Python while and if problems

Question

This is suposed to check if a number is even or odd by dividing it by 2 but it is always looking at the same number

 a= [1,2,5,7,8];
    i=0
    c=a[i]/2.0;
    d=0;
    b=len(a);
    while i < b:
            if (c).is_integer():
                        i=i+1;
                        d=d +1;         
            else:
                        i=i +1;
    print(d);

The answer given to me is 0 but if the first number on the list even the answer given is 5.

Apparently it is always detecting the number a not integer, so is there any way to check if the number has or not decimals?

You divide an integer with a float so the result is a float. You also do not need semicolons `;` in Python. It would also be good to give an example case of your code running. — Vasilis G., Jan 14 '18 at 16:28
I'd recommend you use e.g. http://pythontutor.com/ to investigate the behaviour. This is not useful to other SO users, as the code appears to have no point nor any specific problem. — jonrsharpe, Jan 14 '18 at 16:29
Are you expecting `c` to change as `i` changes over time? It doesn't. — chepner, Jan 14 '18 at 16:37
The whole loop could be replaced with `d = sum(i%2 == 1 for i in a)`. — chepner, Jan 14 '18 at 16:38

Justin Minsk · Answer 1 · 2018-01-15T20:10:39.450

You are not moving through each instance of c, so currently you are just looking at the first element over and over.

a= [1,2,5,7,8]
i=0
d=0
b=len(a);
while i < b:
    c=a[i]/2.0
        if (c).is_integer():
                i=i+1
                d=d +1         
        else:
                i=i +1
print(d)

You also do not need the ; after each line that is more of a java thing. You could also look into for loops.

a = [1, 2, 5, 7, 8]
for i in a:
    if (c).is_integer():
            i=i+1
            d=d +1
print(d)

This uses less variables and is a better format. You should look up a style guide for python that might help you a lot. Here is a link to one python style guide.

wlhe · Answer 2 · 2018-01-14T16:36:24.383

0

I guess you want to iterate the list a, then you need to put c=a[i]/2.0; into while loop

a = [1, 2, 5, 7, 8]
i = 0
d = 0
b = len(a)
while i < b:
    c = a[i] % 2
    if (c == 0):
        i = i + 1
        d = d + 1
    else:
        i = i + 1
print(d)

edited Jan 14 '18 at 16:36

answered Jan 14 '18 at 16:33

wlhe

61
3

1

@PatrickArtner `is_integer()` checks if a floating point value is equivalent to an integer; it's not checking Python types. – chepner Jan 14 '18 at 16:36
@chepner my bad, thanks for pointing it out – Patrick Artner Jan 14 '18 at 16:37
yes, i noticed, so i think he want to do if(even_number) @chepner – wlhe Jan 14 '18 at 16:39

Python while and if problems

2 Answers2