Is there any module or function in python I can use to convert a decimal number to its binary equivalent? I am able to convert binary to decimal using int('[binary_value]',2), so any way to do the More on stackoverflow.com

So this is a pretty complicated topic for a beginner called recursion. I'll try my best to explain what's going on. The last line print(n % 2, end="") simply prints the remainder after n is divided by 2. This will be either 1 or 0 since 2 goes into all larger numbers atleast once. By default python puts a new line character after every print and the 'end=""' simply says 'print nothing at the end' instead of the normal 'print \n at the end' (\n indicates a newline) Now for the fun part. Let's step through what happens with n=10. In binary we know this is represented by '1010'. When n is 10 we see it is greater than 1, so we execute line 3 (dec2bin(n//2)). This basically pauses what we're currently doing for now and starts over with a different value. You can think of this like a stack of boxes. Originally we just have our one box, when we call dec2bin again we put another box on top. We can't keep working on the bottom box anymore because it's covered. We need to finish the top box so we can remove it and get to the bottom box again. I should mention that // means integer division. 3//2 is 1 because 2 goes into 3 one whole time. So in our stack of boxes (aka our call stack) we will have: dec2bin(5) dec2bin(10) So let's work through what happens when we call dec2bin(5). It's important to note that 5 is '101' in binary which is the same as 10 ('1010') except the last 0 disappears because of the division. 5 is greater than 1 so we execute line 3 and call dec2bin again! Gotta put another box on top of our call stack: dec2bin(2) dec2bin(5) dec2bin(10) Note that 2 in binary is '10', which is the same as ten in binary except drop the last two digits because we did division twice. Working through dec2bin(2) we get another call to dec2bin: dec2bin(1) dec2bin(2) dec2bin(5) dec2bin(10) Printed: nothing Now this is where it gets interesting. Working through dec2bin(1) we have n=1. In this case n is NOT greater than 1 (line 2). So we continue on and print 1%2 which is 1. Now the function finishes and we're done with this call of dec2bin(1) so we can remove it from the call stack: dec2bin(2) dec2bin(5) dec2bin(10) Printed: '1' Now dec2bin(2) we paused on line 3, but line 3 just finished up so we can continue onto line 4 and print 2%2 which is 0. We've printed '10' so far. Again the function finishes up and we remove it from the call stack. dec2bin(5) dec2bin(10) Printed: '10' dec2bin(5) was paused on line 3, it continues now. We print 5%2 which is 1. The function finishes. It gets removed from the call stack. dec2bin(10) Printed: '101' We've finished up most of the pile back down to our original call. It (finally) runs its line 4 and prints 10%2, which is 0. We've printed '1010'. The function finishes up and we're all done. Printed: '1010' I hope that I haven't just confused you further, recursion is a tough topic to understand especially if not explained in a way that makes sense to you. More on reddit.com

I have a list of decimal numbers as integers, and I want to convert them into binary. The output must be without 0b prefix and also it must be 4 digits like: lista = [1,2,3,4,5,6,7,8,9,10] listres... More on stackoverflow.com

Treat them as true and false or multiply it by its appropriate counterpart. Ex. [1,0,0,1] => 1×23 + 0×22 + 0×21 + 1×20 More on reddit.com