I have a problem in Python language that is described in a title.
for slovo in slova:
if pygame.mouse.get_pressed()[0] and slovo["rect"].collidepoint(pygame.mouse.get_pos()):
for i in range (len(randRijec)):
if slovo["name"] in randRijec[i]:
if i == 0:
slovo1 = randRijec[i].upper()
prvoSlovo = 1
...
...
else:
pogresnoBrojac += 1
slova.remove(slovo)
So, even this IF statement is true, ELSE statement is being executed! However, else statement should be skipped if the if statement is fulfilled.
How to fix this issue?
p.s. I've had this problem few times before and I was not able to solve it...
You have a mixture of tabs and spaces in your code:
Running cat -A test.py (on Unix) yields
for slovo in slova:$
if pygame.mouse.get_pressed()[0] and slovo["rect"].collidepoint(pygame.mouse.get_pos()):$
for i in range (len(randRijec)):$
if slovo["name"] in randRijec[i]:$
if i == 0:$
slovo1 = randRijec[i].upper()$
prvoSlovo = 1$
^I^I^I^I^I^I...$
^I^I^I^I^I^I...$
else:$
pogresnoBrojac += 1$
slova.remove(slovo)$
The ^I indicate tabs.
Thus, the else block is not being interpreted as being at the indentation level on which it appears to be.
Your python code should never mix tabs and spaces for indentation. You can check that your script is not mixing tabs and spaces by running python -t script.py.
In Python you must commit to using either only spaces or only tabs for indentation. PEP8 recommends spaces-only indentation.
You can convert tabs to spaces using the reindent.py program.
So, even this IF statement is true, ELSE statement is being executed!
I can assure you that this is not what happens.
I notice that in the outline of your code the if is inside a for loop. Make sure that in your actual code the else is not accidentally lined up with the for instead of the if. I've seen this mistake more than once.
In Python, for-else is a valid construct. For example, the following is perfectly valid Python:
for i in range(10):
if i < 100:
pass
else:
print 'In else clause'
When run, this prints out In else clause.
Contrast this with the following, which doesn't print anything when run:
for i in range(10):
if i < 100:
pass
else:
print 'In else clause'
It's a question from a long time ago and I stumbled upon it as I was troubleshooting the very same issue - the solution was actually pretty silly and most probably was also the case - as it's a for loop it iterates through every list element, if even one of those elements doesn't fulfill the if condition, it will automatically trigger the else - pretty self-evident but easy to miss for beginners.
Well at least that was the problem in my case :)
Next solution fixed my problem:
for slovo in slova:
if pygame.mouse.get_pressed()[0] and slovo["rect"].collidepoint(pygame.mouse.get_pos()):
xy = 0
for i in range (len(randRijec)):
if slovo["name"] in randRijec[i]:
xy = 1
if i == 0:
slovo1 = randRijec[i].upper()
prvoSlovo = 1
break
if i == 1:
slovo2 = randRijec[i].upper()
drugoSlovo = 1
break
slova.remove(slovo)
if xy == 0:
pogresnoBrojac += 1
...
...
...
xy = 1
pygame.display.update()
time.tick(value)
So, I have just added that xy counter that makes my code work how it should work. When IF statement is met, xy becomes 1, if IF statement isn't fulfilled, xy is set to 0 and then this "else" statement executes. At the end of the code, xy is set to 1 again to prevent executing this "else" (if xy == 0) block.
Related
I'll try to make a python program, convert roman to number. I believe my logic for the program is correct, but I get syntax error. maybe someone want's help me to fixed my program.
this my whole program:
bil = int(input())
if(bil<1 or bil>99):
print("no more than 3999")
else:
while(bil>=1000):
print("M")
bil-=1000
if(bil>=500):
elif(bil>500):
elif(bil>=900):
print("CM")
bil-=900
else:
print("D")
while(bil>=100):
if(bil>=400):
print("CD")
bil-400
else:
bil-=100
if(bil>=50):
elif(bil>=90):
print("XC")
bil-=90
else:
print("L")
bil-=50
while(bil>=10):
if(bil>=40):
print("XL")
bil-=40
else:
print("X")
bil-=10
if(bil>=5):
elif(bil==9)
print("IX")
bil-=9
else:
print("V")
bil-=5
while(bil>=1):
if(bil==4):
print("V")
bil-=4
else:
print("I")
bil-=1
I got syntax error in line :
elif(bil>500):
I need your opinion, thank you.
It shouldn't be elif, it should if bil>500. Because, you are trying to create a nested if condition and not an if/elif/else condition. So the final code in that block should be:
if(bil>=500):
if(bil>500):
if(bil>=900):
print("CM")
bil-=900
Also, I don't understand why you are comparing bil>500 two times at the same time. You could remove one if statement there
And there are many such if/elif blocks out there. You need to replace elif with if, if there is an indentation or you need to remove indentation for elif condition and write a condition for the if block too
I have a complex project and I will try to simplify one of the main problem of the project. So there is the simplification:
We can Imagine a while loop like this:
while(condition):
statement1
statement2
statement3
...
statementn
In this loop there n statements, and each statement can be whatever(function, loop, if statement,...) and there is a condition in the loop, this condition i want to check it BEFORE the while loop do it. Because if the condition is respect since the first statement I have to wait until the end of the while to check if the condition is respect... So there is my question is possible to check the condition BEFORE the loop without have a check-function between EACH statements of the whileloop ?
Because in fact, it's work... BUT the code isn't clear, I really think this way we pollute my code and i want to work more efficiently and with a beautiful code, so how can I solve my problem without this constraint ?
PS: I think about event listener like javascript but i found poor information about them on python, but if there is a tool which act like event listener it would be great !
It sounds like you want to clean up all your if-then-break statements into a single function that handles the "checking" of the value of a. For that purpose you could use exceptions:
import random
class ItIsFiveException(Exception): pass
def check(a):
if a == 5:
raise ItIsFiveException
try:
a = 0
while(a != 5):
a = random.randint(1,5); check(a)
a = random.randint(1,5); check(a)
a = random.randint(1,5); check(a)
a = random.randint(1,5); check(a)
except ItIsFiveException:
print("I saw a five!")
You just have to define your own python Exception as a class, and the raise it in your manually-defined check(a) function. Then you can wrap your entire while loop in a try-except block and catch your exception.
I am not sure if I understand you right, but this is what I'd do:
flag = False
while not flag:
for i in range(4):
a = random.randint(1, 5)
if a == 4:
flag = True
break
I don't know exactly what it happens with "a" but if if you can chain the conditions and will stop when the first one fails
while(checkRandom()):
...
def checkRandom(a):
return random.randint(1,5) == 5 and random.randint(1,5)....
If you can loop the generation of random values you can use
while(a!=5):
for item in range(1, 5):
a=random.randint(1,5)
if a==5:
break
I tried this... but it doesn't work
question = input("do you want the program to start? Type Y/y to start: ")
y = TRUE
Y = TRUE
if(question == TRUE):
run statements
else:
what am i doing wrong? this doesn't work.
To answer your specific question, it is not working because of these issues:
TRUE is not a defined variable in python. True is.
question == TRUE won't work. See 1.
run statements isn't real code.
You need to put something in your else: block.
EDIT:
question can never become True in this code. – #adsmith
NOTE:
Just trying to be comprehensive with my coverage.
The boolean is True in Python, not true or TRUE. In any case, this doesn't do what you expect it to. This is what I'd do.
question = input("...")
if question.lower() == 'y': # or `question in ('y','Y'):` or `question.upper() == "Y":` or `question.casefold() == 'y':` or................
do_things
else:
handle_it
What you had written assigns the variables y and Y to some (undefined) variable TRUE. This will trigger a NameError since there is no such variable TRUE. If you had done:
y = True
Y = True
It still wouldn't have done what you wanted, since your input (fed into the variable question) is a string and those are variables. You could have done that with if globals()[question] but that's really bad practice, and COMPLETELY unnecessary in this situation.
As a side note -- there's never a reason to type == True. if foo will evaluate to True or False, which will fulfill the conditional on its own. It just does a needless compare :)
I think you probably want to use code something along these lines:
answer = raw_input("Do you want the program to start? Type Y/y to start: ")
if answer[0].lower() != "y": # first character not a "Y" or "y"?
exit()
rest of program...
1) You have five (5) spaces indenting your if clause. Should follow Generally Accepted Python Practices (GAPP) ;) (Yes, I just made this up. It may or may not become a thing :p) you should use four (4) spaces.
2) Try adding pass after else:
else:
pass
3) In Python, case matters. As such, boolean testing must be True or False (or 1/0 :p)
4) Don't you mean Y/N? Not Y/y?
I have some if statements like:
def is_valid(self):
if (self.expires is None or datetime.now() < self.expires)
and (self.remains is None or self.remains > 0):
return True
return False
When I type this expressions my Vim automatically moves and to new line with this same indent as if line . I try more indent combinations, but validating always says thats invalid syntax. How to build long if's?
Add an additional level of brackets around the whole condition. This will allow you to insert line breaks as you wish.
if (1+1==2
and 2 < 5 < 7
and 2 != 3):
print 'yay'
Regarding the actual number of spaces to use, the Python Style Guide doesn't mandate anything but gives some ideas:
# No extra indentation.
if (this_is_one_thing and
that_is_another_thing):
do_something()
# Add a comment, which will provide some distinction in editors
# supporting syntax highlighting.
if (this_is_one_thing and
that_is_another_thing):
# Since both conditions are true, we can frobnicate.
do_something()
# Add some extra indentation on the conditional continuation line.
if (this_is_one_thing
and that_is_another_thing):
do_something()
put the linebreaks inside brackets
if ((....) and (...)):
You could invert the tests and return False on sub-sets of the test:
def is_valid(self):
if self.expires is not None and datetime.now() >= self.expires:
return False
if self.remains is not None and self.remains <= 0:
return False
return True
This way you can break up the long line of tests and make the whole thing a lot more readable.
Yes, you can use additional parentheses around your boolean tests to allow for newlines in the test, but readability suffers greatly when you have to go across multiple lines.
When Python is first installed, the default setting executes users' code input line-by-line. But sometimes I need to write programs that executes multiple lines at once. Is there a setting in Python where I can change the code execution to one block at once? Thanks
>>> if (n/2) * 2 == n:;
print 'Even';
else: print 'Odd'
SyntaxError: invalid syntax
When I tried to run the above code, I got an invalid syntax error on ELSE
Your indentation is wrong. Try this:
>>> if (n/2) * 2 == n:
... print 'Even'
... else: print 'Odd'
Also you might want to write it on four lines:
>>> if (n/2) * 2 == n:
... print 'Even'
... else:
... print 'Odd'
Or even just one line:
>>> print 'Even' if (n/2) * 2 == n else 'Odd'
One step towards the solution is to remove the semicolon after the if:
if True:; print 'true'; print 'not ok'; # syntax error!
if True: print 'true'; print 'ok'; # ok
You cannot have an else in the same line because it would be ambiguous:
if True: print 't'; if True: print 'tt; else: ... # <- which if is that else for??
It is also clearly stated in the docs that you need a DEDENT before the else statement can start.
Since python 2.5 you can do one line ifs
print ('Even' if n % 2 == 0 else 'Odd')
Still to answer your question you can either:
1. enter the code properly without syntax errors and your blocks will be executed as blocks regardless if they span multiple lines or not, even in interactive shell. See tutorials in dive into python
2. write code in the script and execute that script using either command line or some IDE (idle, eclipse, etc..)
One of the idea behind python is to prefer multiple lines and to aim for uniform formatting of source, so what you try to do is not pythonic, you should not aim to cram multiple statements into single line unless you have a good reason.
print n % 2 == 0 and 'Even' or 'Odd'
:-)