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'
:-)
Related
This question already has answers here:
Print out message only once from the for loop
(4 answers)
Closed 5 years ago.
how do I get my else print to only print once instead of for every row that the string doesn't exist? I tried moving it around by tabbing back a couple of layers but it doesn't work. I understand the logic, I but don't know how to limit it. I'm adding a little bit at a time to my parsing scripts for practice, learning as I go, but this one got me. Thanks!
import csv
# Testing finding something specifical in a CSV, with and else
testpath = 'C:\Users\Devin\Downloads\users.csv'
developer = "devin"
with open (testpath, 'r') as testf:
testr = csv.reader(testf)
for row in testr:
for field in row:
if developer in row:
print row
else:
print developer + " does not exist!"
In Python, you can have an else clause attached to your for loop. For example
>>> for i in range(10):
... if i == 5: break # this causes the else statement to be skipped
... else:
... print 'not found'
...
Note 5 was found so the else statement is not executed
>>> for i in range(10):
... if i == 15: break
... else:
... print 'not found'
...
not found
See the documentation on for statements
A break statement executed in the first suite terminates the loop
without executing the else clause’s suite. A continue statement
executed in the first suite skips the rest of the suite and continues
with the next item, or with the else clause if there is no next item.
See Gibson's answer first. You can do this:
for row in testr:
found = False
for field in row:
if developer in row:
print row
found = True
break
if found: break
else:
print developer + " does not exist!"
You can also omit the found flag (as suggested by Jean-François Fabre in the comment) but this makes a bit hard to understand imo (I had to compile in my head):
for row in testr:
for field in row:
if developer in row:
print row
# We found the developer. break from the inner loop.
break
else:
# This means, the inner loop ran fully, developer was not found.
# But, we have other rows; we need to find more.
continue
# This means, the else part of the inner loop did not execute.
# And that indicates, developer was found. break from the outer loop.
break
else:
# The outer loop ran fully and was not broken
# This means, developer was not found.
print developer, "does not exist!"
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.
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.
I'm doing an assignment in which I have to move a simulated robot across the screen in a loop - I've got that part down, however, between loops, I also have to print the percentage of area covered with each movement - that's my issue.
I googled a bit and even found someone with the same problem, however, I'm not sure if I'm doing it properly.
This code was offered:
percent_complete = 0
for i in range(5):
percent_complete += 20
print('{}% complete'.format(percent_complete))
However, after an error, further googling revealed that only worked with certain versions
so I used this code:
percent_complete = 0
for i in range(5):
percent_complete += 20
print '% complete' % (percent_complete)
And, at the very least, it now executes the code, however, the output when printing is the following:
Here we go!
hello
omplete
hello
(omplete
hello
<omplete
hello
Pomplete
hello
domplete
What is the cause of this? I assume because one of the codes had to be edited, the other parts do as well, but I'm not sure what needs to be done.
for i in range(5):
percent_complete += 20
print '%d complete' % (percent_complete)
You were missing the d specifier.
The first version only works in Python 3 because it uses print as a function. You're probably looking for the following:
percent_complete = 0
for i in xrange(5):
percent_complete += 20
print '{0} complete'.format(percent_complete)
Your other code doesn't do what you intend to do because it now display the number as a string. What you want is that the number is properly converted to a string first and then displayed in the string. The function format does that for you.
You can also use Ansari's approach which explicitly specifies that percent_complete is a number (with the d specifier).
To add to/correct above answers:
The reason your first example didn't work isn't because print isn't a function, but because you left out the argument specifier. Try print('{0}% complete'.format(percent_complete)). The 0 inside the brackets is the crucial factor there.
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 :)