How can I; if var is integer then execute

Question

I am trying to create a decimal to binary converter. The user inputs their value, and the amount is divided by two each time and added to the invertedbinary list. The amount is then converted back into an integer, to be divided by two again and so on.

value = int(input("Please enter the decimal value to be converted to binary."))
    invertedbinary = []
    while value >= 1:
        value = (value/2)
        invertedbinary.append(value)
        value = int(value)
        print (invertedbinary)
    for n,i in enumerate(invertedbinary):
        if i == isinstance(invertedbinary,int):
            invertedbinary[n]=0
        else:
            invertedbinary[n]=1
    print (invertedbinary)

Let's say I input the number seventeen. This is the output:

[8.5]
[8.5, 4.0]
[8.5, 4.0, 2.0]
[8.5, 4.0, 2.0, 1.0]
[8.5, 4.0, 2.0, 1.0, 0.5]
[1, 1, 1, 1, 1]

So we can tell that from the last line of ones, my isinstance attempt did not work. What I want to be able to do, is that if the amount is anynumber.5 then to display it as a 1, and if it is a whole number to display as a zero. So what it should look like is [1, 0, 0, 0, 1]. Ones for each float value, and zeroes for the integers.

What can I use instead of is instance to achieve this?

For anyone wondering, I've called it invertedbinary because when printed invertedbinary needs to be flipped around and then printed as a string to display the correct binary value.

I changed this to [tag:python-3.x] as this requires `int / int` gives `float` behaviour; [see this question](https://stackoverflow.com/questions/19507808/python3-integer-division). — Ken Y-N, Oct 23 '16 at 23:36
You don't want `i == isinstance(invertedBinary,int)` you just want `isinstance(invertedBinary,int)` I'm pretty sure... — juanpa.arrivillaga, Oct 23 '16 at 23:55
Although, not actually in this case, because you care if the number is a whole number, not an `int` type necessarily. — juanpa.arrivillaga, Oct 23 '16 at 23:57
Anyway, in this case you should be using the modulus operator: `%` — juanpa.arrivillaga, Oct 24 '16 at 00:09

score 1 · Accepted Answer · answered Oct 23 '16 at 23:28

1

You can always check wether the round value is equal to the value...

if (round(x) == x):
    # x is int
else:
    # x is float/double

answered Oct 23 '16 at 23:28

Yotam Salmon

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Wow! Thanks so much for that. It's now working as intended! – August Williams Oct 23 '16 at 23:31
@AugustWilliams Sometimes, old and obvious works best ;-) – Yotam Salmon Oct 23 '16 at 23:33
@KenY-N What's the problem with 8.49999999999999999? – Yotam Salmon Oct 23 '16 at 23:34
I am fairly new to Python so I don't understand the example you have provided @KenY-N . However, in this specific program the floats will always be .5. – August Williams Oct 23 '16 at 23:36
@KenY-N But that wasn't the condition. `round(8.4999999) == 8.499999` is `False`, so it won't be treated as an int value – Yotam Salmon Oct 23 '16 at 23:38
@AugustWilliams Please [see this question](https://stackoverflow.com/questions/26091689/python-floating-point-number-comparison) for starters. – Ken Y-N Oct 23 '16 at 23:38
@YotamSalmon Ack! Too little coffee this morning. Let me delete myself being an idiot... – Ken Y-N Oct 23 '16 at 23:40

How can I; if var is integer then execute

1 Answers1