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.
Hi everyone first time poster long term reader.
my problem is I want an else statment to loop inside a for loop.
I want the else statment to loop until the if statment above it is met?
can anyone tell me where I am going wrong I have tried so many diffrent ways including while loops inside if statments cant get my head round this ?
edit changed the code to a while loop not on prunes suggestion but cant escape the while loop
for url in results:
webdriver.get(url)
try:
liked = webdriver.find_elements_by_xpath("//span[#class=\"glyphsSpriteHeart__filled__24__red_5 u-__7\" and #aria-label=\"Unlike\"]")
maxcount = 0
while not liked:
sleep(1)
webdriver.find_element_by_xpath('//span/button/span').click()
numberoflikesgiven += 1
maxcount += 1
print'number of likes given : ',numberoflikesgiven
sleep(2)
webdriver.find_element_by_link_text('Next').click()
if maxcount >= 10:
print('max count reached .... moving on.')
continue
else:
print ('picture has already been liked...')
continue
Yes, you very clearly wrote an infinite loop:
while not liked:
... # no assignments to "liked"
You do not change the value of liked anywhere in the loop. You do not break the loop. Thus, you have an infinite loop. You need to re-evaluate liked on each iteration. A typical way to to this is to duplicate the evaluation at the bottom of the loop
sleep(2)
webdriver.find_element_by_link_text('Next').click()
liked = webdriver.find_elements_by_xpath(
"//span[#class=\"glyphsSpriteHeart__filled__24__red_5 u-__7\" \
and #aria-label=\"Unlike\"]")
Also note that your continue does nothing, as it's at the bottom of the loop. Just let your code reach the bottom naturally, and the whilewill continue on its own. If you intended to go to the next iteration offor url in results, then you need tobreakthewhile` loop instead.
I have a very simple problem: I have made a while loop, and in the middle of it, set the initial condition to false. This, however, does not stop the loop and runs entirely through, (somewhat obviously,) until it attempts to unsuccessfully go through again. Here is a simplified construction of what I have.
while(a):
print("hi")
a = False
print("bye")
This returns:
hi
bye
Again I would like to only return hi; I want the loop to continually check if its satisfied.
Any help greatly appreciated.
Use:
return, or break
while a:
print('hi')
a = False
if not a:
break
print('bye')
In a function or loop, when something is returned, the function or loop terminates. You could also return True, or break which is a specific way to 'break' out of a loop.
Since the condition is true to start with (by design), the body of the loop will execute at least once. The fact you do something in the body of the loop to make the loop condition false doesn't stop the current iteration. It just means there won't be a next iteration after this one is done.
So if you want to get out in the middle of the current iteration, then you need to use break, return, or something more sophisticated like #inspectorG4dget's suggestion.
while(a):
print("hi")
a = False
if a:
print("bye")
OR
while(a):
for s in ["hi", "bye"]:
if a:
print(s)
if someCondition:
a = False
For my software major work I have to create a program. In summary, the high scores list needs to be sorted before it can be written to file. To do this, I am using a bubble sort and I can't use the inbuilt sort function. The text file that the data is being read from is stored in a nested list. The text file looks like this:
NameOne
10
NameTwo
15
NameThree
9
This is the bubble sort code I have but does not work:
b_not_sorted = True
while b_not_sorted:
counter = 0
b_not_sorted = False
for counter in range(len(highest_scores) - 1):
if highest_scores[counter] < highest_scores[counter + 1]:
b_not_sorted = True
highest_scores[counter], highest_scores[counter+1] = highest_scores[counter+1], highest_scores[counter]
counter = counter + 1
I need the scores to be sorted from highest to lowest. Any help would be greatly appreciated and you will be credited appropriately in my program credits :). Thanks.
Here's a hint:
Check how many times your outer while loop is running. It should be running more than once, correct? What will always happen that causes the loop to exit, no matter what?
Try going through the code line by line and seeing what happens at every point.
The statement b_not_sorted = False at the end of the outer loop results in the outer loop exiting after executing only once. You need to move that statement to another part of your code. Try changing the name of b_not_sorted to I_still_need_to_go_through_the_list in your head:
Obviously in the first line:
while I_still_need_to_go_through_the_list:
it should be True, since you haven't gone over the list at all. You don't know if it's in order or not.
and after the line:
if highest_scores[counter] < highest_scores[counter + 1]:
Of course then we still need to make another pass, since we just made a change to the list and need to make sure no further changes are needed.
But what if no changes are made? I_still_need_to_go_through_the_list should be False then. Hmmm. If we put I_still_need_to_go_through_the_list = False right before the for loop, then it will be False unless we make changes to the list, which is exactly what we want.
You're doing b_not_sorted = False right after the first iteration, but it shouldn't be there! The algorithm just stops before it finishes the sorting.
You should instead do b_not_sorted = True only if highest_scores[counter] < highest_scores[counter + 1]
Also, the swapping code can look much nicer in Python. Instead of using temp_var just do this:
highest_scores[counter], highest_scores[counter+1] = highest_scores[counter+1], highest_scores[counter]
Python style guide suggests that you shoudn't write == True or == False in if statements. Do it like this:
while b_not_sorted:
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.