Related
I'm working on a card number check code, for now I created a function that asks for the card number and checks if it is 8 digits or not (it has to be 8) and then calls another function that will do the math and check if the card is valid. For this function:
Starting from the rightmost digit, form the sum of every other digit. For example, if the card number is 1234 5678, then you form the sum 8 + 6 + 4 + 2 = 20
Double each of the digits that were not included in the preview step and then add all digits of the resulting numbers. For example, the digits that were not included are 7 5 3 1, we double the, 14 10 6 2, and then we sum each digit, 1 + 4 + 1 + 0 + 6 + 2 = 14
Add the sums of the two steps, 20 + 14 = 34, if the last digit is 0 then the card is valid, otherwise it is not valid (which is our case)
My problem is that I don't know how to iterate and get the sum of every other digit or double the other number which were not included in step 2. My thought was to use a while loop but then what?
EDIT: since some answers used lists... we didn't study lists yet, so I should not use it, we are only allowed to use sample stuff, loops, functions, etc.. even sum(map()) we didn't study
That is my code for now (its not much but just thought put it anyway)
def getCard():
CardInput = int(input("Enter your 8 digit credit card number: "))
if len(CardInput) == 8:
CardCheck(CardInput)
else:
print("Invalid Input: Should be exactly 8 digits!")
getCard()
def CardCheck(CardNumber):
Position = 0
Sum = 0
DoubleSum = 0
FinalSum = 0
while CardNumber >= 0:
Position += 1
So, the ugly way of doing is, you can write a for loop and use indexing for access specific elements
for i in range(len(CardInput)):
# it will iterate from 0 to 7
print(CardInput[i]) # here you get ith element
if i % 2 == 1:
print("I am other!") # you can sum your things here into another variable
Or with while:
while position < len(CardInput):
print(CardInput[position])
position += 1
It assumes CardInput is str, so I recommend to not convert it earlier.
However pythonic way would be
sum(map(int, CardInput[1::2])))
CardInput[1::2] returns list of every second element starting from second (0 is first).
map converts every element to in.
sum sums elements.
prompt = "Enter the eight-digit number: "
while True:
number = input(prompt)
if len(number) == 8 and number.isdigit():
break
prompt = "Oops, try again: "
first_digits = number[1::2] # If the user entered '12345678', this will be the substring '2468'
first_sum = sum(map(int, first_digits)) # Take each digit (character), map it to a single-digit integer, and take the sum of all single-digit integers
second_digits = number[0::2] # If the user entered '12345678', this will be the substring '1357'
doubled_ints = [int(char) * 2 for char in second_digits] # Take each digit (character), turn it into an integer, double it, and put it in a list.
second_sum = sum(map(int, "".join(map(str, doubled_ints)))) # Merge all integers in 'doubled_ints' into a single string, take each character, map it to a single digit integer, and take the sum of all integers.
total_sum = first_sum + second_sum
total_sum_last_digit = str(total_sum)[-1]
is_valid_card = (total_sum_last_digit == '0')
if is_valid_card:
print("Your card is valid (total sum: {})".format(total_sum))
else:
print("Your card is NOT valid (total sum: {})".format(total_sum))
def getCard():
CardInput = input("Enter your 8 digit credit card number: ")
if len(CardInput) == 8:
CardCheck(CardInput)
else:
print("Invalid Input: Should be exactly 8 digits!")
getCard()
def CardCheck(CardNumber):
list_CardNumber = [x for x in "25424334"]
Sum = sum(int(x) for x in list_CardNumber[1:8:-2])
DoubleSum = 2*sum(int(x) for x in list_CardNumber[0:8:-2])
FinalSum = Sum + DoubleSum
if str(FinalSum)[-1] == "0":
print("Valid Input")
else:
print("Invalid Input")
To get you started, you should check out enumerate(), it'll simplify things if you're just going to use loops by giving you easy access to both the index and value every loop.
step1 = 0
for i, x in enumerate(number):
if i % 2:
print('index: '+ str(i), 'value: '+ x)
step1 += int(x)
print('step 1: ', step1)
Output:
index: 1 value: 2
index: 3 value: 4
index: 5 value: 6
index: 7 value: 8
step 1: 20
You can use:
# lets say
CardNumber = '12345678'
# as mentioned by kosciej16
# get every other digit starting from the second one
# convert them to integers and sum
part1 = sum(map(int, CardNumber[1::2]))
# get every other digit starting from the first one
# convert them to integers and double them
# join all the digits into a string then sum all the digits
part2 = sum(map(int,''.join(list(map(lambda x: str(int(x)*2), CardNumber[0::2])))))
result = part1 + part2
print(result)
Output:
34
Edit:
Only with loops you can use:
# lets say
CardNumber = '12345678'
total_sum = 0
for idx, digit in enumerate(CardNumber):
if idx % 2 == 1:
total_sum += int(digit)
else:
number = str(int(digit)*2)
for d in number:
total_sum += int(d)
print(total_sum)
output:
34
Since you need to iterate over the digits, it's actually easier IMO if you leave it as a string, rather than converting the input to an int; that way you can just iterate over the digits and convert them to int individuall to do math on them.
Given an 8-digit long string card, it might look like this, broken into steps:
even_sum = sum(int(n) for n in card[1::2])
double_odds = (2 * int(n) for n in card[0::2])
double_odd_sum = sum(int(c) for do in double_odds for c in str(do))
All together with some logic to loop while the input is invalid:
def get_card() -> str:
"""Returns a valid card number, or raises ValueError."""
card = input("Enter your 8 digit credit card number: ")
if len(card) != 8 or not card.isdecimal():
raise ValueError("Invalid input: Should be exactly 8 digits!")
card_check(card)
return card
def card_check(card: str) -> None:
"""Raises ValueError if card checksum fails, otherwise returns None."""
even_sum = sum(int(n) for n in card[1::2])
double_odds = (2 * int(n) for n in card[::2])
double_odd_sum = sum(int(c) for do in double_odds for c in str(do))
if (even_sum + double_odd_sum) % 10:
raise ValueError("Card checksum failed!")
while True:
try:
print(f"{get_card()} is a valid card number!")
except ValueError as e:
print(e)
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 have written this code for finding the numerological value of a name. Is there any way in which i can shorten the code, or nest one loop inside the other?
alphabets = {"A":1,"I":1,"J":1, "Q":1,"Y":1,"B":2,"K":2,"R":3,"C":3,"G":3,"L":3,"S":3,"D":4,"M":4,"T":4,"H":5,"E":5,"N":5,"X":5,"U":6,"V":6,"W":6,"O":7, "Z":7,"P":8,"F":8}
word =input("your numerology score is :") #since i am using python 3 to code this
def digit_sum(n):
#prepare a list of numbers in n convert to string and reconvert
numbers=[]
for digit in str(n):
numbers.append(int(digit))
# add up the total of numbers
total=0
for number in numbers:
total += number
return total
def numerology(word):
total = 0
for letter in word.upper():
total += alphabets[letter]
total = digit_sum(total)
return total
print (numerology(word))
To understand what is meant by numerological value, please see https://en.wikipedia.org/wiki/Numerology#Latin_alphabet_systems.
alphabets = {"A":1,"I":1,"J":1, "Q":1,"Y":1,"B":2,"K":2,"R":3,"C":3,"G":3,"L":3,"S":3,"D":4,"M":4,"T":4,"H":5,"E":5,"N":5,"X":5,"U":6,"V":6,"W":6,"O":7, "Z":7,"P":8,"F":8}
name = "this is a sample name"
digits = str(sum([alphabets[l] for l in name.upper() if l in alphabets.keys()]))
numerological_value = int(digits) % 9
if numerological_value == 0:
numerological_value = 9
print(numerological_value)
Comprehensions allow you to have a short hand for creating various data types.
In this case you want to build generators.
This is as you don't need to build every number before reducing the list, to a single number.
Wrapping one in sum can allow you to significantly shorten digit_sum. Which can become:
def digit_sum(n):
return sum(int(digit) for digit in str(n))
You can also change numerology to be a little shorter, if you combine the addition and assignment.
def numerology(word):
total = 0
for letter in word.upper():
total = digit_sum(total + alphabets[letter])
return total
If you want, you can use functools.reduce to make this span a total of six lines.
def numerology(word):
return functools.reduce(
lambda a, b: sum(int(digit) for digit in str(a + b)),
(alphabets[letter] for letter in word.upper()),
0
)
One-liner for laughs.
alphabets = {"A":1,"I":1,"J":1, "Q":1,"Y":1,"B":2,"K":2,"R":3,"C":3,"G":3,"L":3,"S":3,"D":4,"M":4,"T":4,"H":5,"E":5,"N":5,"X":5,"U":6,"V":6,"W":6,"O":7, "Z":7,"P":8,"F":8}
print("Your numerology score is "+ str((int(sum([alphabets[l] for l in input("Type your name:").upper() if l in alphabets.keys()])) % 9) or 9))
this one !!!
a= input()
b= 0
c= len(a)
d= {"a":1, "b":2, "c":3, "d":4, "e":5, "f":6, "g":7, "h":8, "i":9, "j":10, "k":11, "l":12, "m":13, "n":14, "o":15, "p":16, "q":17, "r":18, "s":19, "t":20, "u":21, "v":22, "w":23, "x":24, "y":25, "z":26, " ":0}
for i in range(c):
b= b+d[a[i]]
print(b)
gematria
i'm sure you can chain the reducer functions a bit better but...
const charMap = {A:1, J:1, S:1, B:2, K:2, T:2, C:3, L:3, U:3, D:4,
M:4, V:4, E:5, N:5, W:5, F:6, O:6, X:6, G:7, P:7, Y:7, H:8,
Q:8, Z:8, I:9, R:9};
let name = prompt ("Type your name in CAPS"); //for example: TOM
let wordScore = Array.from(name).reduce((nameScore, element) => {
let curValue = charMap[element]
return (nameScore + curValue)
},0)
let finalScore = Array.from(String(wordScore), Number).reduce((score, element) => {
return score > 10 ? score + element : score
})
alert(finalScore)
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 am trying to write a function that return the biggest number formed by the digits from an input integer number.
So if the input = 123584
output should be = 854321
My code is -
def maxNumber(inputNumber):
x = len(str(inputNumber))
max_number = []
result= []
while(x>0):
max_number.append(inputNumber%10)
inputNumber = inputNumber/10
x -= 1
while(x<(len(str(max_number)))):
result.append(max(max_number))
x += 1
return result
print maxNumber(1238675)
and off-course the output is not as I want. Please help. I am eager to learn all possible way to do it.
def maxNumber(inputNumber):
return int(''.join(sorted(str(inputNumber), reverse=True)))
The biggest number is formed by sorting the digits in descending order. This can be achived using the rverse=True parameter to sorted():
def max_digit_permutation(n):
return int("".join(sorted(str(n), reverse=True)))
This is more reliable than most answers given so far ;-)
def max_number(n):
s = str(n)
digits = sorted(s, reverse=n>0)
return int(''.join(digits))
print max_number(231)
print max_number(-231)
print max_number(+231)
And good point - I missed the option of doing it with number alone - here it is for completeness. :)
from math import *
def max_number(n):
digit_count = int(log(abs(n+1),10)) + 1
digits = sorted([(n / 10 ** (x - 1) % 10) for x in range(digit_count,0,-1) ], reverse=True)
return reduce(lambda x, y:10*x + y, digits)
print max_number(1000)
print max_number(999)
print max_number(2345128)
print max_number(231)
sort the string of number, reverse it, join it and convert to int
>>> x=123584
>>> int(''.join(sorted(str(x))[::-1]))
854321
You could just treat the number as a list of single digits and then sort the list in decreasing order.
What about something like this:
num = str(123584)
int(''.join(sorted(num, reverse=True)))