Re: [Tutor] Terminology question
On 14 May 2015 at 16:47, Alan Gauld wrote: > Rather than printing the messages you could re-raise > the error but with the error message attached as a string. > I'm a little unclear how you catch the string you create in a raise, in the caller. I tried an example from the docs but it didn't work. Could you provide a simple example? Sometimes the docs are heavy slogging if you don't already know what's what ;') -- Jim ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Terminology question
On 15/05/2015 20:02, Jim Mooney Py3.4.3winXP wrote: On 14 May 2015 at 16:47, Alan Gauld wrote: Rather than printing the messages you could re-raise the error but with the error message attached as a string. I'm a little unclear how you catch the string you create in a raise, in the caller. I tried an example from the docs but it didn't work. Could you provide a simple example? Sometimes the docs are heavy slogging if you don't already know what's what ;') Please help us to help you, if you were to show the example from the docs that "didn't work" we could easily clarify the situation for you. -- My fellow Pythonistas, ask not what our language can do for you, ask what you can do for our language. Mark Lawrence ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
[Tutor] Help!
Hi, I'm a beginner and I've reached a roadblock. I'm trying to create a
simple guessing game program just to practice creating loops, however it is
not working out as planned.
print ("Lets play a game")
import random
# Generate random number from 1-10
rand_value = random.randint(1,10)
guess = input("Guess a number from 1-10 ")
if guess == rand_value:
print ("Congratulations! You guessed it!")
while guess != rand_value:
input ("try again")
else:
print ("Sorry, you're wrong.")
input ("\n\nBetter luck next time. Press the enter key to exit.")
The goal is for the program to keep running until someone successfully
guesses the right number, however I think I've accidentally created an
infinite loop. Any help will be greatly appreciated.
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[Tutor] Connecting Py 2.7 with visa
Greetings I am trying to write a test executive program using python 2.7 on a
windows 7 computer. I want to connect to a Keithley 2100 voltmeter using
National Instruments VISA. I am having trouble installing pyvisa. All the
documentation refers to using 'pip' and a command line "$ pip install pyvisa" .
What interface or console is this? "$" prompt looks like a Linux command line.
How do we do this with windows?
Do I have to do this? Or can I use the native visa module in Python 2.7? Using
the help() and dir() features I can get some basic information about visa, but
the functions have changed, like visa.ResourceManager.open_resource is not
working. I really liked this function... Are there any examples of how to use
this new visa? I have some working code below that uses pyvisa, can it be
converted?
def update_current():
import visa
rm = visa.ResourceManager()
rm.list_resources()
current_1_ma = ""
exe_check = "PASS"
try:
dut_data = open("dut_data.txt", "w")
except:
exe_check = "FAIL"
try:
ki2100 = rm.open_resource('USB0::0x05E6::0x2100::1148525::INSTR')
device_id = ki2100.query("*IDN?")
except:
exe_check = "FAIL"
try:
dut_current_amps = (float(ki2100.query("MEASure:CURRent:DC?")))
dut_current_ma = dut_current_amps * 1000.0
current_1_ma = "%6G" % dut_current_ma
except:
exe_check = "FAIL"
new_line = "Litepoint_Data_Format" + "\r\n"
dut_data.write(new_line)
new_line = "CURRENT_1_MA=" + current_1_ma + "\r\n"
dut_data.write(new_line)
new_line = "EXE_CHECK=" + exe_check + "\r\n"
dut_data.write(new_line)
dut_data.close()
return
if __name__ == "__main__":
update_current()
print "Dumping dut_data.txt"
with open('dut_data.txt') as dut_data:
for line in dut_data:
print line,
if 'str' in line:
break
dut_data.close()
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Re: [Tutor] Help!
On Fri, May 15, 2015 at 07:35:30PM -0400, Fredrick Barrett wrote:
> print ("Lets play a game")
>
> import random
>
> # Generate random number from 1-10
> rand_value = random.randint(1,10)
> guess = input("Guess a number from 1-10 ")
Here you guess once.
> if guess == rand_value:
> print ("Congratulations! You guessed it!")
If you guessed correctly, the game ends.
> while guess != rand_value:
> input ("try again")
> else:
> print ("Sorry, you're wrong.")
You ask the user to try again, but you pay no attention to their reply.
You just completely ignore their guesses, apart from the very first one.
Also, and by the way, although Python does have a "while...else"
construct, it's a bit misleading and I don't think it is useful in this
case.
Experiment with something like this instead:
rand_value = random.randint(1,10)
guess = 0 # Guaranteed not to match.
while guess != rand_value:
guess = input("Guess a number between one and ten: ")
print("Congratulations!")
Notice that each time through the loop, I set guess to a new value?
--
Steve
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Re: [Tutor] Connecting Py 2.7 with visa
On 15/05/2015 23:12, Wilson, Pete wrote:
Greetings I am trying to write a test executive program using python 2.7 on a windows 7 computer. I
want to connect to a Keithley 2100 voltmeter using National Instruments VISA. I am having trouble
installing pyvisa. All the documentation refers to using 'pip' and a command line "$ pip
install pyvisa" . What interface or console is this? "$" prompt looks like a Linux
command line. How do we do this with windows?
It's a Windows command line prompt. You can get this by hitting the
Windows logo key with 'R' and then entering cmd. However the
simplest way for you I think is to download this
https://sites.google.com/site/pydatalog/python/pip-for-windows
Do I have to do this? Or can I use the native visa module in Python 2.7? Using
the help() and dir() features I can get some basic information about visa, but
the functions have changed, like visa.ResourceManager.open_resource is not
working. I really liked this function... Are there any examples of how to use
this new visa? I have some working code below that uses pyvisa, can it be
converted?
Please help us to help you. Stating "is not working" is less than
useless, please show us exactly what happens. Cut and paste any output,
don't rely on typing it as this often results in further errors that
just confuse the issue.
def update_current():
import visa
rm = visa.ResourceManager()
rm.list_resources()
current_1_ma = ""
exe_check = "PASS"
try:
dut_data = open("dut_data.txt", "w")
except:
exe_check = "FAIL"
Don't use bare excepts as it's asking for trouble. Much better to leave
out the error handling to start with and just let your programming
errors bubble up as stack traces. Then add in appropriate things to
catch. In the above FileNotFoundError amongst others seems suitable.
--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.
Mark Lawrence
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Re: [Tutor] Terminology question
On Fri, May 15, 2015 at 12:02:44PM -0700, Jim Mooney Py3.4.3winXP wrote:
> On 14 May 2015 at 16:47, Alan Gauld wrote:
>
> > Rather than printing the messages you could re-raise
> > the error but with the error message attached as a string.
> >
>
> I'm a little unclear how you catch the string you create in a raise, in the
> caller. I tried an example from the docs but it didn't work.
What does "didn't work" mean? Did your computer crash? Catch fire? A
completely different error got printed? Something else?
> Could you
> provide a simple example? Sometimes the docs are heavy slogging if you
> don't already know what's what ;')
I'm not entirely sure what part you are having trouble with. I assume
you know how to catch an exception:
alist = []
try:
value = alist[2]
except IndexError:
pass # Just pretend it didn't happen...
and how to inspect its error message:
alist = []
try:
value = alist[2]
except IndexError as err:
print(err.args[0])
We can also *replace* the error message with one of our own. The args
attribute is a tuple, normally the error message is in the zeroeth
position as above. So we just have to replace args with a tuple of our
own and re-raise the exception:
try:
value = alist[2]
except IndexError as err:
err.args = ("Er, wot?",) + err.args[1:]
raise
An alternative is to replace the exception with a completely different
exception:
try:
value = alist[2]
except IndexError:
raise RuntimeError("er, wot?")
In Python 2, this suppresses the IndexError and raises RuntimeError
instead. However, this hides bugs in the case that the second exception
isn't deliberate, but a bug. So starting with Python 3, exceptions are
chained, which means that if an exception occurs inside an "except"
block, Python will display both exceptions and not just the most recent:
Traceback (most recent call last):
File "", line 2, in
IndexError: list index out of range
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "", line 4, in
RuntimeError: er, wot?
This is a great boon for debugging accidental bugs, but doesn't help
much if you actually do intend to replace one exception with another. So
starting from Python 3.3 you can suppress the chaining:
try:
value = alist[2]
except IndexError:
raise RuntimeError("er, wot?") from None
which gives just a single exception:
Traceback (most recent call last):
File "", line 4, in
RuntimeError: er, wot?
Does this help?
--
Steve
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Re: [Tutor] Terminology question
On Thu, May 14, 2015 at 03:43:30PM -0700, Jim Mooney Py3.4.3winXP wrote:
> I noticed that if I call a function that throws an error, I can catch it
> from the caller, instead of catching it in the function. Is this is what is
> known as "errors bubbling up?" Also, is this how you're supposed to do it?
Consider a chain of functions:
main() calls setup();
setup() calls process_arguments();
process_arguments() calls read_config();
read_config() calls open("somefile");
and open("somefile") raises an exception (perhaps the file cannot be
found, or you don't have permission to read it).
Each function in the chain gets a chance to catch the exception:
open <- read_config <- process_arguments <- setup <- main
If open doesn't catch the exception, we say that the exception "bubbles
up" the chain until it is caught, or if none of them catch it, then it
bubbles all the way up and finally reaches the Python interpreter, which
prints a stack trace and exits.
[...]
> def get_input():
> inp = input("Enter minimum, maximum, rows, and columns, separated by
> commas: ")
> inps = inp.split(',')
> try:
> minimum = int(inps[0])
> maximum = int(inps[1])
> rows = int(inps[2])
> columns = int(inps[3])
> return minimum, maximum, rows, columns
This is not best practice. There are too many things which might cause
an exception. Ideally, a try block should only contain *one* thing which
might go wrong. Here you have at least four. By treating them all the
same, you lose the opportunity to give the user information about
*which* one they screwed up.
> except ValueError as err:
> print("non-integer entered", err)
> return None
This is bad practice, for two reasons.
You might not think so yet, but Python tracebacks give you a lot of
useful debugging information. Here, you flush that useful information
down the toilet and replace it with a not-very helpful message
"non-integer entered" (plus a bit more stuff). In a trivial script that
might not matter, but in a big script, you may have *no idea* where this
non-integer was entered. There could be thirty places where it might
happen. How do you know which one it is?
The second reason is that you really should only catch exceptions that
you can do something about. If you can't fix the problem, there's no
point in catching the exception[1]. In this case, if the user doesn't
enter a valid set of values, there's no way you can proceed. So why
bother delaying the inevitable? Just let the exception exit the program.
This will also look after all the little things which distinguish a
professional quality application from a jerry-built beginner's toy. For
example: error messages should print to stderr, not stdout; the
application should exit with a non-zero result code. If this means
nothing to you, don't be too concerned about it, just understand that
Python will do the right thing provided you don't get in its way, as you
just did by catching the exception.
[1] Well, there is one other reason: in an application aimed at
non-programmers, you might choose to catch the exception, log the
traceback somewhere where you can get to it for debugging, and display a
"friendly" (i.e. non-technical, non-threatening, useless) message to the
user. But as a beginner yourself, you shouldn't do this until you have
become an expert in reading tracebacks. Otherwise you're just hiding
debugging info that you need.
--
Steve
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