I want to make a function what must have 2 loops:
must check that at least 2 characters (letters) are inserted
if the two characters are in the form that the variable must receive at the end, the variable receives the correct orthographic form.
Here is my code (but it throws me in infinite loop) [Please be indulgent im a begginer (: ] :
def sede():
while True:
import re
var_sede_0 = input("Sede : ")
if re.match("([a-zA-Z]+.*?){2,}", var_sede_0):
while True:
if var_sede_0.lower() in 'torino (to)':
var_sede_0 = 'Torino (TO)'
break
elif var_sede_0.lower() in 'reggio calabria (rc)':
var_sede_0 = 'Reggio Calabria (RC)'
break
else:
print("sbagliato")
continue
break
else:
print("formato sbagliato")
continue
return var_sede_0
var_sede = sede()
Your inner loop is the problem:
while True:
if var_sede_0.lower() in 'torino (to)':
var_sede_0 = 'Torino (TO)'
break
elif var_sede_0.lower() in 'reggio calabria (rc)':
var_sede_0 = 'Reggio Calabria (RC)'
break
else:
print("sbagliato")
continue
Consider that nowhere in this loop do you take in new input, so var_sede_0 can never possibly change. Given that, if the first two predicates (if statements) evaluate to false, then you will be in an infinite loop printing out sbagliato. You can probably simply remove the while True (the inner one) and get to what you want.
It is key to note that break and continue really only affect the immediate while context that they are in - they will not 'speak to' the outer while loop.
You also probably do not want to be importing re on every loop. Move import re to the top of your file.
The following code will gather data from an API and the try/except clause will help to handle several errors (from authentication, index, anything).
There's only one error (an authentication error) that I'm using the while True to repeat the API call to make sure I get the data and it will after a try or two. However if by any means I get another error, it'll be infinitely looping and I can't break it so it goes to the next iteration. I tried to create a counter and if the counter reaches to a number then (pass or continue or break) but it's not working.
## Create a array to loop to:
data_array_query = pd.date_range(start_date,end_date,freq='6H')
#This is my idea but is not working
#Create a counter
counter = 0
#Loop through the just created array
for idx in range(len(data_array_query)-1):
## If counter reaches move on to next for loop element
while True:
if counter>=5:
break
else:
try:
start_date = data_array_query[idx]
end_date = data_array_query[idx+1]
print('from',start_date,'to',end_date)
df = api.query(domain, site_slug, resolution, data_series_collection, start_date=str(start_date), end_date=str(end_date), env='prod', from_archive=True, phase='production').sort_index()
print(df.info())
break
except Exception as e:
print(e)
counter +=1
print(counter)
So the output of running this code for a couple of days show that when it runs 5 times (that's the counter max I set up) it does break but it breaks the whole loop and I only want it to move to the next date.
Any help will be appreciated,
You need to use a break statement to get out of a while True loop. pass and continue work for for loops that have a fixed number of iterations. While loops can go on forever (hence the different names)
Does or condition work in a while loop in python? I can't seem to make it work. This is the sample of how my code works.
newslot = 3
moved = False
while newslot > 0 or moved != True:
enabled = query something on the database where slot = newslot
if enabled:
print 'do something here'
moved = True
else:
newslot-=1
print 'slot disabled'
So when the newslot gets to value of zero it still proceeds to go inside the while loop.
I seem to be missing something here.
or is working as should be expected. A while loop will continue until its condition is false. If its condition is two separate conditions connected with an or, it will only be false when the two conditions are both false.
Your loop will continue repeating until moved is false and newslot is <= 0. I'm guessing you actually want to use and in this case, as you want the loop to stop once either condition is met.
Is the line of else: correct/necessary in this Python program?
from random import randrange
for n in range(10):
r = randrange(0,10) # get random int in [0,10)
if n==r: continue # skip iteration if n=r
if n>r: break # exit the loop if n>r
print n
else:
print "wow, you are lucky!\n"
if n<9:
print "better luck next time\n
From the documentation:
Loop statements may have an else clause; it is executed when the loop terminates through exhaustion of the list (with for) or when the condition becomes false (with while), but not when the loop is terminated by a break statement
So yes, it's correct in your example. Although I've never been a fan of it, using an else clause on a loop makes code confusing at first, I'd rather use a boolean flag for achieving the same effect. IMHO else should be used only for conditionals.
from random import randrange
for n in range(10):
r = randrange(0,10)
if n=r: continue # there should be ==
if n>r: break
print n # should be "\n" ?
else:
print "wow, you are lucky!\n" # Yes, you are! Your python interpreter can make miracles :). Try to run me.append(SexyChick(10)) and let's see what happens!
for...else construct is actually valid. else branch is executed if the loop has not been terminated by break. Look here.
What sorts of methods exist for prematurely exiting an if clause?
There are times when I'm writing code and want to put a break statement inside of an if clause, only to remember that those can only be used for loops.
Lets take the following code as an example:
if some_condition:
...
if condition_a:
# do something
# and then exit the outer if block
...
if condition_b:
# do something
# and then exit the outer if block
# more code here
I can think of one way to do this: assuming the exit cases happen within nested if statements, wrap the remaining code in a big else block. Example:
if some_condition:
...
if condition_a:
# do something
# and then exit the outer if block
else:
...
if condition_b:
# do something
# and then exit the outer if block
else:
# more code here
The problem with this is that more exit locations mean more nesting/indented code.
Alternatively, I could write my code to have the if clauses be as small as possible and not require any exits.
Does anyone know of a good/better way to exit an if clause?
If there are any associated else-if and else clauses, I figure that exiting would skip over them.
(This method works for ifs, multiple nested loops and other constructs that you can't break from easily.)
Wrap the code in its own function. Instead of break, use return.
Example:
def some_function():
if condition_a:
# do something and return early
...
return
...
if condition_b:
# do something else and return early
...
return
...
return
if outer_condition:
...
some_function()
...
from goto import goto, label
if some_condition:
...
if condition_a:
# do something
# and then exit the outer if block
goto .end
...
if condition_b:
# do something
# and then exit the outer if block
goto .end
# more code here
label .end
(Don't actually use this, please.)
while some_condition:
...
if condition_a:
# do something
break
...
if condition_b:
# do something
break
# more code here
break
You can emulate goto's functionality with exceptions:
try:
# blah, blah ...
# raise MyFunkyException as soon as you want out
except MyFunkyException:
pass
Disclaimer: I only mean to bring to your attention the possibility of doing things this way, while in no way do I endorse it as reasonable under normal circumstances. As I mentioned in a comment on the question, structuring code so as to avoid Byzantine conditionals in the first place is preferable by far. :-)
may be this?
if some_condition and condition_a:
# do something
elif some_condition and condition_b:
# do something
# and then exit the outer if block
elif some_condition and not condition_b:
# more code here
else:
#blah
if
For what was actually asked, my approach is to put those ifs inside a one-looped loop
while (True):
if (some_condition):
...
if (condition_a):
# do something
# and then exit the outer if block
break
...
if (condition_b):
# do something
# and then exit the outer if block
break
# more code here
# make sure it is looped once
break
Test it:
conditions = [True,False]
some_condition = True
for condition_a in conditions:
for condition_b in conditions:
print("\n")
print("with condition_a", condition_a)
print("with condition_b", condition_b)
while (True):
if (some_condition):
print("checkpoint 1")
if (condition_a):
# do something
# and then exit the outer if block
print("checkpoint 2")
break
print ("checkpoint 3")
if (condition_b):
# do something
# and then exit the outer if block
print("checkpoint 4")
break
print ("checkpoint 5")
# more code here
# make sure it is looped once
break
Generally speaking, don't. If you are nesting "ifs" and breaking from them, you are doing it wrong.
However, if you must:
if condition_a:
def condition_a_fun():
do_stuff()
if we_wanna_escape:
return
condition_a_fun()
if condition_b:
def condition_b_fun():
do_more_stuff()
if we_wanna_get_out_again:
return
condition_b_fun()
Note, the functions don't HAVE to be declared in the if statement, they can be declared in advance ;) This would be a better choice, since it will avoid needing to refactor out an ugly if/then later on.
There is another way which doesn't rely on defining functions (because sometimes that's less readable for small code snippets), doesn't use an extra outer while loop (which might need special appreciation in the comments to even be understandable on first sight), doesn't use goto (...) and most importantly let's you keep your indentation level for the outer if so you don't have to start nesting stuff.
if some_condition:
...
if condition_a:
# do something
exit_if=True # and then exit the outer if block
if some condition and not exit_if: # if and only if exit_if wasn't set we want to execute the following code
# keep doing something
if condition_b:
# do something
exit_if=True # and then exit the outer if block
if some condition and not exit_if:
# keep doing something
Yes, that also needs a second look for readability, however, if the snippets of code are small this doesn't require to track any while loops that will never repeat and after understanding what the intermediate ifs are for, it's easily readable, all in one place and with the same indentation.
And it should be pretty efficient.
Here's another way to handle this. It uses a single item for loop that enables you to just use continue. It prevents the unnecessary need to have extra functions for no reason. And additionally eliminates potential infinite while loops.
if something:
for _ in [0]:
# Get x
if not x:
continue
# Get y
if not y:
continue
# Get z
if not z:
continue
# Stuff that depends on x, y, and z
Effectively what you're describing are goto statements, which are generally panned pretty heavily. Your second example is far easier to understand.
However, cleaner still would be:
if some_condition:
...
if condition_a:
your_function1()
else:
your_function2()
...
def your_function2():
if condition_b:
# do something
# and then exit the outer if block
else:
# more code here
So here i understand you're trying to break out of the outer if code block
if some_condition:
...
if condition_a:
# do something
# and then exit the outer if block
...
if condition_b:
# do something
# and then exit the outer if block
# more code here
One way out of this is that you can test for for a false condition in the outer if block, which will then implicitly exit out of the code block, you then use an else block to nest the other ifs to do something
if test_for_false:
# Exit the code(which is the outer if code)
else:
if condition_a:
# Do something
if condition_b:
# Do something
The only thing that would apply this without additional methods is elif as the following example
a = ['yearly', 'monthly', 'quartly', 'semiannual', 'monthly', 'quartly', 'semiannual', 'yearly']
# start the condition
if 'monthly' in b:
print('monthly')
elif 'quartly' in b:
print('quartly')
elif 'semiannual' in b:
print('semiannual')
elif 'yearly' in b:
print('yearly')
else:
print('final')
There are several ways to do it. It depends on how the code is implemented.
If you exit from a function, then use return; no code will be executed after the return keyword line. Then the example code is:
def func1(a):
if a > 100:
# some code, what you want to do
return a*a
if a < 100:
# some code, what you want to do
return a-50
if a == 100:
# some code, what you want to do
return a+a
If you exit from a loop, use break. No code will be executed after break keyword. Then the example code is, for while and for loops:
a = 1
while (True):
if (a == 10):
# some code, what you want to do
break
else:
a=a+1
print("I am number", a)
for i in range(5):
if i == 3:
break
print(i)
If you exit from the basic conditional, then you can use the exit() command directly. Then code after the exit() command will not be executed.
NB: This type of code is not preferable. You can use a function instead of this. But I just share the code for example.
The example code is:
if '3K' in FILE_NAME:
print("This is MODIS 3KM file")
SDS_NAME = "Optical_Depth_Land_And_Ocean"
elif 'L2' in FILE_NAME:
print("This is MODIS 10KM file")
SDS_NAME = "AOD_550_Dark_Target_Deep_Blue_Combined"
exit()
else:
print("It is not valid MODIS file")
I scrolled through, and nobody mentioned this technique. It's straightforward compared to other solutions and if you do not want to use a try-except statement, this may be your best option.
#!/usr/bin/python3
import sys
foo = 56
if (foo != 67):
print("ERROR: Invalid value for foo")
sys.exit(1)
use return in the if condition will returns you out from the function,
so that you can use return to break the the if condition.