if statements, comparing lists, python

Question

what am i doing wrong with my if statement that it doesn't recognise if an element in a is equal to 0? what i am attempting to print is for ever 0 the program prints . and for ever 1 #. cheers.

a=[0,0,1,0,1,1,0,1,1,0,0,0,0,1]
print(a)
for i in range(len(a)):
    if a[i]==[0]:
        print('.', end='')
    else:
        print('#', end='')
print()

bash:

[0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1]
##############

Sven Marnach · Accepted Answer · 2012-07-06 13:13:50Z

3

You probably want

if a[i] == 0:

instead of

if a[i] == [0]:

You want to compare the items to the integer value 0, not to the single-element list [0].

answered Jul 6, 2012 at 13:13

Sven Marnach

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3 Comments

JensD Over a year ago

oh yes, thankyou. before when i kept on testing i kept getting not iterable. Im new to programing so its hard to remember what i trying before.

JensD Over a year ago

this raises a rhetorical question. why does a[1] return an integer while a[1:2] return a list. i suppose it doesnt matter why, but its interesting to note and the reason for my confusion. cheers

Sven Marnach Over a year ago

@JensD: A slicing a[i:j] returns a sublist of length j - i. If this difference is one, you get a sublist of length 1. It would be inconsistent if you would get the item itself.

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