I write many print(...) lines in the code to verify the middle results.
These are handy when you need them, but when everything works well, I need to commit/delete each one of them. Sometimes I found that I may need them again... then I have to go through each print(...) line again.
So is there a switch, that could activate & deactivate those tmp-print lines? While not influencing other print lines that I always want to activate.
Ps: No one wants to do it with if in each line:
if I_want_tmp_print:
print("one of my 100 tmp print lines")
Sometimes if I print enough and I want the debugging 'toggleable' I will write my own function such as:
def prnt(string):
print(string)
return
So then wherever I wan to check my code I will use my function, prnt, instead. And it will display all of the output for me. When I no longer want the output I comment out the print(string) line so nothing is outputted.
If you only want to stop commenting out certain sections of code I will block my code calling a similar function but with a reference number indicating where I have used this in my code.
def prnt(string, idx):
if idx = 1:
print(string)
elif idx = 2:
print(string)
elif idx = 3:
print(string)
...
return
So then say I found out that where I was testing in the idx=1 range and I was calling prnt('string to print', 1) was working as I wanted it to, I can just comment out that print line in my function.
The last five or so lines of my code have an if statement and that doesn't print in my code.
I tried to change the numbers, and add new conditions, however, it still does not print. I even tried to remove the rounding for the if statement and it still does not print. I'm confused as to why the if statement at the bottom does not print.
Because your logic in the if elif statements is the same...
change one.
Edit:
RoundedExpectedReturnE > ARE == ARE < RoundedExpectedReturnE:
Started learning python and trying to learn try blocks. However I am utterly confused to why this program creates an unident error (does not match other identation levels).
while True:
try:
yes=(int(input("how smart is neervesh")))
if neervesh == 0:
print("yes is smart")
elif neervesh ==1:
print("yeas")
else:
print("kill me now")
except ValueError:
print("dont be retarded like yes")
Your try is indented by 3 spaces, your except by 4. You should stick to one indentation width (usually 4).
I always use the program 'SublimeText'. This text editor gives you the possibility to change your indentation on the fly. You can either convert your indentation to tabs or spaces. This way, you'll likely prevent future
indentation errors.
This Python script to process a string is suffering from Syntax error in line 24, else:.
Any ideas as to what it might be?
j=raw_input("Enter a string: ")
import os
def addtoClipBoard(text):
command = 'echo ' + text.strip() + '| clip'
os.system(command)
def parse(string):
result=""
lineList=string.split("\n")
for i in range(len(lineList)):
h=lineList[i].split("#")
if len(h)<2:
continue
if len(h)>2:
count=0
for x in range(len(h)):
if x==len(h)-1:
continue
re0=count+len(h[x])+(x*1)
re1=i+1
re3=str(re1)+"-"+str(re0)
result+=str(re3)+", "
count+=len(h[x])
else:
re0=len(h[0])
re1=i+1
re3=str(re1)+"-"+str(re0)
result+=str(re3)+", "
result =result[:-2]
addtoClipBoard(result)
print result
parse(j)
It's probably a problem with the indentation level of the else, make sure to align it at the same level of the corresponding if. Please use a good IDE or text editor to help you catch this kind of errors. As it is, it's nearly impossible to se what you intended to do with the code.
Python uses indentation to define what is inside of a block of code (for,if,elif,while,with) You want to keep everything at the same indentation within the block and then go back a level to write your else statement
Also it doesn't look like your else: statement even belongs there, there is no if statement preceding it that would imply the need for an else are you sure that else is what you want?
Here is my suggestion without knowledge of what you want your code to do, based on observation of patterns
for x in range(len(h)):
if x==len(h)-1:
continue
if (###My condition goes here###):
re0=count+len(h[x])+(x*1)
re1=i+1
re3=str(re1)+"-"+str(re0)
result+=str(re3)+", "
count+=len(h[x])
else:
re0=len(h[0])
re1=i+1
re3=str(re1)+"-"+str(re0)
result+=str(re3)+", "
result =result[:-2]
addtoClipBoard(result)
print result
parse(j)
I have no idea how to fix this. I've tried retyping the program.
I get an unexpected indentation error for the last main function.
resident = 81
nonresident = 162
def main():
# initialize counters and total tuition
resident_counter = 0
nonresident_counter = 0
total_tuition = 0
print("Name \tCode\tCredits\tTuition")
print
try:
# open the data file
infile = open('enroll.txt', 'r')
# read the first value from the file
student_name = infile.readline()
# continue reading from file until the end
while student_name != '':
# strip the new line character and print the student's name
student_name = student_name.rstrip('\n')
print(student_name, end='\t')
# read the code type, strip the new line, and print it
code = infile.readline()
code = code_type.rstrip('\n')
print(code_type, end='\t')
# read the number of credits, strip the new line, and print it
credits = infile.readline()
credits = int(credits)
print(format(credits, '3.0f'), end='\t')
# check the room type and compute the rental amount due
# increment the appropriate counter
if code_type == "R" or room_type == "r":
payment_due = credits * resident
resident_counter += 1
elif code_type == "N" or room_type == "n":
payment_due = credits * nonresident
nonresident_counter += 1
elif code_type != "R" or code_type != "r" or code_type != "N" or code_type != "n":
payment_due = 0
# accumulate the total room rent
tuition += payment_due
# print the appropriate detail line
if payment_due == 0:
print('invalid code')
else:
print('$', format(tuition, '8,.2f'))
# get the next studen't name
student_name = infile.readline()
# close the input file
infile.close()
# print the counters and payment total amount
print
print('total number of resident students: ', resident_counter)
print('total number of nonresident: ', nonresident_counter)
print
print('total students: ', end='')
print('$', format(tuition, ',.2f'))
# execute the main function
main()
You don't have an except clause to match the try.
Despite what everyone else is saying, if you have a try, you don't have to have an except: you must have an except or a finally.
That may seem nitpicky, but it's pretty plausible that the code you were copying was actually using finally (e.g., to make sure some cleanup code always gets run—given that it's doing C-/Java-style manual cleanup).
At any rate, adding an except clause is not the right answer. If you're actually looking to handle or ignore errors, then yes, add an except clause. If you're looking for somewhere to add cleanup code that gets run even on an error, add a finally clause instead. If you don't want either, just remove the try line.
you are mixing room type and code type in your if else statement. You are also missing your except statement. This should be your last two lines before calling main.
It should read:
except IOError:
print('an error occurred trying to open or read enroll.txt')
You are using the try statement. It means that you MUST have an except block
try:
some code
except:
pass
It is simplification, but will solve your problem.
The other answers have (correctly) pointed out that you need to have an except to go with your try: (or not use a try altogether). I'm going to try to explain how you could go about debugging similar issues in the future, since these can be very annoying to debug.
Your error message most likely looked something like this:
File "test.py", line 74
main()
^
IndentationError: unexpected unindent
At first glance, that doesn't seem too helpful, and it's definitely not very clear.
The fact that you "tried retyping the program" sounds like you saw IndentationError and thought, "this means my indents are messed up; maybe there's an extra space somewhere or a mixture of spaces and tabs or something." Those are definitely valid IndentationErrors, and retyping the program would fix them, if you didn't make the same typing error again. (On a side note, I believe that any text editor worth using will have the option to convert all of your tabs to 4 spaces to avoid a lot of indent headaches.)
However, looking at the error text, we see "unexpected unindent". To me this sounds kind of like gibberish, but what it means is that, when python was looking at your program, it ran into a line that it thought should be indented more than it was. (Unfortunately, this same kind of error can be called "unexpected unindent" or "expected an indented block", and it can even show up as a plain old SyntaxError: invalid syntax.)
So knowing what the error message means, we can look back through the code to see why python thought that main() was indented strangely- though we have to be careful because python isn't always right about where the problem actually is. Here would be something like my thought process:
There's a function definition right before main() is called, maybe that function wants main() to be inside it? No that isn't it, because python only really expects at least one line of code after a function definition.
What else "wants" code to be indented after it? if, elif, and while all do, but all of them have code after them as well.
The only other "indenter" is try:. At this point, you either know that try: needs to be followed by a except: or a finally: and you fix it, or you don't. If you don't know that, than you could still see that try: is followed by an indented block, and, given that it is a likely culprit for you error, decide that it's time to look up what try: does.
Well I hope that helps troubleshooting problems like this in the future, and check out abarnert's answer and my link for more on handling errors with try/except/else/finally statements.
As noted by LevLevitsky and david above, there's no except clause. The easiest way to solve this is adding a line like the following to your code before the call to main:
except: pass
The second comment I have to your code is more stylistic, but you could add the following, before your call to main and after the except clause:
if __name__ == '__main__':
main()
Besides the missing except or finally or the extra try... If you have a look at the edited code, you can see the strangely mixed color blocks in the indentation. It means that you are mixing tabs and spaces which can lead to problems like bad indentation. The reason is that the interpreter thinks differently about how the tabs should be interpreted (for example 8 vs. 4 column tab stops).
You must not mix tabs and spaces for the indentation. I personally would recommend to use only the spaces. Any decent editor is capable to expand TAB key to spaces and/or Untabify the existing mixture of tabs and spaces.