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

up vote
7
down vote

favorite

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.

asked 3 hours ago

David Board

878

add a commentÂ |Â

up vote
7
down vote

favorite

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.

asked 3 hours ago

David Board

878

add a commentÂ |Â

up vote
7
down vote

favorite

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.

asked 3 hours ago

David Board

878

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.

python exception index-error

asked 3 hours ago

David Board

878

asked 3 hours ago

David Board

878

asked 3 hours ago

David Board

878

asked 3 hours ago

David Board

878

asked 3 hours ago

David Board

878

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
5
down vote

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.

edited 3 hours ago

answered 3 hours ago

Thierry Lathuille

6,07982630

that's very good indeed!
â€“Â Jean-FranÃ§ois Fabre
3 hours ago

Also, the assignment is never reached in the first. You cannot differentiate the third case where both would raise an error.
â€“Â schwobaseggl
3 hours ago

@schwobaseggl That's right, as the right hand side is evaluated first, it is where the error would be raised.
â€“Â Thierry Lathuille
3 hours ago

good spot! and i totally agree this should not be relied on though as an implementation detail
â€“Â Chris_Rands
2 hours ago

add a commentÂ |Â

up vote
2
down vote

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.

edited 3 hours ago

answered 3 hours ago

Olivier MelanÃ§on

11.5k11436

add a commentÂ |Â

up vote
1
down vote

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)
 node = ast.parse(line.strip(), filename)
 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(node)

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

edited 13 mins ago

answered 50 mins ago

blhsing

14.5k2630

Nice. Perhaps you could edit this answer so that everything in the except block is in a function that can be used whenever the developer needs to unpick what caused an IndexError?
â€“Â David Board
19 mins ago

Sure. Edited as suggested then.
â€“Â blhsing
13 mins ago

add a commentÂ |Â

up vote
0
down vote

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

answered 3 hours ago

Chirila Alexandru

1,12431736

add a commentÂ |Â

up vote
0
down vote

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')

answered 3 hours ago

Xnkr

178211

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
5
down vote

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.

edited 3 hours ago

answered 3 hours ago

Thierry Lathuille

6,07982630

that's very good indeed!
â€“Â Jean-FranÃ§ois Fabre
3 hours ago

Also, the assignment is never reached in the first. You cannot differentiate the third case where both would raise an error.
â€“Â schwobaseggl
3 hours ago

@schwobaseggl That's right, as the right hand side is evaluated first, it is where the error would be raised.
â€“Â Thierry Lathuille
3 hours ago

good spot! and i totally agree this should not be relied on though as an implementation detail
â€“Â Chris_Rands
2 hours ago

add a commentÂ |Â

up vote
5
down vote

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.

edited 3 hours ago

answered 3 hours ago

Thierry Lathuille

6,07982630

that's very good indeed!
â€“Â Jean-FranÃ§ois Fabre
3 hours ago

Also, the assignment is never reached in the first. You cannot differentiate the third case where both would raise an error.
â€“Â schwobaseggl
3 hours ago

@schwobaseggl That's right, as the right hand side is evaluated first, it is where the error would be raised.
â€“Â Thierry Lathuille
3 hours ago

good spot! and i totally agree this should not be relied on though as an implementation detail
â€“Â Chris_Rands
2 hours ago

add a commentÂ |Â

up vote
5
down vote

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.

edited 3 hours ago

answered 3 hours ago

Thierry Lathuille

6,07982630

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.

edited 3 hours ago

answered 3 hours ago

Thierry Lathuille

6,07982630

edited 3 hours ago

answered 3 hours ago

Thierry Lathuille

6,07982630

answered 3 hours ago

Thierry Lathuille

6,07982630

answered 3 hours ago

Thierry Lathuille

6,07982630

that's very good indeed!
â€“Â Jean-FranÃ§ois Fabre
3 hours ago

Also, the assignment is never reached in the first. You cannot differentiate the third case where both would raise an error.
â€“Â schwobaseggl
3 hours ago

@schwobaseggl That's right, as the right hand side is evaluated first, it is where the error would be raised.
â€“Â Thierry Lathuille
3 hours ago

good spot! and i totally agree this should not be relied on though as an implementation detail
â€“Â Chris_Rands
2 hours ago

add a commentÂ |Â

that's very good indeed!
â€“Â Jean-FranÃ§ois Fabre
3 hours ago

Also, the assignment is never reached in the first. You cannot differentiate the third case where both would raise an error.
â€“Â schwobaseggl
3 hours ago

@schwobaseggl That's right, as the right hand side is evaluated first, it is where the error would be raised.
â€“Â Thierry Lathuille
3 hours ago

good spot! and i totally agree this should not be relied on though as an implementation detail
â€“Â Chris_Rands
2 hours ago

that's very good indeed!
â€“Â Jean-FranÃ§ois Fabre
3 hours ago

Also, the assignment is never reached in the first. You cannot differentiate the third case where both would raise an error.
â€“Â schwobaseggl
3 hours ago

@schwobaseggl That's right, as the right hand side is evaluated first, it is where the error would be raised.
â€“Â Thierry Lathuille
3 hours ago

good spot! and i totally agree this should not be relied on though as an implementation detail
â€“Â Chris_Rands
2 hours ago

add a commentÂ |Â

up vote
2
down vote

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.

edited 3 hours ago

answered 3 hours ago

Olivier MelanÃ§on

11.5k11436

add a commentÂ |Â

up vote
2
down vote

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.

edited 3 hours ago

answered 3 hours ago

Olivier MelanÃ§on

11.5k11436

add a commentÂ |Â

up vote
2
down vote

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.

edited 3 hours ago

answered 3 hours ago

Olivier MelanÃ§on

11.5k11436

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.

edited 3 hours ago

answered 3 hours ago

Olivier MelanÃ§on

11.5k11436

edited 3 hours ago

answered 3 hours ago

Olivier MelanÃ§on

11.5k11436

answered 3 hours ago

Olivier MelanÃ§on

11.5k11436

answered 3 hours ago

Olivier MelanÃ§on

11.5k11436

add a commentÂ |Â

up vote
1
down vote

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)
 node = ast.parse(line.strip(), filename)
 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(node)

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

edited 13 mins ago

answered 50 mins ago

blhsing

14.5k2630

Nice. Perhaps you could edit this answer so that everything in the except block is in a function that can be used whenever the developer needs to unpick what caused an IndexError?
â€“Â David Board
19 mins ago

Sure. Edited as suggested then.
â€“Â blhsing
13 mins ago

add a commentÂ |Â

up vote
1
down vote

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)
 node = ast.parse(line.strip(), filename)
 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(node)

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

edited 13 mins ago

answered 50 mins ago

blhsing

14.5k2630

Nice. Perhaps you could edit this answer so that everything in the except block is in a function that can be used whenever the developer needs to unpick what caused an IndexError?
â€“Â David Board
19 mins ago

Sure. Edited as suggested then.
â€“Â blhsing
13 mins ago

add a commentÂ |Â

up vote
1
down vote

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)
 node = ast.parse(line.strip(), filename)
 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(node)

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

edited 13 mins ago

answered 50 mins ago

blhsing

14.5k2630

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)
 node = ast.parse(line.strip(), filename)
 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(node)

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

edited 13 mins ago

answered 50 mins ago

blhsing

14.5k2630

edited 13 mins ago

answered 50 mins ago

blhsing

14.5k2630

answered 50 mins ago

blhsing

14.5k2630

answered 50 mins ago

blhsing

14.5k2630

Nice. Perhaps you could edit this answer so that everything in the except block is in a function that can be used whenever the developer needs to unpick what caused an IndexError?
â€“Â David Board
19 mins ago

Sure. Edited as suggested then.
â€“Â blhsing
13 mins ago

add a commentÂ |Â

Nice. Perhaps you could edit this answer so that everything in the except block is in a function that can be used whenever the developer needs to unpick what caused an IndexError?
â€“Â David Board
19 mins ago

Sure. Edited as suggested then.
â€“Â blhsing
13 mins ago

Nice. Perhaps you could edit this answer so that everything in the except block is in a function that can be used whenever the developer needs to unpick what caused an IndexError?
â€“Â David Board
19 mins ago

Sure. Edited as suggested then.
â€“Â blhsing
13 mins ago

add a commentÂ |Â

up vote
0
down vote

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

answered 3 hours ago

Chirila Alexandru

1,12431736

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up vote
0
down vote

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

answered 3 hours ago

Chirila Alexandru

1,12431736

add a commentÂ |Â

up vote
0
down vote

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

answered 3 hours ago

Chirila Alexandru

1,12431736

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

answered 3 hours ago

Chirila Alexandru

1,12431736

answered 3 hours ago

Chirila Alexandru

1,12431736

answered 3 hours ago

Chirila Alexandru

1,12431736

answered 3 hours ago

Chirila Alexandru

1,12431736

add a commentÂ |Â

up vote
0
down vote

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')

answered 3 hours ago

Xnkr

178211

add a commentÂ |Â

up vote
0
down vote

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')

answered 3 hours ago

Xnkr

178211

add a commentÂ |Â

up vote
0
down vote

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')

answered 3 hours ago

Xnkr

178211

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')

answered 3 hours ago

Xnkr

178211

answered 3 hours ago

Xnkr

178211

answered 3 hours ago

Xnkr

178211

answered 3 hours ago

Xnkr

178211

add a commentÂ |Â

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