how to convert decimal to binary by using repeated division in python?
i know i have to use a while loop, and use modulus sign and others {%} and {//} to do this...but i need some kind of example for me to understand how its done so i can understand completely.
CORRECT ME, if I'm wrong:
number = int(input("Enter a numberto convert into binary: "))
result = ""
while number != 0:
remainder = number % 2 # gives the exact remainder
times = number // 2
result = str(remainder) + result
print("The binary representation is", result)
break
Thank You
Making a "break" without any condition, makes the loop useless, so the code only executes once no matter what.
-
If you don't need to keep the original number, you can change "number" as you go.
If you do need to keep the original number, you can make a different variable like "times".
You seem to have mixed these two scenarios together.
-
If you want to print all the steps, the print will be inside the loop so it prints multiple times.
If you only want to print the final result, then the print goes outside the loop.
while number != 0:
remainder = number % 2 # gives the exact remainder
number = number // 2
result = str(remainder) + result
print("The binary representation is", result)
-
The concatenation line:
Putting the print inside the loop might help you see how it works.
we can make an example:
the value in result might be "11010" (a string, with quotes)
the value in remainder might be 0 (an integer, no quotes)
str(remainder) turns the remainder into a string = "0" instead of 0
So when we look at the assignment statement:
result = str(remainder) + result
The right side of the assignment operator = is evaulated first.
The right side of the = is
str(remainder) + result
which, as we went over above has the values:
"0" + "11010"
This is string concatenation. It just puts one string on the end of the other one. The result is:
"0 11010"
"011010"
That is the value evaluated on the right side of the assignment statement.
result = "011010"
Now that is the value of result.
B_Number = 0
cnt = 0
while (N != 0):
rem = N % 2
c = pow(10, cnt)
B_Number += rem * c
N //= 2
# Count used to store exponent value
cnt += 1
return B_Number
Related
Given a number N, the function should convert the number to binary form, count the number of consecutive zero (the binary gap), and return the maximum binary gap. For example, 9 = 1001, the binary gap of length 2. The number 529 = 1000010001, has 2 binary gaps with length 4 and 3. If the number has 2 or more binary gaps, the function should return the maximum binary gap i.e. 4 in the case of N = 529.
I tried this function:
def solution(N):
binaryN = bin(N)[2:]
n = len(binaryN)
binaryGap = []
for i in range(n):
if binaryN[i] == 0 and binaryN[i + 1] == 0:
m = len(binaryN)
else:
return 0
binaryGap = binaryGap.append(m)
return max(binaryGap)
The function returns 0 for all values of N which is incorrect. How do I debug/improve the code to produce the accurate result?
Check out the below code. It would solve your problem.
The code is self-explanatory, yet let me know in-case of any doubts.
The Code:
import sys
num = int(sys.argv[1])
# Function to get the binary gap.
def binaryGapFinder(num):
binnum = bin(num).replace("0b", "") # binnum is binary form of the given number.
i = 0
x = 0
x_list = []
while i <= len(binnum)-1:
if binnum[i] == "0":
x += 1
if i == len(binnum)-1: # This loop will also consider if binary form is ending with 0. for example: 12 -> 1100
x_list.append(x)
else:
x_list.append(x)
x = 0
i += 1
return f"The Number: {num}\nIt's Binary Form: {binnum}\nMaximum Consecutive 0's: {max(x_list)}"
print(binaryGapFinder(num))
The Output:
python3 /the/path/to/your/script/binarygap.py 529
The Number: 529
It's Binary Form: 1000010001
Maximum Consecutive 0's: 4
python3 /the/path/to/your/script/binarygap.py 12
The Number: 12
It's Binary Form: 1100
Maximum Consecutive 0's: 2
python3 /the/path/to/your/script/binarygap.py 512
The Number: 512
It's Binary Form: 1000000000
Maximum Consecutive 0's: 9
There's a few issues here worth mentioning to aid you. (Just a side note to start with is that, in Python, it's recommended/best practice to use all lower case for variable names, so I'll replace them in my examples below.)
The bin() built in function returns a string. So you should be checking for equality with "0" (or '0') instead of an integer. e.g.
if binaryN[i] == "0" and binaryN[i + 1] == "0":
With Python you don't need to bother with checking for lengths of strings (or any other iterables) to use in a for loop in scenarios like this. e.g. You can replace:
n = len(binaryN)
for i in range(n):
with the more "Pythonic" way:
for bit in binary_number:
You can then use the variable bit (call it whatever you want of course, bearing in mind that good variable names make code more readable) instead of binary_number[index]. In this case, with each iteration of the for loop, bit will be replaced with the next character in the binary_number string.
From there on in your code:
m = len(binaryN)
will always be the same value, which is the total length of the string binaryN. e.g. 4 for '1001'.) This is not what you intended.
The first statement in your else block of code return 0 will terminate your function immediately and return 0 and thus your binaryGap = binaryGap.append(m) code will never, ever execute as it's unreachable due to that preceding return stopping any further execution of code in that suite.
You've got the right idea(s) and heading towards the right track for a solution but I don't think your code, even when the issues above are corrected, will match all possible binary numbers you may encounter. So, another possible alternative (and yet roughly sticking with the solution I think that you had in mind yourself) would be something like this which I hope will help you:
def solution(n):
binary_no = bin(n)[2:]
binary_gaps = []
gap_counter = 0
for bit in binary_no:
if bit == "0":
gap_counter += 1
else:
# Encountered a 1 so add current count of 0's -- if any -- to list and reset gap_counter
if gap_counter > 0:
binary_gaps.append(gap_counter)
gap_counter = 0
else:
# A for else suite (block of code) is run when all iterables have been exhausted.
if gap_counter > 0:
binary_gaps.append(gap_counter)
if binary_gaps: # If there is at least one element in the list
if len(binary_gaps) > 1:
return max(binary_gaps)
else:
return binary_gaps[0]
else:
# The list is empty, so no gaps were found at all. i.e. Binary number was all 1's.
return 0
print(solution(529))
The outline:
Write a function named collatz() that has one parameter named number. If the number is even, then collatz() should print number // 2 and return this value. If the number is odd, then collatz() should print and return 3 * number + 1. Then write a program that lets the user type in an integer and that keeps calling collatz() on that number until the function returns the value 1.
my code:
def collatz(number):
if number % 2 == 0:
print(number // 2)
return number // 2
elif number % 2 == 1:
result = 3 * number + 1
print(result)
return result
n = input("Give me a number")
while n != 1:
collatz(int(n))
the output keeps infinitely printing the n value, and I can't figure out why. Please enlighten me
Your loop should update the value of n:
while n != 1:
n = collatz(int(n))
Otherwise, your loop has no progression, hence the infinite loop.
Edit: Here's an explanation as to why this is necessary.
Think about it logically. If you enter the body of the loop, then n must not be equal to 1. The loop continues to execute until n becomes 1. The only way for this to happen is for you to change (i.e. update) n so that it eventually reaches 1 and the loop can terminate.
Hello I have to reorder a string, I am banned from using other types and str methods
So my problem is that I could not figure out how to end my code to get it work with any string
I tried to compare the results with sorted() to check and I am stuck at the first exchange
My code:
i = 0
s1 = "hello"
s2 = sorted(s1)
while (i<len(s1)):
j=i+1
while (j<=len(s1)-1):
if (s1[i] > s1[j]):
s1 = s1[0:i] + s1[j] + s1[i]
j+=1
i+=1
print(s1)
print(s2)
I tried to add + s1[len(s1):] at the end of the operation but
I only had found the result for a single string(that I was testing) adding thisI am really stuck, how can I make it work for all the strings with different lenghts??
Thanks
You're not reconstructing the string correctly when doing s1 = s1[0:i] + s1[j] + s1[i] as you're replacing one character for the other but you omit to actually interchange the two and to add the remains of the splitted string to the end of the new string.
Given what your code looks like, I would do it like this:
i = 0
s1 = "hello"
s2 = sorted(s1)
while i < len(s1):
j = i + 1
while j <= len(s1)-1:
if s1[i] > s1[j]:
s1 = s1[0:i] + s1[j] + s1[i+1:j] + s1[i] + s1[j+1:len(s1)]
j += 1
i += 1
print("".join(s2))
# > 'ehllo'
print(s1)
# > 'ehllo'
Please tell me if anything is unclear!
I am banned from using other types and str methods
Based upon your criteria, your request is impossible. Just accessing the elements of a string requires string methods.
The technique that you are using is very convoluted, hard to read and is difficult to debug. Try running your code in a debugger.
Now given that you are allowed to convert a string to a list (which requires string methods), redesign your code to use simple, easy to understand statements.
The following code first converts the string into a list. Then loops thru the list starting at the beginning and compares each following character to the end. If any character is less then the current character, swap. As you step thru the string, the character swaps will result in a sorted list. At the end convert the list back to a string using join().
msg = 'hello'
s = list(msg)
for i in range(len(s) - 1):
for j in range(i + 1, len(s)):
if s[i] <= s[j]:
continue
# swap characters
s[i], s[j] = s[j], s[i]
print(msg)
print(''.join(s))
I am trying to convert a string of varchar to ascii. Then i'm trying to make it so any number that's not 3 digits has a 0 in front of it. then i'm trying to add a 1 to the very beginning of the string and then i'm trying to make it a large number that I can apply math to it.
I've tried a lot of different coding techniques. The closest I've gotten is below:
s = 'Ak'
for c in s:
mgk = (''.join(str(ord(c)) for c in s))
num = [mgk]
var = 1
num.insert(0, var)
mgc = lambda num: int(''.join(str(i) for i in num))
num = mgc(num)
print(num)
With this code I get the output: 165107
It's almost doing exactly what I need to do but it's taking out the 0 from the ord(A) which is 65. I want it to be 165. everything else seems to be working great. I'm using '%03d'% to insert the 0.
How I want it to work is:
Get the ord() value from a string of numbers and letters.
if the ord() value is less than 100 (ex: A = 65, add a 0 to make it a 3 digit number)
take the ord() values and combine them into 1 number. 0 needs to stay in from of 65. then add a one to the list. so basically the output will look like:
1065107
I want to make sure I can take that number and apply math to it.
I have this code too:
s = 'Ak'
for c in s:
s = ord(c)
s = '%03d'%s
mgk = (''.join(str(s)))
s = [mgk]
var = 1
s.insert(0, var)
mgc = lambda s: int(''.join(str(i) for i in s))
s = mgc(s)
print(s)
but then it counts each letter as its own element and it will not combine them and I only want the one in front of the very first number.
When the number is converted to an integer, it
Is this what you want? I am kinda confused:
a = 'Ak'
result = '1' + ''.join(str(f'{ord(char):03d}') for char in a)
print(result) # 1065107
# to make it a number just do:
my_int = int(result)
I have tried to write a small program that converts a decimal to binary that doesn't use the inbuilt function that do that. My program won't convert anything over 12287. 12288 just spits out an infinite loop. Where have I gone wrong? why can't I get above 12287?
while (number != 1) or (number != 0):
a = number // 2
b = number % 2
number = a
if b == 0:
output = "0" + output
if number == 1:
output = "1" + output
break
else:
output = "1" + output
You just need to change the condition from:
while (number != 1) or (number != 0):
To:
while number != 0:
Other than that, your code looks OK. Here is a more streamlined version of the same basic idea:
output = ""
while number:
number, b = divmod(number, 2)
output = str(b) + output
For 12288, number eventually becomes 0. This means while condition is always True.
Note that b becomes 0, so number will stay at 0.