Python gives a "Syntax Error" on "else:" - python

My Error
Python throws a syntax error pointed at the last "e" of "else:", preceded by an if statement and inside a while loop.
My Objective
Test if certain parameters are true, if true then go to the beginning of the loop and if not true then perform certain statements and increment a value.
My Source Code
from random import randint
def returnDigRoot(num):
digs = []
while len(str(num)) != 1:
num = str(num)
for each in num:
digs.append(each)
num = int(num)
digs = [int(i) for i in digs]
num = sum(digs)
return(num)
def rnum():
return(randint(1,99999))
ran_nums = []
sols = []
it = 1
The problem area is here
while it <= 3:
print("Generating numbers")
current = randint(1,99999)
print("randomly intializing the 'current' int value")
print("testing if the digital root is greater than 6")
if returnDigRoot(current) > 6:
print("going back to start of loop")
continue
print("testing if it isnt")
else:
ran_nums.append(current)
print("append 'current' to ran_nums")
sols.append(returnDigRoot(current))
print("appending its digital root to sols")
it += 1
print("incrementing the iterator variable")
My Research
I looked at many questions on StackOverflow and other sites and could not find a solution to my problem; most problems people had with else statements were related to tabbing errors, preceding errors (which I checked for), no preceding if statement, or multiple else statements.
Thanks in advance for any help.

print("testing if it isnt") needs to be indented. As it stands, your code doesn’t really relate the if with the else because of the indentation. It’s like writing something like this in C:
if(<condition>)
{
<action>
}
prinf(...)
else
{
<action>
}
Just align the print line with the rest of the code under the if statement.

The line:
print("testing if it isnt")
isn't indented correctly. You can't have anything in between an if block and the else block.

Your statement:
print("testing if it isnt")
is indented to the wrong level; that makes the else: that follows an independent statement, which is syntactically wrong. Probably you meant that print statement to follow the else and be indented one level.

This is most likely a indentation/space/tabbing issue since I copy pasted the code and I don't get any errors. Though I am on Python 2.7.10. (Repasting it here to ensure you can copy paste the same and try):
from random import randint
def returnDigRoot(num):
digs = []
while len(str(num)) != 1:
num = str(num)
for each in num:
digs.append(each)
num = int(num)
digs = [int(i) for i in digs]
num = sum(digs)
return(num)
def rnum():
return(randint(1,99999))
ran_nums = []
sols = []
it = 1
while it <= 3:
current = randint(1,99999)
if returnDigRoot(current) > 6:
continue
else: # this is where the error is pointed
ran_nums.append(current)
sols.append(returnDigRoot(current))
it += 1
On an unrelated note, that while loop will take a long time to exit since the exit criteria is very small (two current <=36 only will cause exit).

Related

Trying to understand indents in Python

I am relatively new to python, and have been working on a problem to find the prime numbers between two inputs. I have a solution that works (helped somewhat by online searching too), but am unsure on why the else statement shown below should not be at the same tab setting as the if statement. If it is, though, it doesn't work correctly. Can anyone clarify this for me?
My code is here:
n1 = int(input("Enter the lower number: "))
n2 = int(input("Enter the higher number: "))
for num in range(n1, n2 + 1):
if num > 1:
for i in range(2, num):
if num % i == 0:
break
else:
print(num)
You're seeing Python's (rather unique) for:else: pattern, to execute something when a break is not encountered within the for suite:
When the items are exhausted, the suite in the else clause, if present, is executed, and the loop terminates.
A break statement executed in the first suite terminates the loop without executing the else clause’s suite. [...]

Get stuck in a 'while True' loop python

I just learned about break and return in Python.
In a toy code that I wrote to get familiar with the two statements, I got stuck in a loop, but I don't know why. Here is my code:
def break_return():
while True:
for i in range(5):
if i < 2:
print(i)
if i == 3:
break
else:
print('i = ', i)
return 343
break_return()
I'm new to programming, any suggestions will be appreciated.
With the for-else construct you only enter the else block if the for loop does not break, which your for loop always does because i inevitably becomes 3 with your range generator. Your infinite while loop is therefore never able to reach the return statement, which is only in the said else block.
nvm I'm super wrong here
First of all, when you define a function in Python, any code that belongs in the function should be in the same indentation block. With this in mind, your code would look like this:
def break_return():
while True:
for i in range(5):
if i < 2:
print(i)
if i == 3:
break
else:
print('i = ', i)
return 343
break_return()
The next problem I see is that your else statement isn't correctly formatted with an if statement. If you mean for it to go on the 2nd if statement, your code would look like this:
def break_return():
while True:
for i in range(5):
if i < 2:
print(i)
if i == 3:
break
else:
print('i = ', i)
return 343
break_return()
This is only formatting. But in this example, the code would only run once because it immediately returns and exits the function.
I think this may be a better example of using both break and return:
def break_return(value):
for i in range(5):
print(i)
if i == 3:
break #This exits the for loop
if i == 4:
print("This won't print!")
#Won't print because the loop "breaks" before i ever becomes 4
return value * 2 #Returns the input value x 2
print(break_return(30)) #Display the return value of break_return()
This demonstrates how break exits a for loop and how return can return a value from the function.
The output of the code above is:
0 #Value of i
1 #Value of i
2 #Value of i
3 #Value of i
60 #The value returned by the function
Glad to hear you're learning Python! It's a lot of fun, and super useful.

Can't figure out how to make my funtion not return None in Python

I have a program I am trying to make which will either show all the factors of a number or say it is prime. It's a simple program but I have one main issue. Once it prints all of the factors of an inputted number, it always returns none. I have tried multiple ways to get rid of it but nothing works without screwing something else up. The code is below.
def mys(x):
x = input("Enter a number: ")
for i in range(2,x):
r = x % i
if r == 0:
print(i)
print(mys(x))
That code is just for printing the factors but that is where the problem lies. The results I get after entering a number, in this case 20, are as follows:
2
4
5
10
None
No matter what I do, I can't get the None to not print.
So if you don't want the return value of mys (None) not printed, then don't print it:
mys(x)
In python, a function that has no return statement always returns None.
I guess what you are trying to do is calling the mys function, and not printing it.
Note that you should remove x parameter, because it is asked inside of the function.
def mys():
x = input("Enter a number: ")
for i in range(2,x):
r = x % i
if r == 0:
print(i)
mys()
It would be better not to include user input and printing in your function. It would make it easier to test and to reuse:
def mys(x):
result = []
for i in range(2,x):
r = x % i
if r == 0:
result.append(i)
return result
x = input("Enter a number: ")
print(mys(x))

Unexpected EOF while parsing in Credit check (Luhn’s algorithm)

EDIT: Reminder, before you ask a stupid question on Stackoverflow, be sure to check all braces, semicolons and parentheses!
ORIGINAL POST:
I'm switching from C to python and have a hopefully rather basic question: Why does this return an unexpected EOF while parsing error?
A little background:
This is supposed to check if a credit card number is valid according to Luhn’s algorithm.
number = input("Number: ", end="")
numArray = []
for i in number:
numArray.append(int(i))
firstTime = 0;
secondTime = 0;
cycle2 = 0
for cycle in range(15):
if(cycle % 2 != 0):
firstTime += numArray[cycle]*2
else:
secondTime += numArray[cycle2]
cycle2 += 1
print("{} and {}".format(firstTime, secondTime)
That's because you're missing a trailing ) in the call to print, and the Python interpreter doesn't expect that:
print("{} and {}".format(firstTime, secondTime)
^
Fix it to print("{} and {}".format(firstTime, secondTime))

Else clause on Python while statement

I've noticed the following code is legal in Python. My question is why? Is there a specific reason?
n = 5
while n != 0:
print n
n -= 1
else:
print "what the..."
Many beginners accidentally stumble on this syntax when they try to put an if/else block inside of a while or for loop, and don't indent the else properly. The solution is to make sure the else block lines up with the if, assuming that it was your intent to pair them. This question explains why it didn't cause a syntax error, and what the resulting code means. See also I'm getting an IndentationError. How do I fix it?, for the cases where there is a syntax error reported.
The else clause is only executed when your while condition becomes false. If you break out of the loop, or if an exception is raised, it won't be executed.
One way to think about it is as an if/else construct with respect to the condition:
if condition:
handle_true()
else:
handle_false()
is analogous to the looping construct:
while condition:
handle_true()
else:
# condition is false now, handle and go on with the rest of the program
handle_false()
An example might be along the lines of:
while value < threshold:
if not process_acceptable_value(value):
# something went wrong, exit the loop; don't pass go, don't collect 200
break
value = update(value)
else:
# value >= threshold; pass go, collect 200
handle_threshold_reached()
The else clause is executed if you exit a block normally, by hitting the loop condition or falling off the bottom of a try block. It is not executed if you break or return out of a block, or raise an exception. It works for not only while and for loops, but also try blocks.
You typically find it in places where normally you would exit a loop early, and running off the end of the loop is an unexpected/unusual occasion. For example, if you're looping through a list looking for a value:
for value in values:
if value == 5:
print "Found it!"
break
else:
print "Nowhere to be found. :-("
Allow me to give an example on why to use this else-clause. But:
my point is now better explained in Leo’s answer
I use a for- instead of a while-loop, but else works similar (executes unless break was encountered)
there are better ways to do this (e.g. wrapping it into a function or raising an exception)
Breaking out of multiple levels of looping
Here is how it works: the outer loop has a break at the end, so it would only be executed once. However, if the inner loop completes (finds no divisor), then it reaches the else statement and the outer break is never reached. This way, a break in the inner loop will break out of both loops, rather than just one.
for k in [2, 3, 5, 7, 11, 13, 17, 25]:
for m in range(2, 10):
if k == m:
continue
print 'trying %s %% %s' % (k, m)
if k % m == 0:
print 'found a divisor: %d %% %d; breaking out of loop' % (k, m)
break
else:
continue
print 'breaking another level of loop'
break
else:
print 'no divisor could be found!'
The else-clause is executed when the while-condition evaluates to false.
From the documentation:
The while statement is used for repeated execution as long as an expression is true:
while_stmt ::= "while" expression ":" suite
["else" ":" suite]
This repeatedly tests the expression and, if it is true, executes the first suite; if the expression is false (which may be the first time it is tested) the suite of the else clause, if present, is executed and the loop terminates.
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 goes back to testing the expression.
The else clause is only executed when the while-condition becomes false.
Here are some examples:
Example 1: Initially the condition is false, so else-clause is executed.
i = 99999999
while i < 5:
print(i)
i += 1
else:
print('this')
OUTPUT:
this
Example 2: The while-condition i < 5 never became false because i == 3 breaks the loop, so else-clause was not executed.
i = 0
while i < 5:
print(i)
if i == 3:
break
i += 1
else:
print('this')
OUTPUT:
0
1
2
3
Example 3: The while-condition i < 5 became false when i was 5, so else-clause was executed.
i = 0
while i < 5:
print(i)
i += 1
else:
print('this')
OUTPUT:
0
1
2
3
4
this
My answer will focus on WHEN we can use while/for-else.
At the first glance, it seems there is no different when using
while CONDITION:
EXPRESSIONS
print 'ELSE'
print 'The next statement'
and
while CONDITION:
EXPRESSIONS
else:
print 'ELSE'
print 'The next statement'
Because the print 'ELSE' statement seems always executed in both cases (both when the while loop finished or not run).
Then, it's only different when the statement print 'ELSE' will not be executed.
It's when there is a breakinside the code block under while
In [17]: i = 0
In [18]: while i < 5:
print i
if i == 2:
break
i = i +1
else:
print 'ELSE'
print 'The next statement'
....:
0
1
2
The next statement
If differ to:
In [19]: i = 0
In [20]: while i < 5:
print i
if i == 2:
break
i = i +1
print 'ELSE'
print 'The next statement'
....:
0
1
2
ELSE
The next statement
return is not in this category, because it does the same effect for two above cases.
exception raise also does not cause difference, because when it raises, where the next code will be executed is in exception handler (except block), the code in else clause or right after the while clause will not be executed.
I know this is old question but...
As Raymond Hettinger said, it should be called while/no_break instead of while/else.
I find it easy to understeand if you look at this snippet.
n = 5
while n > 0:
print n
n -= 1
if n == 2:
break
if n == 0:
print n
Now instead of checking condition after while loop we can swap it with else and get rid of that check.
n = 5
while n > 0:
print n
n -= 1
if n == 2:
break
else: # read it as "no_break"
print n
I always read it as while/no_break to understand the code and that syntax makes much more sense to me.
thing = 'hay'
while thing:
if thing == 'needle':
print('I found it!!') # wrap up for break
break
thing = haystack.next()
else:
print('I did not find it.') # wrap up for no-break
The possibly unfortunately named else-clause is your place to wrap up from loop-exhaustion without break.
You can get by without it if
you break with return or raise → the entire code after the call or try is your no-break place
you set a default before while (e.g. found = False)
but it might hide bugs the else-clause knows to avoid
If you use a multi-break with non-trivial wrap-up, you should use a simple assignment before break, an else-clause assignment for no-break, and an if-elif-else or match-case to avoid repeating non-trival break handling code.
Note: the same applies to for thing in haystack:
Else is executed if while loop did not break.
I kinda like to think of it with a 'runner' metaphor.
The "else" is like crossing the finish line, irrelevant of whether you started at the beginning or end of the track. "else" is only not executed if you break somewhere in between.
runner_at = 0 # or 10 makes no difference, if unlucky_sector is not 0-10
unlucky_sector = 6
while runner_at < 10:
print("Runner at: ", runner_at)
if runner_at == unlucky_sector:
print("Runner fell and broke his foot. Will not reach finish.")
break
runner_at += 1
else:
print("Runner has finished the race!") # Not executed if runner broke his foot.
Main use cases is using this breaking out of nested loops or if you want to run some statements only if loop didn't break somewhere (think of breaking being an unusual situation).
For example, the following is a mechanism on how to break out of an inner loop without using variables or try/catch:
for i in [1,2,3]:
for j in ['a', 'unlucky', 'c']:
print(i, j)
if j == 'unlucky':
break
else:
continue # Only executed if inner loop didn't break.
break # This is only reached if inner loop 'breaked' out since continue didn't run.
print("Finished")
# 1 a
# 1 b
# Finished
The else: statement is executed when and only when the while loop no longer meets its condition (in your example, when n != 0 is false).
So the output would be this:
5
4
3
2
1
what the...
Suppose you've to search an element x in a single linked list
def search(self, x):
position = 1
p =self.start
while p is not None:
if p.info == x:
print(x, " is at position ", position)
return True
position += 1
p = p.link
else:
print(x, "not found in list")
return False
So if while conditions fails else will execute, hope it helps!
The better use of 'while: else:' construction in Python should be if no loop is executed in 'while' then the 'else' statement is executed. The way it works today doesn't make sense because you can use the code below with the same results...
n = 5
while n != 0:
print n
n -= 1
print "what the..."
As far as I know the main reason for adding else to loops in any language is in cases when the iterator is not on in your control. Imagine the iterator is on a server and you just give it a signal to fetch the next 100 records of data. You want the loop to go on as long as the length of the data received is 100. If it is less, you need it to go one more times and then end it. There are many other situations where you have no control over the last iteration. Having the option to add an else in these cases makes everything much easier.

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