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I wrote a script to perform a connectivity search in a binary image (the possible values for the array are 0 and 1). For each pixel, the code looks at how many neighbours have intensity 1, and if there are at least 5 neighbours with I=1, it assigns 1 as a value of the considered pixel. I want the code to repeat the process until no new pixels are assigned intensity 1. At present the code is not performing the connectivity search iteratively; do you have any suggestion on how to fix this?
pixel_count = 0
pixel_counter = [0] * 100
for p in range(1, 100):
if p < 3:
continue
else:
if pixel_counter[p-1] > pixel_counter[p-2]:
continue
else:
break
for q in range(1, ran_x-1):
for r in range(1, ran_y-1):
counter = 0
if neighbours_mask_1[q,r] == 1:
counter = counter +1
if neighbours_mask_2[q,r] == 1:
counter = counter +1
if neighbours_mask_3[q,r] == 1:
counter = counter +1
if neighbours_mask_4[q,r] == 1:
counter = counter +1
if neighbours_mask_5[q,r] == 1:
counter = counter +1
if neighbours_mask_6[q,r] == 1:
counter = counter +1
if neighbours_mask_7[q,r] == 1:
counter = counter +1
if neighbours_mask_8[q,r] == 1:
counter = counter +1
if counter > 5:
mask_1[q,r] = 1
pixel_count = pixel_count + 1
print pixel_count
else:
mask_1[q,r] = 0
pixel_counter[p] = pixel_count
This section of the code:
for p in range(1, 100):
...
if pixel_counter[p-2] > pixel_counter[p-1]:
continue
else:
break
... dead code ...
will slurp all of the execution, the part where I've marked dead code, which contains your counters are never executed because they're unreachable.
I'm not quite sure what your trying to do there.
To answer the question in the title: the easiest way to exit a nested loop is to just move the loops into a function and return from that. E.g.
def f():
for i in range(10):
for j in range(10):
if i + j == 9: return
Related
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This code in under is will let the user enter the number of elements of list and the user need enter the elements ( User will need type number_of_elements_of_list times) and then count how many positive number, negative number, zero number. The code is like this:
the_number_of_elements_of_the_list = int(input())
number_list = []
for i in range(1, the_number_of_elements_of_the_list + 1):
number_list.append(int(input()))
positive_number_count = 0
negative_number_count = 0
zero_number_count = 0
if number_list == 0:
zero_number_count += 1
elif number_list == 1 or number_list > 1:
positive_number_count += 1
elif number_list < 0:
negative_number_count += 1
print(positive_number_count, negative_number_count, zero_number_count)
The code have a problem: The list can not compare like that. It will be error but i don't know how to compare the elements of list. Can you help me solve this problem?
Firstly, as teambob pointed out, add indent to the for-loop block.
Secondly, as DarkKnight pointed out, put the count variables outside the for-loop.
Thirdly, for each iteration, in order to use that value alone, use number_list[i-1] instead of number_list. (The index is i-1 rather than i because the range in your code starts from 1 rather than 0)
The final code would look like:
the_number_of_elements_of_the_list = int(input())
positive_number_count = 0
negative_number_count = 0
zero_number_count = 0
number_list = []
for i in range(1, the_number_of_elements_of_the_list + 1):
number_list.append(int(input()))
if number_list[i-1] == 0:
zero_number_count += 1
elif number_list[i-1] == 1 or number_list[i-1] > 1:
positive_number_count += 1
elif number_list[i-1] < 0:
negative_number_count += 1
print(positive_number_count, negative_number_count, zero_number_count)
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so I'm new to learning def function() in python. And I'm implementing a code that counts the number of steps to reduce an integer to zero.
I have defined a function that makes a list of the integer input by the user, and another function that counts the steps.
the problem is in the second function: currentlynumberOfSteps only takes the first input, but it needs to take all user inputs listed by the first function UserEntryList
def UserEntryList ():
integerEntry = input("Please enter: ")
integerEntry = integerEntry.split()
listOfInt = []
for i in integerEntry:
try:
listOfInt.append(int(i))
except ValueError:
continue
return(listOfInt)
def numberOfSteps():
counter = 0
listofnumbers = UserEntryList()
for i in listofnumbers:
while i > 0:
if i % 2 == 0:
i /= 2
else:
i -= 1
counter += 1
return counter
desired output:
Please enter a set of space-separated positive integers: 10 15 59
[(10, 5), (15, 7), (59, 10)]
First off, couple of things:
always use snake_case in Python, instead of camelCase or PascalCase. More on naming conventions in Python here
return statement in Python should not be enclosed by brackets, unless you're returning a tuple. More on this here.
in the function numberOfSteps(), you have entered the return statement within the for loop. this will not work, as the program will terminate after processing the first item in your loop. So, append your results from each iteration to a list/dict (dict example is shown below.) and return it in the end after the for loop has completed.
def numberOfSteps():
counter = 0
listofnumbers = UserEntryList()
counter_dict = {}
for i in listofnumbers:
temp = i
while i > 0:
if i % 2 == 0:
i /= 2
else:
i -= 1
counter += 1
counter_dict[temp] = counter
return counter_dict
And to answer your question, you can just do append this at the end of your code:
if __name__ == "__main__":
counter = numberOfSteps()
print("Counter = " + str(counter))
Here is a proper way of defining the main function and call sub-function from main function. You will not need to pass any arguments to any functions since you are directly calling UserEntryList from numberOfSteps, which gets executed from the main function.
def UserEntryList ():
integerEntry = input("Please enter: ")
integerEntry = integerEntry.split()
listOfInt = []
for i in integerEntry:
try:
listOfInt.append(int(i))
except ValueError:
continue
return(listOfInt)
def numberOfSteps():
counter = 0
listofnumbers = UserEntryList()
for i in listofnumbers:
while i > 0:
if i % 2 == 0:
i /= 2
else:
i -= 1
counter += 1
return counter
def main():
counter = numberOfSteps()
print("Counter = " + str(counter))
if __name__ == "__main__":
main()
def prime_count(a, b):
for i in range(a,b):
if i % 2 != 0:
return sum(i)
else:
return 0 "
I am a beginner programmer, I have been practicing on some challenges on Edabit, small problems but that require some thinking (for me).
I am trying to count how many prime numbers are on a and b, they are already integers and there is no user input.
I am not very good at loops yet and thought this was going to be a good challenge to practice, what am I doing wrong? I keep getting int object is not iterable
If reference need, here is the link for the challenge.
Link : https://edabit.com/challenge/6QYwhZstMuHYtZRbT
it is an interesting problem that you are solving. You will have to run two-loops here. One to iterate from a to b and the inner loop to check if each element in a -->b is prime or not (the inner loop should run from 2 to j in [a,b).
def primecheck(a,b):
printlist=[]
if a > 1:
for i in range(a,b+1):
for j in range(2,i):
if i % j == 0:
break
else:
printlist.append(i)
if len(printlist) == 0:
return 0
else:
return len(printlist), printlist
Try this out.
As people say in the comment section, calling sum() is what's causing an error.
But, even if you somehow got that part right, you wouldn't quite get what you want. Maybe you were just trying a simple for loop to check if numbers are odd...?
Anyway, I normally like using Sieve of Eratosthenes to generate prime numbers because it's simple.
def sieve_of_eratosthenes(start, end):
if start > end:
raise AssertionError
# end = end + 1 # If end is inclusive, then uncomment this line.
if end < 2:
return 0
sieve = [i for i in range(2, end)] # Initialize an array of number from 2 to end.
size = len(sieve)
p = 2 # Initial prime.
count = 0
# This block implements Sieve of Eratosthenes.
while count < size:
for i in range(count, size):
num = sieve[i]
if num != p and num % p == 0:
sieve[i] = 0
if count == size-1:
break
count += 1
while sieve[count] == 0:
count += 1
if count == size-1:
break
p = sieve[count] # Update the next prime.
# This block calculates the numbers of primes between start and end.
num_of_primes = 0
for p in sieve:
if p == 0 or p < start:
continue
num_of_primes += 1
print(sieve)
return num_of_primes
sieve_of_eratosthenes(1, 100) # This should return 25.
I am trying to put together a simple program which could work out n prime numbers. I would like to do this by using a nested for loop, where one would go through the numbers, and another would divide that number by all of the numbers up to it to see if it would be divisible by anything.
The problem I am having is that in the main for loop, I need to start it at 2, seeing as 1 would mess up the system and I don't want it to be considered a prime. For the loop to have a starting number however, it also needs an ending number which is difficult in this instance as it is hard to generate the largest prime that will be needed prior to the loop working.
Here's the program that I am using right now. Where I have marked X is where I need to somehow put an ending number for the For Loop. I guess it would be much simpler if I let the For Loop be completely open, and simply take out anything that '1' would produce in the loop itself, but this feels like cheating and I want to do it right.
check = 0
limit = int(input("Enter the amount of Prime Numbers"))
for i in range(2,X):
check = 0
if i > 1:
for j in range(2,i):
if (i % j) == 0:
check = 1
if check == 0:
print (i)
Thanks for your help!
You can step through an unlimited amount of numbers using a generator object.
Insert the following somewhere near the top of your code:
def infinite_number_generator(initial_value=2):
""" Generates an infinite amount of numbers """
i = initial_value
while True:
yield i
i += 1
What this does is it creates a function for constructing generator objects that "pause" whenever they reach the yield statement to "yield" whatever value is specified by the yield command, and then continue to execute from the next line beneath the yield statement.
Python's own range function is itself an example of a generator, and is roughly equivalent to (ignoring the step argument and other peculiarities)
def range(start, end):
i = start
while i < end:
yield i
i += 1
So your program would then look like this:
def infinite_number_generator(initial_value=2):
""" Generates an infinite amount of numbers """
i = initial_value
while True:
yield i
i += 1
check = 0
limit = int(input("Enter the amount of Prime Numbers"))
for i in infinite_number_generator():
check = 0
for j in range(2,i):
if (i % j) == 0:
check = 1
if check == 0:
print (i)
if i == limit:
break
I should also point out that the code you provided is buggy - it will never stop printing because there's no checking whether you've found your limit number of primes yet or not.
This should do what you want.
check = 0
limit = int(input("Enter the amount of Prime Numbers"))
counter = 0
i = 2
while counter < limit:
check = 0
if i > 1:
for j in range(2,i):
if (i % j) == 0:
check = 1
if check == 0:
counter += 1
print (i)
i += 1
In your code you start i with 2 and always increment by 1, so the i will always remain greater than 1, therefore the test if i > 1 is useless.
For efficiency you can stop the check at the square of i or i/2 (no divisors in [i/2 + 1, i[ ).
you can update your code as follow:
n = int(input("Enter the amount of Prime Numbers: "))
FoundPrimes = 0
i = 2
while FoundPrimes < n:
isPrime = True
for j in range(2,1 + i//2):
if (i % j) == 0:
isPrime = False
if isPrime:
FoundPrimes += 1
print(i, end = '\t')
i += 1
I am building a prime generator (I know, another one and probably not a very good one at that, but that is for another question.) I am using a dictionary for my collection of primes and ignoring composites by dividing by the previous primes. However it doesn't seem to be iterating properly in the final stage of the function and I get a number of incorrect results. isWhole is a self explanatory call to another function. This is my code where x = the number of primes to be generated:
def prime_generator(x):
count = 2
counter = 2
p = {1: 2}
while len(p) <= x:
if count % 2 == 0:
count += 1
continue
test1 = (math.sqrt(count))
if isWhole(test1) == True:
count += 1
continue
for k, a in p.items():
if count % a == 0:
break
else:
p[ counter ] = count
counter += 1
break
count += 1
return p
Your design intent is not entirely clear, but you may be intending to have the else clause apply to the for loop rather than the if statement. Try un-indenting the entire else clause so that it only runs if the loop terminates without hitting a break.