I have a task to make an output program like this
if input : 5
then output : 2,6,10,14,18
Output must as many of total number as the input
My previous code like this
n = 5
num = 2
i = 1
while i <= n:
if num % 2 == 0:
if num % 4 != 0:
print(num)
i = i+1
num = num+1
But my output was just 2 numbers, i should get 5 numbers.
2,6
Can anyone help me?
Put the i+1 under the second if condition
if num % 4 != 0:
print(num)
i = i + 1
As you can see your output is following the Arithmetic progression
#### first number
a = 2
#### diffrence between the seq numbers
d = 4
#### number of terms in sequence
n = 5
### running loop n times
### a + (n-1) * d ( geeting the nth didgit in AP `enter code here`)
for i in range(1, n+1):
print(2 + (i-1)*d , end=',')
How do I include the user input value in the very first place in the output?
here is my code below:
seq = []
n = int(input("\nEnter a number (greater than 1): "))
while (n > 1):
if n % 2 == 0:
n = n // 2
else:
n = 3 * n + 1
seq.append(n)
print()
print(*seq)
So when I entered 6, it was printed like this:
3 10 5 16 8 4 2 1
My entered value (which MUST be included) is missing.
Please help!
In your current code, you add n to seq at the end of every iteration. To add the initial value of n, simply do seq.append(n) before entering the while loop:
seq = []
n = int(input("\nEnter a number (greater than 1): "))
seq.append(n) # this is the addition you need
while (n > 1):
if n % 2 == 0:
n = n // 2
else:
n = 3 * n + 1
seq.append(n)
print()
print(*seq)
There are several ways you can do this. I believe the most logical way is to move your seq.append(n) statement to the first line of your while loop to capture your input. The issue will then be that 1 will be dropped off the end of the list. To fix that, you change your while loop condition to capture the one and add a condition to break out of the while loop:
seq = []
n = int(input("\nEnter a number (greater than 1): "))
while (n > 0):
seq.append(n)
if n == 1:
break
if n % 2 == 0:
n = n // 2
else:
n = 3 * n + 1
print()
print(*seq)
#output:
Enter a number (greater than 1): 6
6 3 10 5 16 8 4 2 1
I tried to do the codewars Sum of Digits/Digital Root problem, where you have to:
Given n, take the sum of the digits of n. If that value has more than one digit, continue reducing in this way until a single-digit number is produced. The input will be a non-negative integer.
So passing through 52 would return 7, as 5 + 2 is 7 and passing through 942 would return 6, as 9 + 4 + 2 = 15 and then 1 + 5 = 6.
I came up with this code:
def digital_root(n):
n_str = str(n)
digit_total = 0
while len(n_str) != 1:
for digit in n_str:
digit_total += int(digit)
n_str = str(digit_total)
return(n_str)
But it only works for 2 digit numbers and it won't work for higher digit numbers, it just endlessly runs. This code is probably a bad way to do it and I've looked over other people's answers and I get their solution but I just don't get why this won't work for higher digit numbers.
You have got your program almost right. The only challenge I see is with resetting the variable digit_total = 0 after each iteration.
def digital_root(n):
n_str = str(n)
while len(n_str) != 1:
digit_total = 0 #move this inside the while loop
for digit in n_str:
digit_total += int(digit)
n_str = str(digit_total)
return(n_str)
print (digital_root(23485))
The output for print (digital_root(23485)) is 4
2 + 3 + 4 + 8 + 5 = 22
2 + 2 = 4
If the digit_total = 0 is not inside the while loop, then it keeps getting added and you get a never ending loop.
While you have a lot of code, you can do this in a single line.
def sum_digits(n):
while len(str(n)) > 1: n = sum(int(i) for i in str(n))
return n
print (sum_digits(23485))
You don't need to create too many variables and get lost in keeping track of them.
Alex,
running a recursive function would always be better than a while loop in such scenarios.
Try this :
def digital_root(n):
n=sum([int(i) for i in str(n)])
if len(str(n))==1:
print(n)
else:
digital_root(n)
This program will ask the user a series of questions about two numbers. These two numbers will be generated randomly between 1 and 10 and it will ask the user 10 times. At the end of these 10 questions the program will display how many the user got correct out of those questions.
Each question should randomly decide between asking for the product, sum, or difference. Separate the question asking into a function, as well as the validating user input.
I tried using with three product, sum or difference in random to generate. I tried to use z = random.randint(1, 4) is to select from 1 is product, 2 is sum, or 3 is difference and then I used with if variable z is 1, then do product math or if var z is 3, then it should be difference like this x / y, but I couldn't figure it finish it up. I have the expected result when I first run with product but it works so I just need to add with sum and difference included.
EXPECTED OUTPUT with product (Some are incorrect for testing with scores):
> python3 rand3.py
What is 3 x 4
Enter a number: 12
What is 3 x 7
Enter a number: 27
What is 6 x 3
Enter a number: 18
What is 7 x 10
Enter a number: 70
What is 9 x 10
Enter a number: 90
What is 9 x 7
Enter a number: 72
What is 5 x 9
Enter a number: 54
What is 6 x 8
Enter a number:
Incorrect Input!
Enter a number: 48
What is 1 x 5
Enter a number: 5
What is 10 x 3
Enter a number: 30
You got 7 correct out of 10
My Work for Product Only (Success):
import random
def askNum():
while(1):
try:
userInput = int(input("Enter a number: "))
break
except ValueError:
print("Incorrect Input!")
return userInput
def askQuestion():
x = random.randint(1, 100)
y = random.randint(1, 100)
print("What is " + str(x) + " x " +str(y))
u = askNum()
if (u == x * y):
return 1
else:
return 0
amount = 10
correct = 0
for i in range(amount):
correct += askQuestion()
print("You got %d correct out of %d" % (correct, amount))
My Currently Work: (I am working to add sum and difference like the expected output
UPDATED: After the expected output works well with product so I am trying to add new random int for z with 1-3 which means I am using with 1 is for product, 2 is for sum and 3 is difference by using if-statement by given random select. I am struggle at this is where I stopped and figure it out how to do math random because I am new to Python over a month now.
import random
def askNum():
while(1):
try:
userInput = int(input("Enter a number: "))
break
except ValueError:
print("Incorrect Input!")
return userInput
def askQuestion():
x = random.randint(1, 10)
y = random.randint(1, 10)
z = random.randint(1, 4)
print("What is " + str(x) + " "+ str(z)+ " " +str(y))
u = askNum()
if (z == 1):
x * y #product
return 1
else if (z == 2):
x + y #sum
return 1
else if (z == 3):
x / y #difference
return 1
else
return 0
amount = 10
correct = 0
for i in range(amount):
correct += askQuestion()
print("You got %d correct out of %d" % (correct, amount))
OUTPUT:
md35#isu:/u1/work/python/mathquiz> python3 mathquiz.py
File "mathquiz.py", line 27
if (z == 1):
^
IndentationError: unexpected indent
md35#isu:/u1/work/python/mathquiz>
With this currently output, I double checked with corrected Python formatting and everything are sensitive, and still the same as running output. Any help would be more appreciated with explanation. (I hope my English is okay to understand since i'm deaf) I have started this since on Saturday, than expected on time to meet.
Your problem is that python 3 does not allow mixing spaces and tabs for indentation. Use an editor that displays the whitespace used (and fix manually) or one that replaces tabs into spaces. It is suggested to use 4 spaces for indentation - read PEP-0008 for more styling tips.
You can make your program less cryptic if you use '+','-','*','/' instead of 1,2,3,4 to map your operation: ops = random.choice("+-*/") gives you one of your operators as string. You feed it into a calc(a,ops,b) function and return the correct result from it.
You can also shorten your askNum and provide the text to print.
These could look like so:
def askNum(text):
"""Retunrs an integer from input using 'text'. Loops until valid input given."""
while True:
try:
return int(input(text))
except ValueError:
print("Incorrect Input!")
def calc(a,ops,b):
"""Returns integer operation result from using : 'a','ops','b'"""
if ops == "+": return a+b
elif ops == "-": return a-b
elif ops == "*": return a*b
elif ops == "/": return a//b # integer division
else: raise ValueError("Unsupported math operation")
Last but not least you need to fix the division part - you allow only integer inputs so you can also only give division problems that are solveable using an integer answer.
Program:
import random
total = 10
correct = 0
nums = range(1,11)
for _ in range(total):
ops = random.choice("+-*/")
a,b = random.choices(nums,k=2)
# you only allow integer input - your division therefore is
# limited to results that are integers - make sure that this
# is the case here by rerolling a,b until they match
while ops == "/" and (a%b != 0 or a<=b):
a,b = random.choices(nums,k=2)
# make sure not to go below 0 for -
while ops == "-" and a<b:
a,b = random.choices(nums,k=2)
# as a formatted text
result = askNum("What is {} {} {} = ".format(a,ops,b))
# calculate correct result
corr = calc(a,ops,b)
if result == corr:
correct += 1
print("Correct")
else:
print("Wrong. Correct solution is: {} {} {} = {}".format(a,ops,b,corr))
print("You have {} out of {} correct.".format(correct,total))
Output:
What is 8 / 1 = 3
Wrong. Correct solution is: 8 / 1 = 8
What is 5 - 3 = 3
Wrong. Correct solution is: 5 - 3 = 2
What is 4 - 2 = 3
Wrong. Correct solution is: 4 - 2 = 2
What is 3 * 1 = 3
Correct
What is 8 - 5 = 3
Correct
What is 4 / 1 = 3
Wrong. Correct solution is: 4 / 1 = 4
What is 8 * 7 = 3
Wrong. Correct solution is: 8 * 7 = 56
What is 9 + 3 = 3
Wrong. Correct solution is: 9 + 3 = 12
What is 8 - 1 = 3
Wrong. Correct solution is: 8 - 1 = 7
What is 10 / 5 = 3
Wrong. Correct solution is: 10 / 5 = 2
You have 2 out of 10 correct.
def askQuestion():
x = random.randint(1, 10)
y = random.randint(1, 10)
z = random.randint(1, 4)
print("What is " + str(x) + " "+ str(z)+ " " +str(y))
u = askNum()
if (z == 1):
x * y #product
return 1
elif (z == 2):
x + y #sum
return 1
elif (z == 3):
x / y #difference
return 1
else:
return 0
Write your block like this your u = askNum() and next if loop should be on same vertical line.
To Generate n number of random number you can use
random.sample(range(from, to),how_many_numbers)
User this as reference for more info on random
import random
low=0
high=4
n=2 #no of random numbers
rand = random.sample(range(low, high), n)
#List of Operators
arithmetic_operators = ["+", "-", "/", "*"];
operator = random.randint(0, 3)
x = rand[0];
y = rand[1];
result=0;
# print(x, operator, y)
if (operator == 0):
result = x + y# sum
elif(operator == 1):
result = x - y# difference
elif(operator == 2):
result= x / y#division
else :
result=x * y# product
print("What is {} {} {}? = ".format(x,arithmetic_operators[operator],y))
The following stores a random number(int)
operator = random.randint(0, 3)
to compare it with the list for operators.
Example: operator = 2
elif(operator == 2):
result= x / y#division
than this code will executed and because operator=2, 3rd element from list(/) will be selected
Output:
What is 3 / 2?
So this is a cs assignment, and I wrote code that works for 2 out of the 9 test cases (which are unknown), but I don't know why it won't work/pass any other ones. I understand I should try to figure out how to do this on my own, and I have tried, but I am really lost and need help!
"Write a file named credit_card.py containing a single function, check. Check accepts a single input – a positive integer. It returns True if the integer represents a valid credit card number. As with all functions that return a bool value, if it does not return True it should return False.
Credit card numbers have what is called a check digit. This is a simple way of detecting common mis-typings of card numbers. The algorithm is as follows:
Form a sum of every other digit, including the right-most digit; so 5490123456789128 (5490123456789128) sums to 4 + 0 + 2 + 4 + 6 + 8 + 1 + 8 = 33.
Double each remaining digit, then sum all the digits that creates it; the remaining digits (5 9 1 3 5 7 9 2) in our example (5490123456789128) double to 10 18 2 6 10 14 18 4, which sums to 1+0 + 1+8 + 2 + 6 + 1+0 + 1+4 + 1+8 + 4 = 37
Add the two sums above (33 + 37 = 70)
If the result is a multiple of 10 (i.e., its last digit is 0) then it was a valid credit card number."
def check(x):
num1 = 0
num2 = 0
if x < 0:
return False
for i in str(x) [1::2]:
num1 += int(i)
return num1
for i in str(x) [0::2]:
num2 += int(int(i * 2) % 10) + int(int(i * 2) / 10)
return num2
check_digit = num1 + num2
if check_digit % 10 == 0:
return True
else:
return False
def check(x):
if x[0] == '-': # x is [str], "x < 0" is wrong
return False
try:
nums = map(int, x)
except Exception:
return False
sum1 = 0
sum2 = 0
for i in nums[1::2]:
sum1 += int(i)
for i in nums[0::2]:
sum2 += ((i * 2) % 10 + (i * 2) / 10) # i is [str], "i * 2" is wrong
check_digit = sum1 + sum2
if check_digit % 10 == 0:
return True
else:
return False