How do you convert base 12 to base 10 in Python 3 and the other way around too? (from 10 to 12)
I wrote this code to make a list of numbers in base 4, but the code never executes and eventually has a timeout error. Any ideas on how to fix it? It also identifies the mode and counts the frequency of the mode. The program should start at 1 and print the next 21 numbers in base 4 and then do the mode stuff.
while num > 0:
if start in yes:
list4.append(start)
start += 1
num -= 1
if start not in yes:
num += 1
new_4 = ''.join([str(n) for n in list4])
print(statistics.mode(new_4))
count_m = new_4.count(statistics.mode(new_4))
print(count_m)
Comment any other questions you may have!
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))
I don't know why it is not working there are no errors or warnings it was suppossed to multiply by 3 and add 1 if the number in odd and if the number is even it is supposed to divide by 2 until it reaches 0
here is my code any help will be appreciated
num = 4
while num != 0:
if num % 2 == 1:
num = (num * 3) + 1
else:
num /= 2
Your loop wil run indefinitely because num wil never be 0. I think you want num != 1. look at this wikipedia page https://en.wikipedia.org/wiki/Collatz_conjecture, And specificly the picture, you can see that 0 is never reached.
I am using a function to get the number from user, and I am trying to use a while loop to separate the digits of a number. And I am trying to add the digits of the number. But my code runs infinitely.
Example : 2345 -> 15
def sumDigits(n):
sum=0
while len(str(n))>0:
a = n%10
n = n//10
sum += a
return sum
print(sumDigits(2345))
Expected: 15
Actual: I had to shut down the jupyter kernel to stop the while loop.
Edit 2: Removed the updated code as it was answered by the community.
This condition len(str(n))>0 can never be false as long as n is an integer, because str(0) is '0', which has a length of 1.
You need to change the looping condition to exit where there is no more digit to sum, which happens when n reaches 0:
def sum_digits(n):
total = 0
while n > 0:
a = n % 10
n = n // 10
total += a
return total
print(sum_digits(2345))
Note: sum is a built-in in python, so naming a variable sum is not advised. Also, method names usually are written in snake_case, so sum_digits is recommended.
def all_sum(number):
total = 0
if number > 0:
for e in str(number):
if e.isdigit():
total += int(e)
else:
pass
return total
a = all_sum(567897)
This should do your work. Instead of doing two arithmetic operations to 'fetch' the digits, better to change the argument to string and just use each digit. Its faster and saves memory (though it's not too memory consuming).
I am trying to identify how many times values exceed the previous input. the following code works for 1,7,9,0 (two times 7, exceeds 1, 9 exceeds 7), but fails on 1,5,2,4,3,0. I get that this is because I set the maximum to 5 and of course 2 and 4 are less than 5. I cannot figure out what to do to "reset" maximum back to 1
a = int(input())
counter = 0
highest = 1
while a != 0:
if a > highest:
highest = a
counter += 1
a=int(input())
# need to reset highest to next input of 'a'
print(counter)
Grateful for your patience. Still an oldie seeing whether I am brain-dead yet. Sorry if this is a dumb question, I just don't see it. Also the course I am following intimates that I can't use anything but if, else, while, no lists or anything
first_num = int(input())
incremental_count = 0
while first_numv != 0:
second_num = int(input())
if second_num != 0 and first_num < second_num:
incremental_count += 1
first_num = second_num
print(incremental_count)
Once I solved it my way, this was the answer given which is cleverer and simpler!
I've come across a puzzling challenge. I have to check if a number contains the same digit multiple times ex. 11, 424, 66 and so on. at first this seems easy enough but i'm having trouble coming up with a logic to check for this. any ideas?
This is what I've got so far. the function takes in a list. (updated)
arr = [[1,20],[1,10]]
for i in arr:
l = list(range(i[0],i[1]))
for num in l:
if num < 11: continue
for c in str(num):
if str(num).count(c) > 1:
# dont know why code is popping off 12 and 13
print(l.pop(num))
If your ultimate goal is simply detecting if there's a double, this function may help:
def has_doubles(n):
return len(set(str(n))) < len(str(n))
The best way I can think about is converting the number to a string and doing a Counter on it
from collections import Counter
a = 98
c = Counter(str(a))
if any(value > 1 for value in c.values()):
print "The number has repeating digits"
#Two-BitAlchemist thanks for the suggestion
looks like you wanted to create your own algorithm probably researching or a student practice well you just have to understand the properties of numbers divided by 10 where 1/10 = 0.1 10/10 = 1 13/10 = 1 reminder 3 13013/10 = 1301 rem 3 hence we can create a function that stores the reminders in an array an check them against the reminder of next number here is the algorithm in python using recursion, you can achieve the same via loops
def countNumber(foundDigits,number):
next_number = int(number/10);
reminder = number % 10;
if(next_number < 1):
for num in foundDigits:
if(num == number or num == reminder):
return True
return False;
foundDigits.append(reminder);
return countNumber(foundDigits,next_number)
example in interpreter could be
digitsFound = list()
countNumber(digitsFound, 435229)
Solved this! I didn't know pop executes based on position not value! remove is a better fit here.
arr = [[1,40],[1,10]]
for i in arr:
l = list(range(i[0],i[1]))
for num in l:
if num < 11: continue
for char in str(num):
if str(num).count(char) < 2: continue
l.remove(num)
break
print(l)
Here is my solution, its simple and works for 2 digit numbers.
nums = list(input().rstrip().split())
def has_doubles(nums):
for number in nums:
if number[0] == number[1]:
print(number)
else:
continue
has_doubles(nums)