hi I m wrtng a very simple Python Program to implement Binary Search.
tup=input("enter tup:")
start=0
length=len[tup]
end=tup[length-1]
mid=(int(start)+int(end))/2
key=input("enter value to search")
def search(start,end,key):
if key==tup[mid]
print mid
else if key<tup[mid]
search(start,mid,key)
else if key>tup[mid]
search(mid,end,key)
else
return(-1)
I get an error as
File "binsearch.py", line 8
if key==tup[mid]
^
SyntaxError: invalid syntax
I believe I m missing something trivial but unable to figure out.! Let me know if u feel there are any other errors. thanks :)
if key==tup[mid]
^
needs a : at the end
|
v
if key==tup[mid]:
Same problem in the rest of the statement:
else if key<tup[mid]
^
search(start,mid,key)
else if key>tup[mid]
^
Aside:
Instead of else if consider using Python's neat elif construct, e.g.,
elif key<tup[mid]:
etc.
You need to end all statements that begin a new block with : (i.e. the statements where you increase the indentation level in the next line)
You need to replace else if X with elif X:
You should use raw_input instead of input as the latter evals whatever the user entered.
return is a statement, not a function, so you do not need () around the return value.
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 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.
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.
I get a syntax error in the following code:
if value[0] == "ta" or "su":
num_var = len(value)
i = 0
while value[i][0] != "-" and i <= num_var:
if i == 0 and value[0][0].isdigit():
f3["var_%s" %i] = VARFD[[value[0].split("/")[1]]
else:
f3["var_%s" %i] = VARFD[[value[0]]
f4["val_%s" %i] = "T"
i += 1
it claims that the syntax error is on line that starts with "else:". What's wrong with it?
Is your supply of new lines limited or why are you writing code like this?
Your error is here, one ] is missing:
VARFD[[value[0].split("/")[1]]
You're missing a square bracket in the
if i == 0 and value[0][0].isdigit(): f3["var_%s" %i] = VARFD[[value[0].split("/")[1]]
line. But Python code really isn't meant to be this densely written. Space and light!
It's as simple as that you're missing an end bracket on the line before the else.
VARFD[[value[0].split("/")[1]]
I suspect the expression should be
VARFD[value[0].split("/")[1]]
It's pretty much sure sign that you should break apart and simplify your code when errors like this show up :)
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'
:-)