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)
Related
I have one code here
and need to change the order of the digits
import math
def sucet_cisel(number):
bla: int = 0
while number > 0:
xyzpremenna = number % 10
bla += xyzpremenna
number = (number - xyzpremenna) / 10
return bla
def digit_root(n):
if n == 0: return 0
return (n - 1) % 9 + 1
if __name__ == '__main__':
n = int(input("od čisla:"))
m = int(input("do čisla:"))
for i in range(1,m + 1):
sucet: int = math.floor(sucet_cisel(n*i))
t=(n*i)*(2)
x=' ';
print(n,"*",i,"=",n*i,(x*4),"*2","=",t,sep='')
they need to add () to this code so that in each result where there are 4 numbers they are moved
therefore print (t) need this script to run at that number
t=(ni)(2)
and the result of this to turn into this code
val = list(str(i))
digit = val.pop(-3)
new = int(''.join(val+[digit]))
od čisla:2554
do čisla:4505
2554*4505=11505770 *2=23011540
23011540
23011405
the script stops at the number I enter where is the problem?
20*1=20 *2=40
20*2=40 *2=80
20*3=60 *2=120
20*4=80 *2=160
20*5=100 *2=200
20*6=120 *2=240
20*7=140 *2=280
20*8=160 *2=320
20*9=180 *2=360
20*10=200 *2=400
20*11=220 *2=440
20*12=240 *2=480
20*13=260 *2=520
20*14=280 *2=560
20*15=300 *2=600
20*16=320 *2=640
20*17=340 *2=680
20*18=360 *2=720
20*19=380 *2=760
20*20=400 *2=800
this makes a code if I give
n = int (input ("from number:"))
m = int (input ("to number:"))
n20
m20
however, if in this script there is i
val = list (page (s))
digit = val.pop (-3)
new = int (''. join (val + [digit]))
does it calculate only one result where is the error?
Very similar to the answer from Tim Roberts, but using slices and format strings.
n = 12345678
s = str(n)
x = int(f"{s[:-5]}{s[::-1][:4]}")
s is '12345678', s[:-5] is '1234', s[::-1] is '87654321', and s[::-1][:4] is '8765'. Put it all together and x is 12348765.
OK, let's rewrite your problem to "given a number of greater than 4 digits, I want that same number but with all permutations of the last 4 digits.
import itertools
def permute(number):
val = str(number)
prefix = val[:-4]
for combo in itertools.permutations(val[-4:]):
yield int(prefix+''.join(combo))
print(list(permute(12345678)))
whenever I run my code I get a TypeError saying "not all arguements converted during string formatting" and I tried using str() around what didn't get converted but I ran into more errors.
Here is my code:
def decimalToBinary(num):
bits = " "
while(num > 0):
bits = str(num%2) + bits
num = num//2
return bits
def binaryToDecimal(bits):
deciNum = 0
powers = 0
for i in reversed(bits):
deciNum = 2 **powers** (bits % 10)
bits /= 10
powers += 1
return deciNum
#program tester
for i in range(135, 146):
x = decimalToBinary(i)
deciNum = binaryToDecimal(x)
print(str(decimal))+ ' is '+ ' in Binary.'
I get this TypeError on the line that says "deciNum = 2 ** powers ** (bits%10)
Try this:
def binaryToDecimal(b_num: str) -> int:
d_num = 0
for i in range(len(b_num)):
digit = b_num.pop()
if digit == '1':
d_num = d_num + 2**i
return d_num
Note that b_num is a string, not an int. So you need to use this function in this way binaryToDecimal('101') (and not in this way binaryToDecimal(101)).
To answer the title of the post and to keep the types consistent with your program:
import math
def binaryToDecimal(bits):
# Initialize integer for number.
num = 0
# For each bit, multiply by power of 2 corresponding to its position.
# Then, add that power of 2 to the total counter.
for i in range(len(bits)):
num += int(bits[i]) * (2 ** (len(bits)-i-1))
# Return integer type.
return str(num)
def decimalToBinary(num):
# Determine how many bits represent the decimal number.
num_of_bits = int(math.log(num, 2)) + 1
bits = ''
# Shift the number over 1 more place to the right in each iteration.
# Then test the sign of the bit with AND.
for i in reversed(range(num_of_bits)):
bits += str(int(1&(num>>i)))
return bits
for i in range(135, 146):
x = decimalToBinary(i)
deciNum = binaryToDecimal(x)
print(str(deciNum)+ ' is '+ str(x)+' in Binary.')
Hey I have this function in python3 , can anyone explain why it is giving 1 as output instead of the number as reverse
def reverse(a , rev):
if a > 0:
d = a % 10
rev = (rev * 10) + d
reverse(a/10 , rev)
return rev
b = input("Enter the Number")
x = reverse(b , 0)
print(x)
You need to:
use integer division (//)
capture the value returned from the recursive call, and return it
convert the string input to number (int())
Corrected script:
def reverse(a, rev):
if a > 0:
d = a % 10
rev = (rev * 10) + d
return reverse(a//10, rev)
return rev
b = input("Enter the Number")
x = reverse(int(b), 0)
print(x)
I'm not sure why you're doing it like that. Seems like the following is easier
def rev(a):
return int(str(a)[::-1])
Anyway, I believe you should use "//" instead of "/" for dividing without the rest in python 3?
My prof wants me to create a function that return the sum of numbers in a string but without using any lists or list methods.
The function should look like this when operating:
>>> sum_numbers('34 3 542 11')
590
Usually a function like this would be easy to create when using lists and list methods. But trying to do so without using them is a nightmare.
I tried the following code but they don't work:
>>> def sum_numbers(s):
for i in range(len(s)):
int(i)
total = s[i] + s[i]
return total
>>> sum_numbers('1 2 3')
'11'
Instead of getting 1, 2, and 3 all converted into integers and added together, I instead get the string '11'. In other words, the numbers in the string still have not been converted to integers.
I also tried using a map() function but I just got the same results:
>>> def sum_numbers(s):
for i in range(len(s)):
map(int, s[i])
total = s[i] + s[i]
return total
>>> sum_numbers('1 2 3')
'11'
Totally silly of course, but for fun:
s = '34 3 542 11'
n = ""; total = 0
for c in s:
if c == " ":
total = total + int(n)
n = ""
else:
n = n + c
# add the last number
total = total + int(n)
print(total)
> 590
This assumes all characters (apart from whitespaces) are figures.
You've definitely put some effort in here, but one part of your approach definitely won't work as-is: you're iterating over the characters in the string, but you keep trying to treat each character as its own number. I've written a (very commented) method that accomplishes what you want without using any lists or list methods:
def sum_numbers(s):
"""
Convert a string of numbers into a sum of those numbers.
:param s: A string of numbers, e.g. '1 -2 3.3 4e10'.
:return: The floating-point sum of the numbers in the string.
"""
def convert_s_to_val(s):
"""
Convert a string into a number. Will handle anything that
Python could convert to a float.
:param s: A number as a string, e.g. '123' or '8.3e-18'.
:return: The float value of the string.
"""
if s:
return float(s)
else:
return 0
# These will serve as placeholders.
sum = 0
current = ''
# Iterate over the string character by character.
for c in s:
# If the character is a space, we convert the current `current`
# into its numeric representation.
if c.isspace():
sum += convert_s_to_val(current)
current = ''
# For anything else, we accumulate into `current`.
else:
current = current + c
# Add `current`'s last value to the sum and return.
sum += convert_s_to_val(current)
return sum
Personally, I would use this one-liner, but it uses str.split():
def sum_numbers(s):
return sum(map(float, s.split()))
No lists were used (nor harmed) in the production of this answer:
def sum_string(string):
total = 0
if len(string):
j = string.find(" ") % len(string) + 1
total += int(string[:j]) + sum_string(string[j:])
return total
If the string is noisier than the OP indicates, then this should be more robust:
import re
def sum_string(string):
pattern = re.compile(r"[-+]?\d+")
total = 0
match = pattern.search(string)
while match:
total += int(match.group())
match = pattern.search(string, match.end())
return total
EXAMPLES
>>> sum_string('34 3 542 11')
590
>>> sum_string(' 34 4 ')
38
>>> sum_string('lksdjfa34adslkfja4adklfja')
38
>>> # and I threw in signs for fun
...
>>> sum_string('34 -2 45 -8 13')
82
>>>
If you want to be able to handle floats and negative numbers:
def sum_numbers(s):
sm = i = 0
while i < len(s):
t = ""
while i < len(s) and not s[i].isspace():
t += s[i]
i += 1
if t:
sm += float(t)
else:
i += 1
return sm
Which will work for all cases:
In [9]: sum_numbers('34 3 542 11')
Out[9]: 590.0
In [10]: sum_numbers('1.93 -1 23.12 11')
Out[10]: 35.05
In [11]: sum_numbers('')
Out[11]: 0
In [12]: sum_numbers('123456')
Out[12]: 123456.0
Or a variation taking slices:
def sum_numbers(s):
prev = sm = i = 0
while i < len(s):
while i < len(s) and not s[i].isspace():
i += 1
if i > prev:
sm += float(s[prev:i])
prev = i
i += 1
return sm
You could also use itertools.groupby which uses no lists, using a set of allowed chars to group by:
from itertools import groupby
def sum_numbers(s):
allowed = set("0123456789-.")
return sum(float("".join(v)) for k,v in groupby(s, key=allowed.__contains__) if k)
which gives you the same output:
In [14]: sum_numbers('34 3 542 11')
Out[14]: 590.0
In [15]: sum_numbers('1.93 -1 23.12 11')
Out[15]: 35.05
In [16]: sum_numbers('')
Out[16]: 0
In [17]: sum_numbers('123456')
Out[17]: 123456.0
Which if you only have to consider positive ints could just use str.isdigit as the key:
def sum_numbers(s):
return sum(int("".join(v)) for k,v in groupby(s, key=str.isdigit) if k)
Try this:
def sum_numbers(s):
sum = 0
#This string will represent each number
number_str = ''
for i in s:
if i == ' ':
#if it is a whitespace it means
#that we have a number so we incease the sum
sum += int(number_str)
number_str = ''
continue
number_str += i
else:
#add the last number
sum += int(number_str)
return sum
You could write a generator:
def nums(s):
idx=0
while idx<len(s):
ns=''
while idx<len(s) and s[idx].isdigit():
ns+=s[idx]
idx+=1
yield int(ns)
while idx<len(s) and not s[idx].isdigit():
idx+=1
>>> list(nums('34 3 542 11'))
[34, 3, 542, 11]
Then just sum that:
>>> sum(nums('34 3 542 11'))
590
or, you could use re.finditer with a regular expression and a generator construction:
>>> sum(int(m.group(1)) for m in re.finditer(r'(\d+)', '34 3 542 11'))
590
No lists used...
def sum_numbers(s):
total=0
gt=0 #grand total
l=len(s)
for i in range(l):
if(s[i]!=' '):#find each number
total = int(s[i])+total*10
if(s[i]==' ' or i==l-1):#adding to the grand total and also add the last number
gt+=total
total=0
return gt
print(sum_numbers('1 2 3'))
Here each substring is converted to number and added to grant total
If we omit the fact eval is evil, we can solve that problem with it.
def sum_numbers(s):
s = s.replace(' ', '+')
return eval(s)
Yes, that simple. But i won't put that thing in production.
And sure we need to test that:
from hypothesis import given
import hypothesis.strategies as st
#given(list_num=st.lists(st.integers(), min_size=1))
def test_that_thing(list_num):
assert sum_numbers(' '.join(str(i) for i in list_num)) == sum(list_num)
test_that_thing()
And it would raise nothing.
I tried running the following code. I tried returning value of j also but it just doesn't work.
def reverse(n):
j=0
while(n!=0):
j=j*10
j=j + (n%10)
n=n/10
print(j)
reverse(45)
Here is a program to reverse a number
def reverse(n):
v = []
for item in reversed(list(str(n))):
v.append(item)
return ''.join(v)
print(reverse("45"))
returns
54
The reverse() function creates an array, adds each digit from the input to said array, and then prints it as plain text. If you want the data from that as an integer then you can replace the return command to this at the end of the function
return int(''.join(v))
Actually, you made one mistake only: for Python 3 you need to use an integer division: n = n // 10.
Here is the correct code without str and list:
def reverse(n):
j = 0
while n != 0:
j = j * 10
j = j + (n%10)
n = n // 10
print(j)
reverse(12345)
Here is the correct code for Python 3:
import sys
def reverse(x):
while x>0:
sys.stdout.write(str(x%10))
x = x//10 # x = x/10 (Python 2)
print() # print (Python 2)
number = 45
int(str(number)[::-1])
a = 1234
a = int("".join(reversed(str(a))))
print a
This will give a = 4321
reversed functions returns an iterable object. If we do :
a = list(reversed(str(a)))
it will return [“3”,”2″,”1″]. We have then joined it and converted into int.