Hello,
I'm a newbie to Python. I have a list which contains integers (about 80,000). I want to find a quick way to get the numbers that occur in the list more than once, and how many times that number is duplicated in the list. I've done this right now by
looping through the list, getting a number, querying the list to find out how many times the number exists, then writing it to a new list. On this many records it takes a couple of minutes. What I am looking for is something in python that can grab this info
without looping through a list.
Thanks!
Paul J. Scipione
GIS Database Administrator
work: 602-371-7091
cell: 480-980-4721
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On Thu, 26 Mar 2009 12:22:27 -0700
[hidden email] wrote: > I'm a newbie to Python. I have a list which contains integers (about 80,000). I want to find a quick way to get the numbers that occur in the list more than once, and how many times that number is duplicated in the list. I've done this right now by looping through the list, getting a number, querying the list to find out how many times the number exists, then writing it to a new list. On this many records it takes a couple of minutes. What I am looking for is something in python that can grab this info without looping through a list. icount = {} for i in list_of_ints: icount[i] = icount.get(i, 0) + 1 Now you have a dictionary of every integer in the list and the count of times it appears. -- D'Arcy J.M. Cain <[hidden email]> | Democracy is three wolves http://www.druid.net/darcy/ | and a sheep voting on +1 416 425 1212 (DoD#0082) (eNTP) | what's for dinner. -- http://mail.python.org/mailman/listinfo/python-list |
In reply to this post by Paul.Scipione
On Thu, 2009-03-26 at 12:22 -0700, [hidden email] wrote:
> Hello, > > I'm a newbie to Python. I have a list which contains integers (about > 80,000). I want to find a quick way to get the numbers that occur in > the list more than once, and how many times that number is duplicated > in the list. I've done this right now by looping through the list, > getting a number, querying the list to find out how many times the > number exists, then writing it to a new list. On this many records it > takes a couple of minutes. What I am looking for is something in > python that can grab this info without looping through a list. > Why not build a histogram? $ cat test.py from random import randint l = list() for i in xrange(80000): l.append(randint(0,10)) hist = dict() for i in l: hist[i] = hist.get(i, 0) + 1 for i in range(10): print "%s: %s" % (i, hist.get(i, 0)) $ time python test.py 0: 7275 1: 7339 2: 7303 3: 7348 4: 7206 5: 7323 6: 7230 7: 7348 8: 7166 9: 7180 real 0m0.533s user 0m0.518s sys 0m0.011s -- http://mail.python.org/mailman/listinfo/python-list |
On Thu, 2009-03-26 at 15:54 -0400, Albert Hopkins wrote:
[...] > $ cat test.py > from random import randint > > l = list() > for i in xrange(80000): > l.append(randint(0,10)) ^^^^^^^^^^^^^^^^^^^^^^^ should have been: l.append(randint(0,9)) > > hist = dict() > for i in l: > hist[i] = hist.get(i, 0) + 1 > > for i in range(10): > print "%s: %s" % (i, hist.get(i, 0)) > > > > $ time python test.py > 0: 7275 > 1: 7339 > 2: 7303 > 3: 7348 > 4: 7206 > 5: 7323 > 6: 7230 > 7: 7348 > 8: 7166 > 9: 7180 > > real 0m0.533s > user 0m0.518s > sys 0m0.011s -- http://mail.python.org/mailman/listinfo/python-list |
On Thu, 26 Mar 2009 16:00:01 -0400
Albert Hopkins <[hidden email]> wrote: > > l = list() > > for i in xrange(80000): > > l.append(randint(0,10)) > ^^^^^^^^^^^^^^^^^^^^^^^ > should have been: > l.append(randint(0,9)) Or even: l = [randint(0,9) for x in xrange(80000)] -- D'Arcy J.M. Cain <[hidden email]> | Democracy is three wolves http://www.druid.net/darcy/ | and a sheep voting on +1 416 425 1212 (DoD#0082) (eNTP) | what's for dinner. -- http://mail.python.org/mailman/listinfo/python-list |
In reply to this post by D'Arcy J.M. Cain
Hi D'Arcy J.M. Cain,
Thank you. I tried this and my list of 76,979 integers got reduced to a dictionary of 76,963 items, each item listing the integer value from the list, a comma, and a 1. I think what this is doing is finding all integers from my list that are unique (only one instance of it in the list), instead of creating a dictionary with integers that are not unique, with a count of how many times they occur. My dictionary should contain only 11 items listing 11 integer values and the number of times they appear in my original list. Thanks, Paul J. Scipione GIS Database Administrator work: 602-371-7091 cell: 480-980-4721 -----Original Message----- From: D'Arcy J.M. Cain [mailto:[hidden email]] Sent: Thursday, March 26, 2009 12:50 PM To: Scipione, Paul (ZP5296) Cc: [hidden email] Subject: Re: Find duplicates in a list and count them ... On Thu, 26 Mar 2009 12:22:27 -0700 [hidden email] wrote: > I'm a newbie to Python. I have a list which contains integers (about 80,000). I want to find a quick way to get the numbers that occur in the list more than once, and how many times that number is duplicated in the list. I've done this right now by looping through the list, getting a number, querying the list to find out how many times the number exists, then writing it to a new list. On this many records it takes a couple of minutes. What I am looking for is something in python that can grab this info without looping through a list. icount = {} for i in list_of_ints: icount[i] = icount.get(i, 0) + 1 Now you have a dictionary of every integer in the list and the count of times it appears. -- D'Arcy J.M. Cain <[hidden email]> | Democracy is three wolves http://www.druid.net/darcy/ | and a sheep voting on +1 416 425 1212 (DoD#0082) (eNTP) | what's for dinner. Email Firewall made the following annotations --------------------------------------------------------------------- --- NOTICE --- This message is for the designated recipient only and may contain confidential, privileged or proprietary information. If you have received it in error, please notify the sender immediately and delete the original and any copy or printout. Unintended recipients are prohibited from making any other use of this e-mail. Although we have taken reasonable precautions to ensure no viruses are present in this e-mail, we accept no liability for any loss or damage arising from the use of this e-mail or attachments, or for any delay or errors or omissions in the contents which result from e-mail transmission. --------------------------------------------------------------------- -- http://mail.python.org/mailman/listinfo/python-list |
On Thu, Mar 26, 2009 at 5:14 PM, <[hidden email]> wrote: Hi D'Arcy J.M. Cain, Not all of the values are 1. The 11 duplicates will be higher. Just iterate through the dict to find all keys with values > 1. >>> icounts {1: 2, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 5, 8: 3, 9: 1, 10: 1, 11: 1} Python 2.x : >>> dups = {} >>> for key, value in icounts.iteritems() : ... if value > 1 : ... dups[key] = value ... >>> dups {8: 3, 1: 2, 7: 5} Python 3.0 : >>> dups = {key:value for key, value in icounts.items() if value > 1} >>> dups {8: 3, 1: 2, 7: 5} -- http://mail.python.org/mailman/listinfo/python-list |
Benjamin Kaplan wrote:
> > > On Thu, Mar 26, 2009 at 5:14 PM, <[hidden email] > <mailto:[hidden email]>> wrote: > > Hi D'Arcy J.M. Cain, > > Thank you. I tried this and my list of 76,979 integers got reduced > to a dictionary of 76,963 items, each item listing the integer value > from the list, a comma, and a 1. I think what this is doing is > finding all integers from my list that are unique (only one instance > of it in the list), instead of creating a dictionary with integers > that are not unique, with a count of how many times they occur. My > dictionary should contain only 11 items listing 11 integer values > and the number of times they appear in my original list. > > > > Not all of the values are 1. The 11 duplicates will be higher. Just > iterate through the dict to find all keys with values > 1. > > >>> icounts > {1: 2, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 5, 8: 3, 9: 1, 10: 1, 11: 1} > > Python 2.x : > >>> dups = {} > >>> for key, value in icounts.iteritems() : > ... if value > 1 : > ... dups[key] = value > ... > >>> dups > {8: 3, 1: 2, 7: 5} > > > Python 3.0 : > >>> dups = {key:value for key, value in icounts.items() if value > 1} > >>> dups > {8: 3, 1: 2, 7: 5} > >>> dups = dict((key, value) for key, value in icounts.iteritems() if value > 1) >>> dups {8: 3, 1: 2, 7: 5} -- http://mail.python.org/mailman/listinfo/python-list |
In reply to this post by D'Arcy J.M. Cain
On Mar 27, 8:14 am, [hidden email] wrote:
> Hi D'Arcy J.M. Cain, > > Thank you. I tried this and my list of 76,979 integers got reduced to a dictionary of 76,963 items, each item listing the integer value from the list, a comma, and a 1. I doubt this very much. Please show: (a) your implementation of D'Arcy's suggestion (b) the code you used that lead you to the conclusion that all counts were 1. See example below. > I think what this is doing is finding all integers from my list that are unique (only one instance of it in the list), instead of creating a dictionary with integers that are not unique, with a count of how many times they occur. My dictionary should contain only 11 items listing 11 integer values and the number of times they appear in my original list. The only way of getting your desired result is to get a dict of counts and then to filter out the ones where the count is greater than one. D'Arcy appears to have presumed that it was not necessary to show the second stage :-) [assuming Python 2.6] >>> list_of_ints = [999, 2, 3, 999, 2, 2, 8, 42, 999, 42, 5] >>> len(list_of_ints) 11 >>> icount = {} >>> for i in list_of_ints: ... icount[i] = icount.get(i, 0) + 1 ... >>> icount {2: 3, 3: 1, 5: 1, 999: 3, 8: 1, 42: 2} >>> len(icount) 6 >>> all(count == 1 for count in icount.itervalues()) False >>> dups = dict((k, v) for k, v in icount.iteritems() if v > 1) >>> dups {2: 3, 42: 2, 999: 3} >>> HTH, John -- http://mail.python.org/mailman/listinfo/python-list |
In reply to this post by D'Arcy J.M. Cain
On Thu, 26 Mar 2009 16:02:20 -0400
"D'Arcy J.M. Cain" <[hidden email]> wrote: or l = ( randint(0,9) for x in xrange(80000) ) > On Thu, 26 Mar 2009 16:00:01 -0400 > Albert Hopkins <[hidden email]> wrote: > > > l = list() > > > for i in xrange(80000): > > > l.append(randint(0,10)) > > ^^^^^^^^^^^^^^^^^^^^^^^ > > should have been: > > l.append(randint(0,9)) > > Or even: > > l = [randint(0,9) for x in xrange(80000)] > -- Josh Dukes MicroVu IT Department -- http://mail.python.org/mailman/listinfo/python-list |
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