I keep receiving this error (TypeError: input expected at most 1 argument, got 2) and I can't figure out why. The goal of the code is to keep asking for a number until that number is between one and a variable called n.
def enterValidNumber(n):
while True:
num = int(input("Input a number from 1 to", n))
if num >= 1 and num <= n:
break
enterValidNumber(17)
Yes, you are right; you are giving two parameters to input. To put n into the prompt string, you can use an f-string:
num = int(input(f"Input a number from 1 to {n}"))
You can always go with j1-lee's answer but I would have another suggestion:
num = int(input("Input a number from 1 to " + str(n)))
this would work well too!
Related
So im new to python and this question should be fairly easy for anybody in here except me obvisouly haha so this is my code
for c in range(0,20):
print("number",c+1)
number = input ("Enter number:\n")
number = int(number)
if (number < 1 or number > 9):
print("try again by inputing a positive number < 10")
c -=1
so as you may think if the number someone inputs is bigger than 9 or smaller than 0 i just want my c to stay where it is so i get 20 positive numbers from 1-9 instead it doesnt do that and it keeps increasing even tho i have the c-=1 thingy down there
First, do not use range(0,20) but range(20) instead.
Second, range returns an iterator. This means, that when you do c-=1 you do not go back in the iterator, you decrease the number returned by the iterator. Meaning that if c=5 and the number you entered in input is 20, c will become 4, but when returning to the start of the loop, c be be 6.
You probably want something of this sort:
c = 0
while c < 20:
print("number",c+1)
number = input ("Enter number:\n")
number = int(number)
if (number < 1 or number > 9):
print("try again by inputing a positive number < 10")
continue
c += 1
an alternative and simple solution to this would be:
i=20
while(i!=0):
number = input("Enter number:\n")
number = int(number)
if (number <1 or number >9):
print("Invalid no try within 0-9 ")
else
print(number)
i-=1
This question already has answers here:
Python Memory Error while iterating to a big range
(3 answers)
Closed 6 years ago.
Sorry I have to ask such a simple question, but I've been trying to do this for a while with no luck, despite searching around.
I'm trying to define a function that will get user input for X, and then add every integer from 0 to X, and display the output.
For example, if the user inputs 5, the result should be the sum of 1 + 2 + 3 + 4 + 5.
I can't figure out how to prompt the user for the variable, and then pass this variable into the argument of the function. Thanks for your help.
def InclusiveRange(end):
end = int(input("Enter Variable: ")
while start <= end:
start += 1
print("The total of the numbers, from 0 to %d, is: %d" % (end, start))
Just delete argument "end" from function header and use your function.
InclusiveRange()
or define code in other way:
def InclusiveRange(end):
while start <= end:
start += 1
print("The total of the numbers, from 0 to %d, is: %d" % (end, start))
end = int(input("Enter Variable: ")
InclusiveRange(end)
Here's an itertools version:
>>> from itertools import count, islice
>>> def sum_to_n(n):
... return sum(islice(count(), 0, n + 1))
>>>
>>> sum_to_n(int(input('input integer: ')))
input integer: 5
15
You should take the user input out of the function, and instead call the function once you have received the user input.
You also need to store the sum in a separate variable to start, otherwise you're only adding 1 each iteration. (I renamed it to index in this example as it reflects it's purpose more).
def InclusiveRange(end):
index = 0
sum = 0
while index <= end:
sum += start
index += 1
print("The total of the numbers, from 0 to %d, is: %d" % (end, sum))
end = int(input("Enter Variable: "))
InclusiveRange(end)
Demo
Instead of using a loop, use a range object, which you can easily send to sum(). Also, you're never actually using the passed end variable, immediately throwing it away and binding end to a new value. Pass it in from outside the function.
def inclusive_range(end):
num = sum(range(end+1))
print("The total of the numbers, from 0 to {}, is: {}".format(end, num))
inclusive_range(int(input("Enter Variable: ")))
You can also use a math formula to calculate the sum of all natural numbers from 1 to N.
def InclusiveRange(end):
''' Returns the sum of all natural numbers from 1 to end. '''
assert end >= 1
return end * (end + 1) / 2
end = int(input("Enter N: "))
print(InclusiveRange(end))
I am planning to create a program which basically lists the Fibbonacci sequnce up to 10,000
My problem is that in a sample script that I wrote I keep getting the error 'int' object is not iterable
My aim is that you enter a number to start the function's loop.
If anyone would help out that would be great
P.S I am a coding noob so if you do answer please do so as if you are talking to a five year old.
Here is my code:
def exp(numbers):
total = 0
for n in numbers:
if n < 10000:
total = total + 1
return total
x = int(input("Enter a number: "), 10)
exp(x)
print(exp)
numbers is an int. When you enter the number 10, for example, the following happens in exp():
for n in 10:
...
for loops through each element in a sequence, but 10 is not a sequence.
range generates a sequence of numbers, so you should use range(numbers) in the for loop, like the following:
for n in range(numbers):
...
This will iterate over the numbers from 0 to number.
As already mentioned in comments, you need to define a range -- at least this is the way Python do this:
def exp(numbers):
total = 0
for n in range(0, numbers):
if n < 10000:
total = total + 1
return total
You can adjust behavior of range a little e.g. interval is using. But this is another topic.
Your code is correct you just need to change :
for n in numbers:
should be
for n in range(0, numbers)
Since you can iterate over a sequence not an int value.
Good going, Only some small corrections and you will be good to go.
def exp(numbers):
total = 0
for n in xrange(numbers):
if n < 10000:
total += 1
return total
x = int(input("Enter a number: "))
print exp(x)
I am new to Python and would like a simple code to be executed however due to my understanding of the syntax I am not able to create it.
I would need to create a range between 1 to 10 and create an input function to search if this number is within this range. My code looks like this:
range=[1,10]
i=0
for i in len(range):
if (i) > 1 and (i) < c:
print ("HH")
However there is an error. Any solutions with explanations?
range is a function so you can either use
if i in range(1,10):
or
if i >= 1 and i < 10:
In your code, you made a variable that happened to be named range but it was actually just a list with two int elements in it.
The following might help you understand a bit better. First of all avoid calling your variable range as this is a Python function. It is used to return a list of numbers. It can take two parameters, a starting number, and an ending number. The ending number is not included in the list, so for 1 to 10 you need range(1,11).
c = int(input("Enter search number: "))
for i in range(1, 11):
if 1 < i < c:
print(i, "HH")
else:
print(i)
This would display:
1
2 HH
3 HH
4 HH
5
6
7
8
9
10
I am not sure which version of Python you are using. If it is earlier than Python 3, then remove the brackets from the print statements.
If you just want to see if the entered number is in a certain range, something like the following would be suitable:
c = int(input("Enter search number: "))
if 1 <= c <= 10:
print("The number {} is in the range 1 to 10".format(c))
else:
print("The number {} is not in range".format(c))
You would then see it display:
Enter search number: 5
The number 5 is in the range 1 to 10
How do I display the number of one's in any given integer number?
I am very new to python so keep this in mind.
I need the user to input a whole number.
Then I want it to spit out how many one's are in the number they input.
i am using python 3.3.4
How would I be able to do this with this following code?
num = int(input ("Input a number containing more than 2 digits"))
count = 0
for i in range (0):
if num(i) == "1":
count = count + 1
print (count)
I don't know what i'm doing wrong
it gives me 'int' object is not callable error
Something like this:
Int is not iterable so you may need to convert into string:
>>> num = 1231112
>>> str(num).count('1')
4
>>>
str(num).count('1') works just fine, but if you're just learning python and want to code your own program to do it, I'd use something like this, but you're on the right track with the code you supplied:
count = 0
for i in str(num):
if i == "1":
count = count + 1 # or count += 1
print(count)
Just to help you, here is a function that will print out each digit right to left.
def count_ones(num):
while num:
print(num % 10) # last digit
num //= 10 # get rid of last digit
num = 1112111
answer = str(num).count("1")
num = int(input (" Input a number to have the number of ones be counted "))
count = 0
for i in str(num):
if i == "1":
count = count + 1 # or count += 1
print (' The number of ones you have is ' + str(count))
So i took the user input and added it to the correct answer since when i tried the answer from crclayton it didn't work. So this works for me.