list in a list.

Your original code with the change:
num1 = [[1,2,3], [1,2,4], [1,2,5]]
num2 = [1,2,3]
for i in num1:
    if (num2 == i):
        print("same")
    else:
        print("unique")
Output:
same
unique
unique
Nothing has change i think