I am using Jupyter Notebook, I keep getting the asterisk that indicates the kernel is busy when I run this specific code:
var = 2
var += 1
var_rem = var % 3
while var_rem == 0:
var += 2
print var
In order to give some context, I am trying to solve the following exercise:
Define a new number variable and choose a value for it. If the
variable + 1 can be divided by three, increase the variable by two.
Test by printing the final value of the variable and varying the
initial value of that same variable.
I have tried restarting the kernel as it was recommended in front of the asterisk issue but it doesn't work. What is specific about this code that the kernel cannot process it? How do I then solve the exercise?
Note: First post around here, I hope it's relevant.
Your code results in an infinite loop. Your variable var_rem does not change its value in the loop, therefore it runs forever (because it remains 0)
You have to recalculate the while condition within the loop.
Based on the statement your logic is wrong. Try this...
var = 2
if ((var + 1) % 3) == 0:
var +=2
print var
else:
print 'Not divisible by 3'
Related
I'm an absolute beginner, and I don't understand why a is not incremented every time I run the code. It always prints a = 0 and a = 1.
a = 0
b = 2
print ("a = {}".format(a))
a = a + 1
print ("a = {}".format(a))
Output
I'm working on Google Colab. Thanks in advance!
EDIT: I just noticed if I keep the two first lines on a separate block like this, the code does increment. Why??
Use two separate cells, one for declaring the variables and one for incrementing:
I might be asking a simple question. I have a python program that runs every minute. But I would like a block of code to only run once the condition changes? My code looks like this:
# def shortIndicator():
a = int(indicate_5min.value5)
b = int(indicate_10min.value10)
c = int(indicate_15min.value15)
if a + b + c == 3:
print("Trade posible!")
else:
print("Trade NOT posible!")
# This lets the processor work more than it should.
"""run_once = 0 # This lets the processor work more than it should.
while 1:
if run_once == 0:
shortIndicator()
run_once = 1"""
I've run it without using a function. But then I get an output every minute. I've tried to run it as a function, when I enable the commented code it sort of runs, but also the processing usage is more. If there perhaps a smarter way of doing this?
It's really not clear what you mean, but if you only want to print a notification when the result changes, add another variable to rembember the previous result.
def shortIndicator():
return indicate_5min.value5 and indicate_10min.value10 and indicate_15min.value15
previous = None
while True:
indicator = shortIndicator()
if previous is None or indicator != previous:
if indicator:
print("Trade possible!")
else:
print("Trade NOT possible!")
previous = indicator
# take a break so as not to query too often
time.sleep(60)
Initializing provious to None creates a third state which is only true the first time the while loop executes; by definition, the result cannot be identical to the previous result because there isn't really a previous result the first time.
Perhaps also notice the boolean shorthand inside the function, which is simpler and more idiomatic than converting each value to an int and checking their sum.
I'm guessing the time.sleep is what you were looking for to reduce the load of running this code repeatedly, though that part of the question remains really unclear.
Finally, check the spelling of possible.
If I understand it correctly, you can save previous output to a file, then read it at the beginning of program and print output only if previous output was different.
a = 1
for i in range(5):
browser.find_element_by_xpath("/html/body/div[6]/div/div/div[2]/div/div/div[1]/div[3]/button").click()
sleep(1)
I want to increase the 1 in div[1] by 1+ every loop, but how can i do that?
i thought i need to add a value, do "+a+" and last of all a "a = a + 1" to increase the value every time, but it didnt worked.
a = 1
for i in range(5):
browser.find_element_by_xpath("/html/body/div[6]/div/div/div[2]/div/div/div["+a+"]/div[3]/button").click()
a = a + 1
sleep(1)
for i in range(1,6):
browser.find_element_by_xpath("/html/body/div[6]/div/div/div[2]/div/div/div["+str(i)+"]/div[3]/button").click()
sleep(1)
you don't need 2 variables, just one variable i in the loop, convert it to string with str() and add it to where you need it, pretty simple. the value of i increases for every iteration of the loop going from 1 to 5 doing exactly what you need.
alternatively to Elyes' answer, you can use the 'global' keyword at the top of your function then a should increment 'correctly'.
You don't really need two variables for this unless you are going to use the second variable for something. However, look at the following code and it will show you that both i and a will give you the same result:
from time import sleep
a = 1
for i in range(1, 6):
path = "/html/body/div[6]/div/div/div[2]/div/div/div[{idx}]/div[3]/button".format(idx=i)
print(path, 'using i')
path = "/html/body/div[6]/div/div/div[2]/div/div/div[{idx}]/div[3]/button".format(idx=a)
a += 1
print(path, 'using a')
sleep(1)
Result:
/html/body/div[6]/div/div/div[2]/div/div/div[1]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[1]/div[3]/button using a
/html/body/div[6]/div/div/div[2]/div/div/div[2]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[2]/div[3]/button using a
/html/body/div[6]/div/div/div[2]/div/div/div[3]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[3]/div[3]/button using a
/html/body/div[6]/div/div/div[2]/div/div/div[4]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[4]/div[3]/button using a
/html/body/div[6]/div/div/div[2]/div/div/div[5]/div[3]/button using i
/html/body/div[6]/div/div/div[2]/div/div/div[5]/div[3]/button using a
You can read up on range here
I'm trying to search a parameter space for a good model. In my code I'm doing something like the following:
configurations = [None] * 9
index = 0
for param1 in range(3):
for param2 in range (3):
#call constructor
configurations[index] = Configuration(param1, param2)
index = index + 1
return configurations
then I use this array elsewhere to build the models.
However, for some reason it doesn't fill the array. I get an error when accessing index 3 or higher. Inspection shows that the array at these indices are still 'None'. while debugging this, I added a print statement to the code:
configurations = [None] * 9
index = 0
for param1 in range(3):
for param2 in range (3):
#call constructor
configurations[index] = Configuration(param1, param2)
index = index + 1
for x in configurations:
print(x.param1)
return configurations
to my surprise, my code now ran smoothly. The array is completely filled. If I comment out those 2 lines: error. uncomment the lines: it works again.
what is going on here? Why is the print statement altering the behavior of my program? is this an overactive GC that's staved off by printing the values?
I found the issue. the return statement was indented with a Tab, but I use 4 spaces everywhere else in my code. This was causing it to skip the most outer for loop. the print statement somehow prevented this. changing the tab to spaces removed the need for the print statement.
I'm trying to increase the count of an integer given that an if statement returns true. However, when this program is ran it always prints 0.I want n to increase to 1 the first time the program is ran. To 2 the second time and so on.
I know functions, classes and modules you can use the global command, to go outside it, but this doesn't work with an if statement.
n = 0
print(n)
if True:
n += 1
Based on the comments of the previous answer, do you want something like this:
n = 0
while True:
if True: #Replace True with any other condition you like.
print(n)
n+=1
EDIT:
Based on the comments by OP on this answer, what he wants is for the data to persist or in more precise words the variable n to persist (Or keep it's new modified value) between multiple runs times.
So the code for that goes as(Assuming Python3.x):
try:
file = open('count.txt','r')
n = int(file.read())
file.close()
except IOError:
file = open('count.txt','w')
file.write('1')
file.close()
n = 1
print(n)
n += 1
with open('count.txt','w') as file:
file.write(str(n))
print("Now the variable n persists and is incremented every time.")
#Do what you want to do further, the value of n will increase every time you run the program
NOTE:
There are many methods of object serialization and the above example is one of the simplest, you can use dedicated object serialization modules like pickle and many others.
If you want it to work with if statement only. I think you need to put in a function and make to call itself which we would call it recursion.
def increment():
n=0
if True:
n+=1
print(n)
increment()
increment()
Note: in this solution, it would run infinitely.
Also you can use while loop or for loop as well.
When you rerun a program, all data stored in memory is reset. You need to save the variable somewhere outside of the program, on disk.
for an example see How to increment variable every time script is run in Python?
ps. Nowadays you can simply do += with a bool:
a = 1
b = True
a += b # a will be 2