This is my code in python3:
import heapq
myQueue = []
n = raw_input()
try:
num = int(n)
if num<=100000 :
arr = input().split()
for i in range(num):
heapq.heappush(myQueue, arr[i])
print(myQueue[0])
except (NameError, ValueError):
print("Wrong Input, N should be under 100000")
except IndexError:
print("Inputs is less than actually required")
except EOFError:
print ("Error: EOF or empty input!")
I am trying to implement priority queue.
But I am facing this EOF Error while solving this problem on GUVI.
Output:
Error: EOF or empty input!
The tried to catch the error using except EOFError, but that will just make my program run but does not solve input problem right.
I even tried to run this piece of code on Sublime text editor as well as Vs code,
where it runs just fine, Correct Output.
I don't understand, is there a problem in my code or That Online Platform.
I even tried to search the answer on their Q&A platform of GUVI, I found similar question but no one has answered it.
And this is not just for this piece of Code but I found the same error for many before.
Could ANYONE help me, Please!
Thanking you in advance.. :)
EOF error occurs if the input is not provided.
In case of most of the online compilers you need to provide the inputs before running the code.
With that said, while your are trying to access the input via raw_input() there will be no input supplied resulting in the above error.
To avoid this, error supply inputs before running your code like below
Also, I can notice that you are using raw_input() and input(). Please note that raw_input() can be used if you are using Python 2 and input() if you are using python 3 correspondingly.
Related
Im fairly new to coding, and Im practicing along the Automate the boring stuff with python book when I came across the collatz sequence problem.
As you guys can see from my code, in the 1st time, I typed "puppy" and it said "NameError" and ended the run altogether without returning to the input phase. In the 2nd time, I typed "89", and then "puppy", and it printed out the valueerror text along with the result from last time.
How can I fix this for both the times? Thank you guys.
Basically, the value of 268 came from the last calculation and wasn't overwritten due to the exception caught with "puppy". To fix that, simply use the continue keyword to loop back to the "try" condition:
except ValueError :
print('You must enter an integer')
continue
This piece of code is supposed to find account balance after withdraw from bank(fee=0.5).
wd,ac=raw_input().split(" ")
wd=int(wd)
ac=float(ac)
if(ac<wd):
print(ac)
elif(wd%5==0):
if(ac>wd+0.50):
print(ac-wd-0.50)
else:
print(ac)
else:
print(ac)
I got a Runtime NZEC Error while submitting on codechef. I am newbie and had searched for resolve and had used split(" ") in place of int(input()), which I had tried previously, but the problem still occurs.
Geeksforgeeks says:
"In python, generally multiple inputs are separated by commas and we read them using input() or int(input()), but most of the online coding platforms while testing gives input separated by space and in those cases int(input()) is not able to read the input properly and shows error like NZEC"
Given that I've tried to account for that... What is wrong with my code?
It looks like an error with your raw_input statement. Remember that, in python 3, raw_input doesn't exist any more. If you changed it from raw_input to input, then the code works just fine.
I am getting the above error in python.
w=[]
h=[]
l=int(input())
t=int(input())
for i in range(t):
n=raw_input()
w.append(n)
n=int(input())
h.append(n)
for i in range(t):
if w<l and h<l:
print ("UPLOAD ANOTHER")
elif w>l and h>l and w!=h:
print ("CROP IT")
elif w>=l and h>=l and w==h:
print("ACCEPTED")
You probably given a empty input or '\n' for the n=int(input()) part. Try to give the number as input.
put this line in your terminal
x = int(input()). and give empty input, (just tap enter). vola..
SyntaxError: unexpected EOF while parsing
You might be running the script on some online IDE. This also appeared to me while I was submitting my solution. The best way to deal with this is don't run your script on online ide rather directly submit it.
If the answer is correct it would get accepted else show the wrong answer.
This may be repeated, but none of the existing answers solved my problem.
So, I'm using Python 2.7, and I get this error (title) whenever I try this:
number = int(raw_input('Number : '))
I tried this in Sublime Text 2, compileronline.com and in codecademy; it fails in the first 2 of this sites. It works on codecademy and in the terminal compiler, but I can't understand exactly why it is failing.
The issue here is that Sublime text 2's console doesn't support input.
To fix this issue, you can install a package called SublimeREPL. SublimeREPL provides a Python interpreter that takes in input.
And as for compileronline.com, you need to provide input in the "STDIN Input" field on the lower right of the website.
try:
value = raw_input()
do_stuff(value) # next line was found
except (EOFError):
break #end of file reached
This seems to be proper usage of raw_input when dealing with the end of the stream of input from piped input.
Refer this post
One of my early courses in the University I attend, was some basic training in Python 3 years ago. Now I was looking for a program that could help me resize some Grid stuff and I found something that could help me in Python. I reinstalled Python to my PC and found my old editor. However when I run the code I get an invalid syntax error that I can't understand. This is the part of the code that the error appears in :
def downsize(mode, cell_size, inpath, outpath):
from VolumeData import fileformats
try:
grid_data = fileformats.open_file(inpath)
except fileformats.Uknown_File_Type, e:
sys.stderr.write(str(e))
sys.exit(1)
reduced = Reduced_Grid(grid_data, mode, cell_size)
from VolumeData.netcdf.netcdf_grid import write_grid_as_netcdf
write_grid_as_netcdf(reduced, outpath)
The exact invalid syntax error is in the "except fileformats.Uknown_File_Type, e:" line. Can you help me ?
If you are using Python 3.x, you cannot use except fileformats.Uknown_File_Type, e. The comma works as an as statement (in the try/except block), so you should replace it with: except fileformats.Uknown_File_Type as e.
The comma works in Python 2.7, but not 3.x. However, the as should work for both.
Reference: Handling errors in Python 3.3
Maybe you misspelled 'Unknown'?