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Say I had the number 123, how could I return 3, 2, 1, 23, 12, and 123. And if I had the number 100, return 100 and 0 only and not 100, 00, 0.
str can change the format of integers and floats to a string. Then, using the set function, you can change this string to the characters of the string. Note that, sets do not contain the duplicated values, there fore 100 would change to something like {"1", "0"}. You can use something like:
myDigit = 100
myDigitList = set(str(myDigit))
for digit in myDigitList:
print(digit)
Output
1
0
You could take advantage of the string properties in python. Then converting it back to numbers:
number = 1234
number_string = str(number)
for i in range(len(number_string)):
print(int(number_string[i]))
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counttonum = 1
countnum = input("[•] Provide A Number :")
while counttonum < countnum:
print("[", counttonum, "] : Number = ", counttonum)
counttonum +=1
I was trying to make a counting tool that counts up to the provided number from the “input()” function.
For example:
providedNumberFromInput = 5
output = 1
2
3
4
5
And it’ll stop if the provided number is reached. Please help me.
You are very close to solution. Problem is that input() returns value as string so you will need to convert it. And also if you want to include entered number use <= instead of <
while counttonum <= int(countnum):
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they need to order with a problem
so I have the text in a certain order of numbers, something like gematria
input [12345] is what we call gematria and what do they need?
they need to line up the digits backwards
[54321]
have a different count and I would need help with that rather than doing twenty different if
def shiftall(s, n):
n %= len(s)
return s[n:] + s[:n]
it didn't help me much it only moves the simple text
For strings:
return s[::-1]
For integers:
return str(s)[::-1]
Note: This would go inside def shiftall(s, n):
Additional note: Now you don't even need the parameter n
If you want to reverse a number, then you can convert it to a string, reverse the string, and then convert it back to a number.
num = 12345
str_num = str(num)
# reverse and convert
num = int(str_num[::-1])
input=[12345,43545436,88888,843546]
def shiftall(s):
d=[]
for i in s:
res=''.join(reversed(str(i)))
d.append(int(res))
return d
print(shiftall(input))
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The prog is to accept 4 digit binary number separated by ','.
Then I have to print those number which is divisible by 3.
Here is my code
binary_numbers = input("Enter Binary Numbers:").split(",")
newlist = [no for no in binary_numbers if int(no)%3 == 0]
print(newlist)
when I am printing newlist I am getting empty. I dont know where is the problem.
If you want to input binary numbers, you should set the base of the int cast to 2.
[no for no in binary_numbers if int(no, 2) % 3 == 0]
binary_numbers = input("Enter Binary Numbers:").split()
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How would you turn a string that looks like this
7.11,8,9:00,9:15,14:30,15:00
into this dictionary entry
{7.11 : [8, (9:00, 9:15), (14:30, 15:00)]}?
Suppose that the number of time pairs (such as 9:00,9:15 and 14:30,15:00 is unknown and you want to have them all as tuple pairs.
First split the string at the commas, then zip cluster starting from the 3rd element and put it into a dictionary:
s = "7.11,8,9:00,9:15,14:30,15:00"
ss = s.split(',')
d = {ss[0]: [ss[1]] + list(zip(*[iter(ss[2:])]*2))}
Output:
{'7.11': ['8', ('9:00', '9:15'), ('14:30', '15:00')]}
If you need to convert it from string to appropiate data types (you'll have to adapt it according to your needs), then after getting the ss list:
time_list = [datetime.datetime.strptime(t,'%H:%M').time() for t in ss[2:]]
d = {float(ss[0]): [int(ss[1])] + list(zip(*[iter(time_list)]*2))}
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Write a shell (text-based) program, called sum_num.py, that asks the user for a semicolon (;) separated list of numbers, and calculated the total. See the example below.
a = str(raw_input("Enter semicolon separated list of integers:"))
b = a.split(";")
c = (a[0:])
print("the total is " + sum(c))
PS C:\Users\ssiva\Desktop> python sum_num.py
Enter semicolon separated list of integers: 3;10;4;23;211;3
The total is 254
This code will convert into integers and sum them
a=input()
b=a.split(';')
sum=0
for num in b:
sum+=int(num)
print(sum)