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How can I find out which index is out of range?

In the event of an IndexError, is there a way to tell which object on a line is 'out of range'?
Consider this code:
a = [1,2,3]
b = [1,2,3]
x, y = get_values_from_somewhere()
try:
a[x] = b[y]
except IndexError as e:
....
In the event that x or y is too large and IndexError gets caught, I would like to know which of a or b is out of range (so I can perform different actions in the except block).
Clearly I could compare x and y to len(a) and len(b) respectively, but I am curious if there is another way of doing it using IndexError.

There is a way, but I wouldn't consider it as very robust. There is a subtle difference in the error messages:
a = [1,2,3]
b = [1,2,3]
a[2] = b[3]
---------------------------------------------------------------------------
IndexError Traceback (most recent call last)
<ipython-input-69-8e0d280b609d> in <module>()
2 b = [1,2,3]
3
----> 4 a[2] = b[3]
IndexError: list index out of range
But if the error is on the left hand side:
a[3] = b[2]
---------------------------------------------------------------------------
IndexError Traceback (most recent call last)
<ipython-input-68-9d66e07bc70d> in <module>()
2 b = [1,2,3]
3
----> 4 a[3] = b[2]
IndexError: list assignment index out of range
Note the 'assignment' in the message.
So, you could do something like:
a = [1,2,3]
b = [1,2,3]
try:
a[3] = b[2]
except IndexError as e:
message = e.args[0]
if 'assignment' in message:
print("Error on left hand side")
else:
print("Error on right hand side")
Output:
# Error on left hand side
Again, I wouldn't trust it too much, it would fail if the message changes in another version of Python.
I had a look at these parts of the source code, these different messages are really the only difference between the two errors.

The IndexError exception does not store information about what raised the exception. Its only data is an error message. You have to craft your code to do so.
a = [1,2,3]
b = [1,2,3]
x, y = get_values_from_somewhere()
try:
value = b[y]
except IndexError as e:
...
try:
a[x] = value
except IndexError as e:
...
I want to add I tried to recover the culprit through inspect.frame and was unable to do so. Thus I suspect there really is no robust way.
Finally, note that this is specific to IndexError as other exceptions may contain the needed information to infer what caused them. By example a KeyError contains the key that raised it.

A more robust approach would be to tap into the traceback object returned by sys.exc_info(), extract the code indicated by the file name and line number from the frame, use ast.parse to parse the line, subclass ast.NodeVisitor to find all the Subscript nodes, unparse (with astunparse) and eval the nodes with the frame's global and local variables to see which of the nodes causes exception, and print the offending expression:
import sys
import linecache
import ast
import astunparse
def find_index_error():
tb = sys.exc_info()[2]
frame = tb.tb_frame
lineno = tb.tb_lineno
filename = frame.f_code.co_filename
line = linecache.getline(filename, lineno, frame.f_globals)
class find_index_error_node(ast.NodeVisitor):
def visit_Subscript(self, node):
expr = astunparse.unparse(node).strip()
try:
eval(expr, frame.f_globals, frame.f_locals)
except IndexError:
print("%s causes IndexError" % expr)
find_index_error_node().visit(ast.parse(line.strip(), filename))
a = [1,2,3]
b = [1,2,3]
x, y = 1, 2
def f():
return 3
try:
a[f() - 1] = b[f() + y] + a[x + 1] # can you guess which of them causes IndexError?
except IndexError:
find_index_error()
This outputs:
b[(f() + y)] causes IndexError

Something like this perhaps?
a = [1,2,3]
b = [2,3,4]
x = 5
y = 1
try:
a[x] = b[y]
except IndexError:
try:
a[x]
print('b caused indexerror')
except IndexError:
print('a caused indexerror')

No, there is no way of deducing which of the two is out of bounds using the IndexError thrown.

Error in python code

i am getting error in the below python code
a=[1,2,3,4,5,6,7,8,9]
c,d=divmod(len(a),2)
i=iter(a).next
print ''.join('%s\t%s\n' % (i(),i())
for i in xrange(c))\
+ ('%s\t\n' % (i()) if b==1
else '')
i need to print output is
1 2
3 4
5
i am getting error:
Traceback (most recent call last):
File "dhsgj.py", line 5, in <module>
for i in xrange(c))\
File "dhsgj.py", line 5, in <genexpr>
for i in xrange(c))\
TypeError: 'int' object is not callable

You do not need to split the array, try to iterate two items at a time.
I've updated your code to make it little bit easier to follow. This should work:
a=[1,2,3,4,5,6,7,8,9]
iterator = iter(a)
for first in iterator:
try:
second = next(iterator)
except StopIteration:
print first
else:
print('%s\t%s\n' % (first, second))

How to check which statement has thrown exception?

I have a statement like this.
I just want to know which of the two assert statement has thrown the exception.
try:
assert re.search("xyz", statement)
assert re.search("abc", statement)
except AssertionError:
print "AssertionError : Expected Error message not found"
Thanks for the answer.

As mentioned in The assert statement docs, you can give an expression after the assertion test expression; that second expression will be passed in the AssertionError. Here's a simple demo:
for n in (-5, 10, 20):
try:
assert 0 <= n, '%d is too low' % n
assert n <= 10, '%d is too high' % n
print('%d is ok' % n)
except AssertionError as err:
print "AssertionError:", err
output
AssertionError: -5 is too low
10 is ok
AssertionError: 20 is too high
That second expression doesn't have to be a string, it can be anything. Since assertions should only be used to verify program logic, not to validate user data, I generally don't bother passing a nicely-formatted string, I just pass a tuple containing the relevant values, and maybe an identifying string. Eg,
assert (a * b > c), ('Bad product', a, b, c)

You could use functions from the traceback module. For example, extract_tb returns a list of tuples (named tuples in Python 3.5 and newer) representing the stack trace entries. Each tuple contains a line number as well as the source text line (if available).
import traceback
try:
assert 1
assert None
except AssertionError as e:
for x in traceback.extract_tb(e.__traceback__, limit=-1):
print(x.lineno, repr(x.line)) # Prints 5 'assert None'

You are able to print the last raised exception with traceback.print_exc(). An example:
>>> import traceback
>>> try:
... a = 1 / 0
... except:
... traceback.print_exc()
...
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
ZeroDivisionError: integer division or modulo by zero
There is also traceback.format_exc() in case you don't want to print.

Maybe you want to differ the exceptions?Then Getting exception details in Python can help you.
Copying the answer here:
import sys
try:
assert re.search("xyz", statement)
assert re.search("abc", statement)
except AssertionError:
type, value, traceback = sys.exc_info()
Then you can print out the info.

list index out of range in my coding

Can anyone help me fix this error I keep getting please. I have tried to look for a solution but I can't find any. Below is the error message and also part of my coding
Please enter your class Y or X or Z: Y
Traceback (most recent call last):
File "/Volumes/LIAM'S USB/DEV6 FINAL.py", line 118, in <module>
score=int(items[1])
IndexError: list index out of range
results={
start=True
while (start):
pupil_class=input("\nPlease enter your class Y or X or Z: ")
if pupil_class == ("Y"):
classfile="Class_Y_results.txt"
elif pupil_class == ("X"):
classfile="Class_X_results.txt"
elif pupil_class == ("Z"):
classfile="Class_Z_results.txt"
f=open(classfile,'r')
for line in f:
items=line.split(',')
name=items[0]
score=int(items[1])
if name in results:
results[name].append(score)
else:
results[name]=[]
results[name].append(score)
f.close()

A certain line in your Class_Y_Results.txt only has one entry (not separated by commas), hence the list returned by items=line.split(',') only has a length of 1 (or maybe 0), causing score=int(items[1]) to throw an IndexError.
Sample:
>>> a = "foo,bar"
>>> b = "foo"
>>> len(a.split(','))
2
>>> len(b.split(','))
1
>>> a.split(',')[1]
'bar'
>>> b.split(',')[1]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: list index out of range

There is probably an empty like in one of your files. This will not contain a comma, so you will not have an item[1], and this produces the error message you see.
Check how many fields you get back from the split to solve this.

getting assigned variables when exception is raised, python exception handeling

I would like to find a way to get the variables I assigned before an exception is made. E.G if the code is
try:
a=b
c=d
e=f
except:
bla bla
And an exception is generated at "e=f," I still want a=b and c=d
Is it possible? I realize I could make this multiple try statements but is there something I can do in one step?

Yes, it is perfectly possible. Below is a demonstration:
>>> try:
... a = 1
... b = 2
... c = 1/0 # This will raise a ZeroDivisionError
... except ZeroDivisionError:
... print 'an error occurred'
...
an error occurred
>>> a # a still exists
1
>>> b # so does b
2
>>> c # only c is undefined
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'c' is not defined
>>>
try/except is designed to execute the code in the try-block normally until an exception is raised. When that happens, the try-block is immediately exited. This means that only the code after the line that raised the exception is ignored.

The best you can do is to limit the number of expressions inside your try block.
If you need to know where the exception is raised, you are probably better off using multiple try...except's like you mentioned in your queestion, since there is no (practical) way to know where the exception was raised.
If the expressions are of the same type, you may want to put them in lists though, and loop over them, like:
vars = [a, b, c]
values = [1, 2, 0]
for i, (var, value) in enumerate(zip(vars, values)):
try:
var /= value
except ZeroDivisionError:
print 'The exception was raised on the {}. iteration'.format(i)
continue

try:
a = "foo"
c = "bar"
e = unknown_function()
except:
pass
print a, c # prints "foo bar"
Both a and c are set, you can simply use their values after the exception is handled. e is not set to anything as an exception was raised when that line was executed.

If I understand you correctly, this will happen by default. Raising an exception does not magically undo everything that has already happened inside the try block. In your example, if an error occurs on the assignment of e, a and c will still have the values they were assigned.
try:
a = b
c = d
e = f # oh noes! error! Try block skips to the end!
except: pass
print (a) #but a and c are still there
print (c)

This is a straight forward method without much of a hassle . This is the best way out when you cant predict what type of error can occur, or multiple errors can occur .
try :
a=10
except :
print "bla A"
try :
b=20
except:
print "bla B"
try:
c=d
except :
print "bla C"

Is it something like this that you want ?
import sys
code = """
b = 100
d = 200
a=b
c=d
e=10/0
g=e
h = 100
"""
for line in filter(None,code.splitlines()):
print line
try:
exec line
except:
sys.excepthook(sys.exc_info()[0],sys.exc_info()[1],None)
result
b = 100
d = 200
a=b
c=d
e=10/0
ZeroDivisionError: integer division or modulo by zero
g=e
NameError: name 'e' is not defined
h = 100