I want to set the while condition to only accept values between 0 and 20?
when I use >20 only, It works but I don't want any values less than 0.
T1 = int(raw_input("Please enter score out of 20: "))
while T1!=(0<T1<20):
print("out of range. Please try again!")
T1 = int(raw_input("Please enter score out of 20: "))
else:
name_array.append(int(T1))
You can use if its >= 1 && <=20 So it could check if its in between 1-20.
Related
I'm trying to create a function that would verify if a value is within chosen boundaries and if it's not, the user must retry until verification.
I wanted to know why my code skips the if condition for any given value and it returns me for the input prompt : "Error: Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error."
A=limit(5pi/6,-pi/2,pi/2) for example jumps directly to the else condition while it's true for the if condition
Here is my code :
function alpha = limit(pos,min,max)
if (pos >= max) && (pos <= min)
alpha=pos;
else
while pos >= max || pos <= min
prompt = 'Enter value between max and min';
alpha = input(prompt);
end
end
end
The function first checks whether pos is both greater than or equal to max and less than or equal to min.
I expect you intend max to be greater than min. Therefore, the if line should be:
if (pos <= max) && (pos >= min)
I assume your val_max/val_min are typos (and should be max/min) Then, your while loop depends on the value of pos. But pos is never updated in the loop, so your loop is just going to iterate forever.
You need to update pos inside the while loop for this to work. Try something like this:
function alpha = limit(pos,min,max)
while pos > max || pos < min
prompt = 'Enter value between min and max';
pos = input(prompt);
end
alpha = pos;
end
i am using memoization to save the last calculated values of fibonacci numbers in a dictionary. Since (i suppose) that we don't need all of the values that were previously calculated in our dictionary so, i want to delete them. specifically i only want to keep only the last two calculated fibonacci numbers, is there a way to do it?
import sys
sys.setrecursionlimit(10000)
cache = {}
def fib(n):
if n in cache:
return cache[n]
elif n <= 2:
value = 1
else:
value = fib(n-1) + fib(n-2)
cache[n] = value
return value
print(fib(1000))
ok i found it.
cache = {}
for x in range(1, 1000001):
if x > 4:
cache.pop(x-3, x-4)
if x <= 2:
value = 1
cache[x] = value
else:
value = cache[x - 1] + cache[x - 2]
cache[x] = value
print(value)
My mat file contains 40,000 rows and two columns. I have to read it line by line
and then get values of last column in a single row.
Following is my code:
for v = 1:40000
firstRowB = data.d(v,:)
if(firstRowB(1,2)==1)
count1=count1+1;
end
if(firstRowB(1,2)==2)
count2=count2+1;
end
end
FirstRowB gets the row checks whether last column equals 1 or 2 and then increases the value of respective count by 1.
But I keep getting this error:
Reference to non-existent field 'd'.
You could use vectorization (it is always convenient especially in Matlab). Taking advantage of the fact that true is one and false is zero, if you just want to count you can do :
count1 = sum ( data.d(:, 2) == 1 ) ;
count2 = sum (data.d(:,2) == 2 ) ;
in fact in general you could define :
getNumberOfElementsInLastColEqualTo = #(numb) sum (data.d(:,end) == numb ) ;
counts =arrayfun( getNumberOfElementsInLastColEqualTo , [1 2 ] );
Hope this helps.
I'm stuck at a point where the computer has to check for all remainders of the number which has been given by the user. Am I supposed to use a For loop for this if yes then how?
Thank you.
yes, you need to use for loop, below function will give you how you can use for loop to check whether number is prime or not.
def prime(n):
flag = 0
for i in range(2,(n/2)+2):
if n%i==0:
flag = 1
return "composite"
if flag==0:
return "prime"
def prime_or_composite(num):
div = 2
while True:
if num == div:
print("It is a prime number!")
break
elif num % div = 0:
print("It is a composite number!")
break
else:
div += 1
This works if the input number is an integer bigger than 2.
i want a code to define the prompted number by user is prime or not . since it's an assignment
i'm not allowed to use ' isprime ' predefined code .
the following approach was not useful :
N = input( 'please enter a positive enteger value = ' ) ;
Quotient = floor(N - (mod(N,2)./2)) ;
for i = 1 : Quotient
if mod(N,i ) == 0
fprintf(' your prompted number is not prime ' ) ;
if mod(N,i) ~= 0
fprintf(' your prompted number is prime ' ) ;
end
end
end
for example if i enter a prime number like 13 it results in this :
your prompted number is prime
but if i enter a Non-prime num like 12 it repeats the ' your prompted number is prime ' message for 10 times .
for i = 1 : Quotient
if mod(N,i ) == 0
That will give you every number since x mod 1 is always zero. In other words, the remainder (when you divide any positive integer by one) is zero, since all of them divide perfectly.
You need to start at 2 rather than 1.
In addition, once you've found out the number is not prime, you should stop the loop since there's no possibility of it becoming prime again after that :-) And, for efficiency, you only need to go up to the square root of the number since, if it has a factor above that, you would have already found the equivalent factor below that.
The pseudo-code for such a beast would be:
set isprime to true
set chkval to 2
while chkval * chkval <= number:
if number mod chkval is zero:
set isprime to false
exit while
end if
increment chkval
end while
if isprime:
say number, " is prime"
else:
say number, " is composite"
Try to find factors and as soon as you find one you know it's not prime:
prime = true
for f = 2:ceil(sqrt(N)) %// Start from 2 as prime numbers DO have 1 as a factor. Anything larger than sqrt(N) will have to have a corresponding factor smaller than this so there is no point checking them
if mod(N,f) == 0
prime = false;
break;
end
end
There are 2 problems with your code. First, as already explained by paxdiablo, you need to start your loop from 2. Secondly you have nested your if statements, and since they are mutually exclusive conditions, the inner condition will never trigger.