[Tutor] IndexError: list index out of range [ Program work fine , but gives this message , guidance requested ]
Hi All I am trying to find out the duplicates in a list , as for beginning I wanted to compare the adjacent elements in list , for this purpose I wrote a script for checking the adjacent elements in a list, if same , it prints the result that , the adjacent numbers are same , it not it will print adjacent numbers are same My program works , but at the end of the program execution it throws error Traceback (most recent call last): File arrayduplinital.py, line 27, in ? if seats[j] seats[j+1]: IndexError: list index out of range I tried my best to find the logic , why this error is coming , I was not able to find the reason , I request your guidance for the reason for this error and how to avoid it I am adding the code which I wrote for this program ** find duplicates in a list (1) Sort the array (2) counter i = 0 j = 0 (3) for j = length if seats[j] seats[j+1] == print value j and j+1 are not same else === print value j , j +1 are same seats = [] i = 0 length = int(raw_input(Enter The Lenght of the array : )) while i length: num = int(raw_input(Enter the Seats : )) seats.append(num) i += 1 print seats seats.sort() j = 0 for j in range(length): if seats[j] seats[j+1]: print The Seat, j , And the seat, j+1 , value, which are, seats[j], and, seats[j+1], are not same else: print The Seats ,j ,And the seats, j+1, Values,, seats[j], and, seats[j+1], are same ~ ___ To help you stay safe and secure online, we've developed the all new Yahoo! Security Centre. http://uk.security.yahoo.com ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] IndexError: list index out of range [ Program work fine , but gives this message , guidance requested ]
Hi, Look at this: i=[1,2,3] i[len(i)] Traceback (most recent call last): File stdin, line 1, in ? IndexError: list index out of range This means that I have tried to access an element that is out of the list, in this case i[3], that is, the 4th element of the list. You are doing the same. You do a j in range(len(seats)) and then you are accessing your list with seats[j] and seats[j+1]. What happens when j = len(seats)-1 and you call seats[j+1]? :-) It happens indeed the same that in the example I gave you first. Hope it helps. Good luck, G John Joseph wrote: Hi All I am trying to find out the duplicates in a list , as for beginning I wanted to compare the adjacent elements in list , for this purpose I wrote a script for checking the adjacent elements in a list, if same , it prints the result that , the adjacent numbers are same , it not it will print adjacent numbers are same My program works , but at the end of the program execution it throws error “Traceback (most recent call last): File arrayduplinital.py, line 27, in ? if seats[j] seats[j+1]: IndexError: list index out of range” I tried my best to find the logic , why this error is coming , I was not able to find the reason , I request your guidance for the reason for this error and how to avoid it I am adding the code which I wrote for this program ** find duplicates in a list (1) Sort the array (2) counter i = 0 j = 0 (3) for j = length if seats[j] seats[j+1] == print value j and j+1 are not same else === print value j , j +1 are same seats = [] i = 0 length = int(raw_input(Enter The Lenght of the array : )) while i length: num = int(raw_input(Enter the Seats : )) seats.append(num) i += 1 print seats seats.sort() j = 0 for j in range(length): if seats[j] seats[j+1]: print The Seat, j , And the seat, j+1 , value, which are, seats[j], and, seats[j+1], are not same else: print The Seats ,j ,And the seats, j+1, Values,, seats[j], and, seats[j+1], are same ~ ___ To help you stay safe and secure online, we've developed the all new Yahoo! Security Centre. http://uk.security.yahoo.com ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] IndexError: list index out of range [ Program work fine , but gives this message , guidance requested ]
--- Guillermo Fernandez Castellanos [EMAIL PROTECTED] wrote: Hi, Look at this: i=[1,2,3] i[len(i)] Traceback (most recent call last): File stdin, line 1, in ? IndexError: list index out of range This means that I have tried to access an element that is out of the list, in this case i[3], that is, the 4th element of the list. You are doing the same. You do a j in range(len(seats)) and then you are accessing your list with seats[j] and seats[j+1]. What happens when j = len(seats)-1 and you call seats[j+1]? :-) It happens indeed the same that in the example I gave you first. Hope it helps. Good luck, Thanks for the advice But I need to display the results What should I do in the for loop for this message to go my for loop is as for j in range(length) : if seats[j] seats[j+1]: print The Seat, j , And the seat, j+1 , value, which are, seats[j], and, seats[j+1], are not same else: print The Seats ,j ,And the seats, j+1, Values,, seats[j], and, seats[j+1], are same Thanks Joseph John ___ Yahoo! Exclusive Xmas Game, help Santa with his celebrity party - http://santas-christmas-party.yahoo.net/ ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] Landscape Printing
Hi, My text file is printing out in portrait. Is there any instruction that I can use so that notepad prints it in landscape? Thanks, John. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] IndexError: list index out of range [ Program work fine , but gives this message , guidance requested ]
John Joseph said unto the world upon 08/01/06 06:36 AM: --- Guillermo Fernandez Castellanos [EMAIL PROTECTED] wrote: Hi, Look at this: i=[1,2,3] i[len(i)] Traceback (most recent call last): File stdin, line 1, in ? IndexError: list index out of range This means that I have tried to access an element that is out of the list, in this case i[3], that is, the 4th element of the list. You are doing the same. You do a j in range(len(seats)) and then you are accessing your list with seats[j] and seats[j+1]. What happens when j = len(seats)-1 and you call seats[j+1]? :-) It happens indeed the same that in the example I gave you first. Hope it helps. Good luck, Thanks for the advice But I need to display the results What should I do in the for loop for this message to go my for loop is as for j in range(length) : if seats[j] seats[j+1]: print The Seat, j , And the seat, j+1 , value, which are, seats[j], and, seats[j+1], are not same else: print The Seats ,j ,And the seats, j+1, Values,, seats[j], and, seats[j+1], are same Thanks Joseph John Joseph, I think Guillermo has already given you the answer :-). But, I'll make it more explicit. I'll also do a rather more. My intent is to give you some idea of how you can go about incrementally improving your code. The code you posted is doing something like: a_list = [7,8,7] for index in range(len(a_list)): print index, a_list[index] print index + 1, len(a_list), a_list[index + 1] 0 7 1 3 8 1 8 2 3 7 2 7 3 3 Traceback (most recent call last): File pyshell#97, line 3, in -toplevel- print index + 1, len(a_list), a_list[index + 1] IndexError: list index out of range Examine the output. When the iteration hits the last member of the list (when it gets to the final element of range(len(a_list)), there is no next element of the list, so the request to print the next element doesn't work. The list length is 3, so I get the same exception as if I'd done: a_list[3] Traceback (most recent call last): File pyshell#16, line 1, in -toplevel- a_list[3] IndexError: list index out of range directly. That suggests *not* asking for an iteration which uses up the list -- leave a next element and the code will work: for index in range(len(a_list) - 1): # Note the difference print index, a_list[index] print index + 1, len(a_list), a_list[index + 1] 0 7 1 3 8 1 8 2 3 7 Of course, as you pointed out in your OP, right now, even with this fix, you will only be testing for sequential duplicates. My test list a_list = [7,8,7] has dupes, but your approach won't find them. Warning: I'm going to go into a number of issues that you might not yet entirely understand. If I lose you, don't worry; just ask about it. Let's both fix the requirement that the duplicates be sequential and start putting the code into functions. Consider this: def dupe_detector_1(sequence): for item in sequence: if sequence.count(item) 1: print %s is duplicated! %item dupe_detector_1(a_list) 7 is duplicated! 7 is duplicated! OK, that might be better, as we can find non-sequential duplicates. But, there are two problems. 1) The duplication warning is duplicated :-) and 2) a_tuple = (4,3,4) dupe_detector_1(a_tuple) Traceback (most recent call last): File pyshell#47, line 1, in -toplevel- dupe_detector_1(a_tuple) File pyshell#28, line 3, in dupe_detector_1 if sequence.count(item) 1: AttributeError: 'tuple' object has no attribute 'count' The function signature seems to indicate it will work for all sequences, but it chokes on tuples. The second problem is easy to fix: def dupe_detector_2(sequence): # coerce to a list as lists have count methods. sequence = list(sequence) for item in sequence: if sequence.count(item) 1: print %s is duplicated! %item dupe_detector_2((1,1,2)) 1 is duplicated! 1 is duplicated! 1 down, 1 to go. def dupe_detector_3(sequence): sequence = list(sequence) seen_dupes = [] for item in sequence: if item in seen_dupes: # if it is there, we already know it is duplicated continue elif sequence.count(item) 1: print %s is duplicated! %item seen_dupes.append(item) dupe_detector_3(a_list) 7 is duplicated! That's much better :-) There remain 2 things I'd want to do differently. First, I'd separate the print logic from the detection logic: def dupe_detector_4(sequence): '''returns a list of items in sequence that are duplicated''' sequence = list(sequence) seen_dupes =
[Tutor] Python on Windows: any way to access shortcut's info?
I would like to know the name of the shortcut (link, I think, in Unix parlance) from which a python program is being invoked. I'd also like to be able to access the directory where the shortcut lives. Toy example: here's a directory structure: C:\ test\ argtest\ arga.py argb.py testdir\ argc.py arga.py contains: import sys,os print ARGS:, sys.argv print WD:, os.getcwd() raw_input() # just to keep the window open argb.py and argc.py are both shortcuts to arga.py When I run arga.py (e.g., by double-clicking on it, I get, as expected: ARGS: ['C:\\test\\argtest\\arga.py'] WD: C:\test\argtest When I run argb.py, I get the same thing. I'd hoped (but didn't really expect) sys.argv would be ['C:\\test\\argtest\\argb.py'] . When I run argc.py, same thing. In this case, I'd hoped sys.argv would be ['C:\\test\\argtest\\testdir\\argc.py']; and os.getcwd() would have given back 'C:\test\argtest\testdir'. Why do I care? Well, I'd like to set up a directory where the shortcut and all its files will live. I'd hoped keying off of the shortcut name might be an easy way to do this. I suspect I'm out of luck: my guess is that windows does all the lookup of what the shortcut points to, and then launches the file pointed to; and by the time Python gets control, the shortcut information is long gone. Anyone know for sure? ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Python on Windows: any way to access shortcut's info?
Terry Carroll wrote: I would like to know the name of the shortcut (link, I think, in Unix parlance) from which a python program is being invoked. I'd also like to be able to access the directory where the shortcut lives. You can set the command line arguments and working directory for a shortcut by opening the properties window on the shortcut, at least in Win2k. So for example in the properties for argc.py just add 'C:\\test\\argtest\\testdir\\argc.py to the Target field and set the Start in field to 'C:\\test\\argtest\\testdir' If arga.py imports any packages in argtest\ you will have to add that dir to sys.path somehow (many options for this). Kent Toy example: here's a directory structure: C:\ test\ argtest\ arga.py argb.py testdir\ argc.py arga.py contains: import sys,os print ARGS:, sys.argv print WD:, os.getcwd() raw_input() # just to keep the window open argb.py and argc.py are both shortcuts to arga.py When I run arga.py (e.g., by double-clicking on it, I get, as expected: ARGS: ['C:\\test\\argtest\\arga.py'] WD: C:\test\argtest When I run argb.py, I get the same thing. I'd hoped (but didn't really expect) sys.argv would be ['C:\\test\\argtest\\argb.py'] . When I run argc.py, same thing. In this case, I'd hoped sys.argv would be ['C:\\test\\argtest\\testdir\\argc.py']; and os.getcwd() would have given back 'C:\test\argtest\testdir'. Why do I care? Well, I'd like to set up a directory where the shortcut and all its files will live. I'd hoped keying off of the shortcut name might be an easy way to do this. I suspect I'm out of luck: my guess is that windows does all the lookup of what the shortcut points to, and then launches the file pointed to; and by the time Python gets control, the shortcut information is long gone. Anyone know for sure? ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
