This code produces one number after the other as it converts decimal to binary.
dec = int(input("Please enter number to convert to decimal: "))
while dec>0:
quoteint = dec/2
rem = dec%2
print (int(rem), end = " ")
dec = int(dec/2)
The output is the following.
1 0 0 1 0
Is there a way to reverse the output order?
Eg. 0 1 0 0 1
It may work if the individual numbers generated are joined to a string and the string is reversed, but I don't know how to go about doing that.
I think the following code will work.
result = []
dec = int(input("Please enter number to convert to decimal: "))
while dec>0:
quoteint = dec/2
rem = dec%2
result.append(rem)
dec = int(dec/2)
result = result[::-1]
print(' '.join(str(r) for r in result))
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))
Code is supposed to sum all numbers given by user input until only a single digit number remains.
Input 19991229 should result in 6, but I'm getting 33 instead, as if the two remaining numbers
aren't summed.
Here's the code:
b_d = input("enter birthdate")
digits = []
for d in b_d:
digits.append(int(d))
print(digits)
life = 0
while (sum(digits)) >= 10 :
for x in digits:
life += x
digits.remove(x)
print(life)
this resolves the issue, but it's too ad hoc and doesn't work on all cases:
r = 0
for y in str(life):
r += int(y)
print(y)
I am currently writing a Python 3 program to convert decimal to binary among other things for a Uni assignment.
I've nailed everything except for this in the first stage (decimal to binary).
dec = int(input("Enter a number: "))
while dec > 0 or dec == 0:
if dec > 0:
rem = dec % 2
dec = dec // 2
print(rem, end = "")
The output gives the binary number correctly, however it is in reverse.
Can you please tell me how to reverse the output or reverse the conversion process or something to correct the output?
EDIT: I cannot use in-built functions such as bin(dec), etc.
Thanks!
The above code is not decimal to binary, instead it is an example of dividend/reminder. You can do that as follows:
dec, rem = divmod(dec, 2)
If you still want to convert decimal to binary do -
bin(dec)
Based on the comment, would this help?
def dec2bin(d):
s = ''
while d>0:
d,r = divmod(d, 2)
s += str(r)
return s[::-1]
>>> dec2bin(6)
'110'
python program to convert the given binary to decimal, octal and hexadecimal number and vice versa.
conversions of all bases with each other.
x = int(input("press 1 for dec to oct,bin,hex \n press 2 for bin to dec,hex,oct \n press 3 for oct to bin,hex,dec \n press 4 for hex to bin,dec,oct \n"))
if x is 1:
decimal =int(input('Enter the decimal number: '))
print(bin(decimal),"in binary.")
print(oct(decimal),"in octal.")
print(hex(decimal),"in hexadecimal.")
if x is 2:
binary = input("Enter number in Binary Format: ");
decimal = int(binary, 2);
print(binary,"in Decimal =",decimal);
print(binary,"in Hexadecimal =",hex(decimal));
print(binary,"in octal =",oct(decimal));
if x is 3:
octal = input("Enter number in Octal Format: ");
decimal = int(octal, 8);
print(octal,"in Decimal =",decimal);
print(octal,"in Hexadecimal =",hex(decimal));
print(octal,"in Binary =",bin(decimal));
if x is 4:
hex = input("Enter number in hexa-decimal Format: ");
decimal = int(hex, 16);
print(hex,"in Decimal =",decimal);
print(hex,"in octal =",oct(decimal));
print(hex,"in Binary =",bin(decimal));
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
option explicit
dim r, res, num
num= cint(inputbox("Enter the number"))
do while(num > 0)
r= num mod 10
num= num\10
res= res & r
loop
msgbox res
Well this is the code, now my question is this works perfectly fine for input 1234, well if the input is 0123 it just prints 321 which is wrong.It needs to print 3210.
I am unable to figure out, tried a lot but in vain, any help would be appreciated
Thanks and Regards
You must decide whether you want to reverse strings or numbers (accidentially represented as decimals). If you want to reverse strings, you should
not convert the (string) input to a number/integer
use string ops: Mid() for reading, concatenation & for building
Added: In (demo/not production) code:
Option Explicit
Function rev(s)
Dim p
For p = Len(s) To 1 Step -1
rev = rev & Mid(s, p, 1)
Next
End Function
Dim s
For Each s In Array("1234", "0123")
WScript.Echo s, rev(s)
Next
output:
1234 4321
0123 3210
str = Inputbox("Enter the number")
rev=""
Set regx = New RegExp
regx.Global = True
regx.IgnoreCase = True
regx.Pattern = ".{1}"
Set colchars= regx.Execute(str)
For i = 0 To colchars.Count-1
rev= colchars.Item(i)&rev
Next
MsgBox rev
String reverse program without using Reverse String function & Mid function.
str=inputbox("Enter the string: ")
str1=len(str)
a=Left(str,1)
for i=1 to str1
str2=Left(str,i)
if len(str2)>1 then
str3=Right(str2,1)&temp
temp=str3
end if
next
msgbox temp&a
Try this:
Dim num, rev
num = inputbox("Enter a number")
If Len(num)=4 Then
rev = rev*10 + num mod 10
num = num/10
num = left(num,3)
rev = rev*10 + num mod 10
num = num/10
num = left(num,2)
rev = rev*10 + num mod 10
num = num/10
num = left(num,1)
rev = rev*10 + num mod 10
msgbox "Reverse Order of the number is "&rev
Else
msgbox "Number, you entered is not a 4 digit number"
End If