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)))
I want to reverse digits in a number in python. Here are my two implementations.
One: convert the number into string and reverse each char in it
number = 2376674032
number_s = str(number)
index = len(number_s) - 1
str_list = []
while index > -1:
str_list.append(number_s[index])
index -= 1
result = int("".join(str_list))
print(result)
Two: using simple mathematics
number = 2376674032
N = 0
K = number
R = number % 10
while K > 0:
N = N*10 + R
K = K // 10
R = K % 10
result = N
print(result)
As I'm pretty new to python programming, so could someone help me with the following questions:
with the first approach, will "".join(str_list) produce a new string with each list element? if so is a better way to concatenate strings in python(something similar to StringBuffer in java)
which of the implementations is better from performance perspective?
You can reverse a string using -1 as the step in a slice. So this works:
number = 2376674032
number_s = str(number)
reverse_s = number_s[::-1]
reversed = int(reverse_s)
you want to reverse a number …..input it as string format , and do this:
number="8374783246837"
revnumber=number[::-1]
Done
a = 1234
a = int("".join(reversed(str(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.
To make the number an integer type, we have to use the int function, as below:
numbers=str(123456)
#or numbers="123456"
print((int(numbers[::-1])))
print((type(int(numbers[::-1]))))
output:
654321
<class 'int'>
We can do this in a single line as well using [::-1]
print(int(str(int(input()))[::-1]))
#here is my answer . you can do it using simple recursion
# count digits recursively
def reverse_digits(n):
# base case
if n == 0:
pass
#recursive case
else:
print(n%10,end='')
return reverse_digits(n//10)
# reverse 123
reverse_digits(123)
````````````````````````````````````````````````````
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?
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)
I'm trying to figure out what's wrong with my code. Could anyone fix it?
def main():
fname = input("Enter filename:")
infile = open(fname, "r")
SD()
def SD():
b= []
a = 5.0
r = len(b)
for n in range(r-1):
b.append((r[n] -a)**2)
m = (float(b)/r)**0.5
print("The standard deviation is", m)
main()
You have a handful of errors:
def SD():
# b is empty
b= []
a = 5.0
#this will always be 0
r = len(b)
# range(r-1) == [] because it is range(-1)
# whole loop is skipped
for n in range(r-1):
b.append((r[n] -a)**2)
# float(b) should throw an error, maybe sum(b) ?
m = (float(b)/r)**0.5
print("The standard deviation is", m)
main()
b is a list; you can't cast a list into a float.
perhaps you want to pass an array to SD()? You should read() the files contents and then perhaps split() that into an array and pass it to SD() as an argument (where you'll call int() on them.)
For a sample:
def SD(numList):
cntN=len(numList)
sumN=0
for i in numList:
sumN+=i
avgVal=float(sumN)/float(cntN)
sumVar=0.0
for i in range(cntN):
sumVar+=float((numList[i]-avgVal)**2)
return ((float(sumVar)/float((cntN-1)))**0.5)