Develop Reference

Exponential-Golomb Decoder in Python (Codewars)

I am trying to pass one of the Codewars challenge ,,Exponential-Golomb Decoder" here: https://www.codewars.com/kata/5e4d8a53b499e20016b018a0/swift
The idea is to take input in binary format:"001000000101111001010001101" -> 00100 000010111 1 00101 0001101
and return list of decoded numbers:[3, 22, 0, 4, 12]
If You give the Sentence of for example 00100 00101, let's see that You start with 2 leading zeros, then hit 1, and so the first resulted number is number of zeros + 1 + number of zeros (00+1+00), let's say the Length of this number is 5, so I wanted to make new sentence like sentence = [length:] and iterate through it, but it is not going to iterate through new sentence, but the old one
Any Ideas? Big thanks
def reccurent():
sentence = input("Enter the number:")
nums = []
number = ""
for i in sentence:
i = int(i)
if i == 0:
number += str(0)
else:
number += str(1)
if len(number) == 1:
nums.append(number)
number = ""
sentence = sentence[i:]
else:
length = len(number) + len(number) - 1
word = str(sentence[:length])
nums.append(word)
number = ""
sentence = sentence[length:]
print(nums)
reccurent()
Result
Enter the number:0011000100
['00110', '1', '0100']
Process finished with exit code 0
Should be
['00110','00100']
Thanks

The following is a modification based on the provided code.
def reccurent():
sentence = input("Enter the number:")
nums = []
number = ""
# Define parameters to save the cursor offset position
offset = 0
for index, i in enumerate(sentence):
# Skip the value before the cursor
if index < offset:
continue
number += i
if i != "0":
if len(number) == 1:
nums.append(number)
offset = index + 1
else:
length = len(number) * 2 - 1
word = sentence[offset: offset + length]
nums.append(word)
offset += length
number = ""
print(nums)
reccurent()
Enter the number:0011000100
['00110', '00100']

Trying to use zfill and increment characters with Python

Hello lovely stackoverflowians!
I am fairly new to programming. Only have been programming a little under 2 months using CS50 which uses C and MITx Python. I went on Codewars and am trying to solve a problem where you basically get an id and then come out with a license plate number like this aaa001...aaa999, aab001...zzz999
if you catch my drift. For some reason my code compiles but when I run it I get this error.
File "/Users/pauliekennedy/Desktop/temp.py", line 9, in find_the_number_plate
a_str = (a_numb.zfill(3), range(0, 10))
AttributeError: 'int' object has no attribute 'zfill'
Because of this I am not able to test my code. If you could help me with this problem I would be much appreciated. As well, if you have anything to say about my code in general, tips, advice how to make it better, and if it will achieve this goal at all. Here is my code thanks again all.
#set number to 1 to start
a_numb = 1
#incrementing loop when 999 go back set back 0
while a_numb <1001:
a_numb += 1
a_str = str(a_numb)
# giving the number 00 or just 0 in front
if a_numb < 100:
a_str = (a_numb.zfill(3), range(0, 10))
#resetting the number back to 1
if a_numb == 999:
a_numb = 1
# Setting i to 0 and incrementing the characters
i = 0
ch = 'a'
ch2 = 'a'
ch3 = 'a'
#looping through alphabet
for i in range(26):
ch = chr(ord(ch) + 1)
print(ch)
if i == 26:
i = 0
if ch == 'z':
ch2 = chr(ord(ch) + 1)
if ch == 'z' & ch2 == 'z':
ch3(ord(ch) + 1)
# Adding results together and returning the string of them all
letter_plate = str(ch3 + ch2 + ch)
plate = str(a_numb) + str(letter_plate)
return plate```

Maybe you could consider using f-string string formatting instead:
def find_the_number_plate(customer_id):
number_part = customer_id % 999 + 1
customer_id //= 999
letter_part = ['a', 'a', 'a']
i = 0
while customer_id:
letter_part[i] = chr(ord('a') + customer_id % 26)
customer_id //= 26
i += 1
return f"{''.join(letter_part)}{number_part:03}"

You could use product from itertools to form the license plate numbers from 3 letters and numbers from 1 to 999 formatted with leading zeros:
from itertools import product
letter = "abcdefghijklmnopqrstuvwxyz"
numbers = (f"{n:03}" for n in range(1,1000))
plates = [*map("".join,product(letter,letter,letter,numbers))]
for plate in plates: print(plate)
aaa001
aaa002
aaa003
aaa004
aaa005
aaa006
aaa007
aaa008
...
If you only need to access a license place at a specific index, you don't have to generate the whole list. You can figure out which plate number will be at a given index by decomposing the index in chunks of 999,26,26,26 corresponding to the available option at each position/chunk of the number.
def plate(i):
letter = "abcdefghijklmnopqrstuvwxyz"
result = f"{i%999+1:03}"
i //= 999
for _ in range(3):
result = letter[i%26] + result
i //= 26
return result
output:
for i in range(10):print(plate(i))
aaa001
aaa002
aaa003
aaa004
aaa005
aaa006
aaa007
aaa008
aaa009
aaa010
plate(2021) # aac024

Convert a number to Excel’s base 26

OK, I'm stuck on something seemingly simple. I am trying to convert a number to base 26 (ie. 3 = C, 27 = AA, ect.). I am guessing my problem has to do with not having a 0 in the model? Not sure. But if you run the code, you will see that numbers 52, 104 and especially numbers around 676 are really weird. Can anyone give me a hint as to what I am not seeing? I will appreciate it. (just in case to avoid wasting your time, # is ascii char 64, A is ascii char 65)
def toBase26(x):
x = int(x)
if x == 0:
return '0'
if x < 0:
negative = True
x = abs(x)
else:
negative = False
def digit_value (val):
return str(chr(int(val)+64))
digits = 1
base26 = ""
while 26**digits < x:
digits += 1
while digits != 0:
remainder = x%(26**(digits-1))
base26 += digit_value((x-remainder)/(26**(digits-1)))
x = remainder
digits -= 1
if negative:
return '-'+base26
else:
return base26
import io
with io.open('numbers.txt','w') as f:
for i in range(1000):
f.write('{} is {}\n'.format(i,toBase26(i)))
So, I found a temporary workaround by making a couple of changes to my function (the 2 if statements in the while loop). My columns are limited to 500 anyways, and the following change to the function seems to do the trick up to x = 676, so I am satisfied. However if any of you find a general solution for any x (may be my code may help), would be pretty cool!
def toBase26(x):
x = int(x)
if x == 0:
return '0'
if x < 0:
negative = True
x = abs(x)
else:
negative = False
def digit_value (val):
return str(chr(int(val)+64))
digits = 1
base26 = ""
while 26**digits < x:
digits += 1
while digits != 0:
remainder = x%(26**(digits-1))
if remainder == 0:
remainder += 26**(digits-1)
if digits == 1:
remainder -= 1
base26 += digit_value((x-remainder)/(26**(digits-1)))
x = remainder
digits -= 1
if negative:
return '-'+base26
else:
return base26

The problem when converting to Excel’s “base 26” is that for Excel, a number ZZ is actually 26 * 26**1 + 26 * 26**0 = 702 while normal base 26 number systems would make a 1 * 26**2 + 1 * 26**1 + 0 * 26**0 = 702 (BBA) out of that. So we cannot use the usual ways here to convert these numbers.
Instead, we have to roll our own divmod_excel function:
def divmod_excel(n):
a, b = divmod(n, 26)
if b == 0:
return a - 1, b + 26
return a, b
With that, we can create a to_excel function:
import string
def to_excel(num):
chars = []
while num > 0:
num, d = divmod_excel(num)
chars.append(string.ascii_uppercase[d - 1])
return ''.join(reversed(chars))
For the other direction, this is a bit simpler
import string
from functools import reduce
def from_excel(chars):
return reduce(lambda r, x: r * 26 + x + 1, map(string.ascii_uppercase.index, chars), 0)
This set of functions does the right thing:
>>> to_excel(26)
'Z'
>>> to_excel(27)
'AA'
>>> to_excel(702)
'ZZ'
>>> to_excel(703)
'AAA'
>>> from_excel('Z')
26
>>> from_excel('AA')
27
>>> from_excel('ZZ')
702
>>> from_excel('AAA')
703
And we can actually confirm that they work correctly opposite of each other by simply checking whether we can chain them to reproduce the original number:
for i in range(100000):
if from_excel(to_excel(i)) != i:
print(i)
# (prints nothing)

Simplest way, if you do not want to do it yourself:
from openpyxl.utils import get_column_letter
proper_excel_column_letter = get_column_letter(5)
# will equal "E"

Sorry, I wrote this in Pascal and know no Python
function NumeralBase26Excel(numero: Integer): string;
var
algarismo: Integer;
begin
Result := '';
numero := numero - 1;
if numero >= 0 then
begin
algarismo := numero mod 26;
if numero < 26 then
Result := Chr(Ord('A') + algarismo)
else
Result := NumeralBase26Excel(numero div 26) + Chr(Ord('A') + algarismo);
end;
end;

You can do it in one line (with line continuations for easier reading). Written here in VBA:
Function sColumn(nColumn As Integer) As String
' Return Excel column letter for a given column number.
' 703 = 26^2 + 26^1 + 26^0
' 64 = Asc("A") - 1
sColumn = _
IIf(nColumn < 703, "", Chr(Int((Int((nColumn - 1) / 26) - 1) / 26) + 64)) & _
IIf(nColumn < 27, "", Chr( ((Int((nColumn - 1) / 26) - 1) Mod 26) + 1 + 64)) & _
Chr( ( (nColumn - 1) Mod 26) + 1 + 64)
End Function
Or you can do it in the the worksheet:
=if(<col num> < 703, "", char(floor((floor((<col num> - 1) / 26, 1) - 1) / 26, 1) + 64)) &
if(<col num> < 27, "", char(mod( floor((<col num> - 1) / 26, 1) - 1, 26) + 1 + 64)) &
char(mod( <col num> - 1 , 26) + 1 + 64)
I've also posted the inverse operation done similarly.

Based on #TheUltimateOptimist's answer, I looked in the openpyxl implementation and found the "actual" algorithm used by openpyxl==3.0.10:
Be warned; it only supports values between 1 & 18278 (inclusive).
def _get_column_letter(col_idx):
"""Convert a column number into a column letter (3 -> 'C')
Right shift the column col_idx by 26 to find column letters in reverse
order. These numbers are 1-based, and can be converted to ASCII
ordinals by adding 64.
"""
# these indicies corrospond to A -> ZZZ and include all allowed
# columns
if not 1 <= col_idx <= 18278:
raise ValueError("Invalid column index {0}".format(col_idx))
letters = []
while col_idx > 0:
col_idx, remainder = divmod(col_idx, 26)
# check for exact division and borrow if needed
if remainder == 0:
remainder = 26
col_idx -= 1
letters.append(chr(remainder+64))
return ''.join(reversed(letters))

How to draw this bow tie pattern using Python 2.7?

I need to draw the following pattern using Python While Loops.
I have spent quite a lot of time and came up with this code which prints it perfectly but this code is so much long and I feel like it is not one of those good codes.
If anybody here can help me out shrinking this code or suggesting a better way to output?
here is the code:
#Question 10, Alternate Approach
temp = 1
pattern = ""
innerSpace = 7
starCount = 1
while temp <= 5:
st = 1
while st <= starCount:
pattern = pattern + "*"
if st != starCount:
pattern = pattern + " "
st = st + 1
sp = 0
if temp == 5:
innerSpace = 1
while sp < innerSpace:
pattern = pattern + " "
sp = sp + 1
st = 1
while st <= starCount:
if temp == 5:
st = st + 1
pattern = pattern + "*"
if st != starCount:
pattern = pattern + " "
st = st + 1
temp = temp + 1
innerSpace = innerSpace - 2
pattern = pattern + "\n"
if temp % 2 == 0:
pattern = pattern + " "
else:
starCount = starCount + 1
starCount = 2
innerSpace = 1
while temp > 5 and temp <= 9:
st = 1
while st <= starCount:
pattern = pattern + "*"
if st != starCount:
pattern = pattern + " "
st = st + 1
sp = 0
while sp < innerSpace:
pattern = pattern + " "
sp = sp + 1
st = 1
while st <= starCount:
pattern = pattern + "*"
if st != starCount:
pattern = pattern + " "
st = st + 1
temp = temp + 1
innerSpace = innerSpace + 2
pattern = pattern + "\n"
if temp % 2 == 0:
starCount = starCount - 1
pattern = pattern + " "
print pattern

Since this looks like an assignment, I'll give you a hint how I would do it.
Take advantage of the symmetry of the bow. It is symmetrical about the horizontal and vertical axis. Therefore, you really only need to solve 1 corner, then copy/mirror the results to get the rest.
This code gives one way of looking at the problem, which is just shifting a initial string (the middle of the bow) to get the desired shape:
m = '*'
size = 4
n = 5 # must be odd
pad = ' ' * n
middle = (m + pad) * size
half = int(n / 2) + 1
print middle
print middle[half*1:]
print middle[half*2:]
print middle[half*3:]
print middle[half*4:]
print middle[half*5:]
print middle[half*6:]
Which yields this:
* * * *
* * *
* * *
* *
* *
*
*
Good luck!

I would use list comprehensions and strings and would exploit the symmetry of the figure.
Not a complete solution, but could be a part of a loop body
In [2]: a = '*' + ' '*8
In [3]: a
Out[3]: '* '
In [24]: result = ''
In [25]: result += a
In [26]: result
Out[26]: '* '
In [27]: result += a[-1::-1]
In [28]: result
Out[28]: '* *'
In [29]: result += '\n'
In [30]: a = ' '+'*' + ' '*7
In [31]: a
Out[31]: ' * '
In [32]: result += a
In [33]: result += a[-1::-1]
In [34]: result += '\n'
In [36]: print result
* *
* *

IMHO you use while loop much as if they where for loops.
I don't think that's what your teacher wants.
The idea behind while is to run until a certain condition is met, not
necessarily when the number of iterations exceed a certain limit.
The condition does not need to be included in the while statement, you can check it later and use the break command to escape the loop
Try for example this:
start = '*'
while True:
print start
if start[0] == '*':
start = ' ' + start
else:
start = '*' + start
if (start == '* * *'):
break
output is just a part of your assignment, think you should be able to work it out to the final, expected result!

Hopefully by this time HW is done. Since I solved this using dynamic programming, I thought I would list solution here.
Observations:
While looking at pattern its observed that bottom half is palindrome of top half. Hence we need to calculate only the top half.
Next we see that for every row count,we have pattern like,
row 1 = 1 , n
row 2 = 2 , n -1
row 3 = 1,3, n-2, n
row 4 = 2, 4 , n-3, n-1
.. and so on.
With iteration index as row count and n as input value we can dynamically calculate remaining values very efficiently.
Source-Code
def get_list(bound, alist):
tmp_list = []
for i in xrange(1,bound + 1):
tmp_list.append(star if i in alist else dot)
return tmp_list
star = "*"
dot = " "
n = 20 #How large of BowTie do you want?
m = (n * 2) - 1
#get top half list
th = []
for idx,k in enumerate(xrange(1,n+1)): #run through 1 - n
row = idx + 1
tmplst = []
if row % 2 != 0:
tmplst.append(i for i in xrange(1,row + 1) if i % 2 != 0)
tmplst.append(i for i in xrange(m, m-row, -1) if i % 2 != 0)
else:
tmplst.append(i for i in xrange(1,row + 1) if i % 2 == 0)
tmplst.append(i for i in xrange(m, m-row, -1) if i % 2 == 0)
#append each row value to top half list.
th.append(sorted(set([j for i in tmplst for j in i])))
#create palindrome of top half which is our bottom half
th = th + th[len(th) -2::-1]
#create list of * and blanks
final = [get_list(m, i) for i in th]
#Print BowTie
for i in final:
print ' '.join(i)

Using a stars and spacing and counting variable
counting=1
star_amount=1
space_amount=6
loop_var=7
while loop_var>0:
loop_var-=1
if space_amount==0:
counting*=-1
stars=" * "*star_amount
spaces=" "*space_amount
print(stars+spaces+stars)
star_amount+=counting
space_amount-= counting*2

How to reverse an int in python?

I'm creating a python script which prints out the whole song of '99 bottles of beer', but reversed. The only thing I cannot reverse is the numbers, being integers, not strings.
This is my full script,
def reverse(str):
return str[::-1]
def plural(word, b):
if b != 1:
return word + 's'
else:
return word
def line(b, ending):
print b or reverse('No more'), plural(reverse('bottle'), b), reverse(ending)
for i in range(99, 0, -1):
line(i, "of beer on the wall")
line(i, "of beer"
print reverse("Take one down, pass it around")
line(i-1, "of beer on the wall \n")
I understand my reverse function takes a string as an argument, however I do not know how to take in an integer, or , how to reverse the integer later on in the script.

Without converting the number to a string:
def reverse_number(n):
r = 0
while n > 0:
r *= 10
r += n % 10
n /= 10
return r
print(reverse_number(123))

You are approaching this in quite an odd way. You already have a reversing function, so why not make line just build the line the normal way around?
def line(bottles, ending):
return "{0} {1} {2}".format(bottles,
plural("bottle", bottles),
ending)
Which runs like:
>>> line(49, "of beer on the wall")
'49 bottles of beer on the wall'
Then pass the result to reverse:
>>> reverse(line(49, "of beer on the wall"))
'llaw eht no reeb fo selttob 94'
This makes it much easier to test each part of the code separately and see what's going on when you put it all together.

Something like this?
>>> x = 123
>>> str(x)
'123'
>>> str(x)[::-1]
'321'

best way is
x=12345
a=str(x)[::-1]\\ In this process i have create string of inverse of integer (a="54321")
a=int(a) \\ Here i have converted string a in integer
or
one line code is
a=int(str(x)[::-1]))

def reverse(x):
re = 0
negative = x < 0
MAX_BIG = 2 ** 31 -1
MIN_BIG = -2 ** 31
x = abs(x)
while x != 0:
a = int(x % 10)
re = re * 10 + a
x = int(x // 10)
reverse = -1 * re if negative else re
return 0 if reverse < MIN_BIG or reverse > MAX_BIG else reverse
this is for 32 - bit integer ( -2^31 ; 2^31-1 )

def reverse_number(n):
r = 0
while n > 0:
r = (r*10) + (n % 10)
print(r)
r *=10
n //= 10
return r
print(reverse_number(123))

You can cast an integer to string with str(i) and then use your reverse function.
The following line should do what you are looking for:
def line(b, ending):
print reverse(str(b)) or reverse('No more'), plural(reverse('bottle'),reverse(str(b))), reverse(ending)

Original number is taken in a
a = 123
We convert the int to string ,then reverse it and again convert in int and store reversed number in b
b = int("".join(reversed(str(a))))
Print the values of a and b
print(a,b)

def reverse_number(n):
r = 0
while n > 0:
r *= 10
r += n % 10
n /= 10
return r
print(reverse_number(123))
This code will not work if the number ends with zeros, example 100 and 1000 return 1

def reverse(num):
rev = 0
while(num != 0):
reminder = num % 10
rev = (rev * 10 ) + reminder
num = num // 10
print ("Reverse number is : " , rev )
num=input("enter number : ")
reverse(int(num))
#/ always results into float
#// division that results into whole number adjusted to the left in the number line

I think the following code should be good to reverse your positive integer.
You can use it as a function in your code.
n = input() # input is always taken as a string
rev = int(str(n)[::-1])
If you are having n as integer then you need to specify it as str here as shown. This is the quickest way to reverse a positive integer

import math
def Function(inputt):
a = 1
input2 = inputt
while(input2 > 9):
input2 = input2/10
a = a + 1
print("There are ", a, " numbers ")
N = 10
m = 1
print(" THe reverse numbers are: ")
for i in range(a):
l = (inputt%N)/m
print(math.floor(l), end = '')
N = N*10
m = m*10
print(" \n")
return 0
enter = int(input("Enter the number: "))
print(Function(enter))

More robust solution to handle negative numbers:
def reverse_integer(num):
sign = [1,-1][num < 0]
output = sign * int(str(abs(num))[::-1])

An easy and fast way to do it is as follows:
def reverse(x: int|str) -> int:
reverse_x = int(''.join([dgt for dgt in reversed(num:=str(x)) if dgt != '-']))
if '-' in num:
reverse_x = -reverse_x'
return reverse_x
First we create a list (using list comprehension) of the digits in reverse order. However, we must exclude the sign (otherwise the number would turn out like [3, 2, 1, -]). We now turn the list into a string using the ''.join() method.
Next we check if the original number had a negative sign in it. If it did, we would add a negative sign to reverse_x.

Easily you can write this class:
class reverse_number:
def __init__(self,rvs_num):
self.rvs_num = rvs_num
rvs_ed = int(str(rvs_num)[::-1])
print(rvs_ed)
You can use it by writing:
reverse_number(your number)

I have written it in a different way, but it works
def isPalindrome(x: int) -> bool:
if x<0:
return False
elif x<10:
return True
else:
rev=0
rem = x%10
quot = x//10
rev = rev*10+rem
while (quot>=10):
rem = quot%10
quot = quot//10
rev = rev*10+rem
rev = rev*10+quot
if rev==x:
return True
else:
return False
res=isPalindrome(1221)