Re: list of the lists - append after search
On Thursday, March 2, 2017 at 9:33:14 PM UTC+5:30, Andrew Zyman wrote:
> Hello,
> please advise.
>
> I'd like search and append the internal list in the list-of-the-lists.
>
> Example:
> ll =[ [a,1], [b,2], [c,3], [blah, 1000] ]
>
> i want to search for the internal [] based on the string field and, if
> matches, append that list with a value.
>
> if internal_list[0] == 'blah':
> ll[ internal_list].append = [ 'new value']
>
> End result:
> ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
>
>
> I came up with the following, but the second stmnt is not correct:
>
> print [x for x in ll if x[0]== 'blah']
> print ll.index([x for x in ll if x[0]=='blah'])
>
> output:
> ValueError: [['blah', 1]] is not in list
>
>
>
>
> thank you
> AZ
list_of_lists = [['a', 1], ['b', 2], ['c', 3], ['blah', 1000]]
for sublist in list_of_lists:
if sublist[0] == 'blah':
sublist.append('new value')
Hope that helps.
--
https://mail.python.org/mailman/listinfo/python-list
Re: list of the lists - append after search
On Thursday, March 2, 2017 at 3:53:25 PM UTC-5, Andrew Zyman wrote:
> On Thursday, March 2, 2017 at 3:31:36 PM UTC-5, Jussi Piitulainen wrote:
> > Andrew Zyman writes:
> >
> > > On Thursday, March 2, 2017 at 2:57:02 PM UTC-5, Jussi Piitulainen wrote:
> > >> Peter Otten <__pete...@web.de> writes:
> > >>
> > >> > Andrew Zyman wrote:
> > >> >
> > >> >> On Thursday, March 2, 2017 at 11:27:34 AM UTC-5, Peter Otten wrote:
> > >> >>> Andrew Zyman wrote:
> > >> >>> .
> > >> >>> .
> > >> >>> > End result:
> > >> >>> > ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
> > >> >>>
> > >> >>> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> > >> >>> >>> for inner in outer:
> > >> >>> ... if inner[0] == "blah":
> > >> >>> ... inner.append("new value")
> > >> >>
> > >> >> thank you. this will do.
> > >> >> Just curious, is the above loop can be done in a one-liner?
> > >> >
> > >> > Ah, that newbie obsession ;)
> > >> >
> > >> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> > >> [inner + ["new value"] if inner[0] == "blah" else inner for inner
> > >> in
> > >> > outer]
> > >> > [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
> > >> >
> > >> > Note that there is a technical difference to be aware of -- matching
> > >> > lists are replaced rather than modified.
> > >>
> > >> I take it you are too sane, or too kind, to suggest the obvious
> > >> solution:
> > >>
> > >> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> > >> >>> [inner.append("new value") for inner in outer if inner[0] == "blah"]
> > >> [None]
> > >> >>> outer
> > >> [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
> > >>
> > >> [snip]
> > >
> > > Arh!!! this is it :)
> > >
> > > I'm sure i'll regret this line of code in 2 weeks - after i
> > > successfully forget what i wanted to achieve :)
> >
> > Jokes aside, you should strive to express your intention in your code.
> > Be kind to your future self. Write the three-line loop if you want
> > in-place modification.
> >
> > I use comprehensions a lot, but not to save lines. I might make Peter's
> > expression above a four-liner to make its structure more visible:
> >
> > res = [ ( inner + ["new value"]
> > if inner[0] == "blah"
> > else inner )
> > for inner in outer ]
> >
> > Maybe.
>
> thank you gentlemen. This is very helpful discussion and i appreciate your
> explanations - helped a lot.
Well, i'll take my earlier statement ( about using lists ) back . 100K
search/expend on list is taking way too much time ( over 4 minutes ). i'll
follow your advice and move the processing to dict.
Thanx again.
--
https://mail.python.org/mailman/listinfo/python-list
Re: list of the lists - append after search
On Thursday, March 2, 2017 at 3:31:36 PM UTC-5, Jussi Piitulainen wrote:
> Andrew Zyman writes:
>
> > On Thursday, March 2, 2017 at 2:57:02 PM UTC-5, Jussi Piitulainen wrote:
> >> Peter Otten <__pete...@web.de> writes:
> >>
> >> > Andrew Zyman wrote:
> >> >
> >> >> On Thursday, March 2, 2017 at 11:27:34 AM UTC-5, Peter Otten wrote:
> >> >>> Andrew Zyman wrote:
> >> >>> .
> >> >>> .
> >> >>> > End result:
> >> >>> > ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
> >> >>>
> >> >>> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> >> >>> >>> for inner in outer:
> >> >>> ... if inner[0] == "blah":
> >> >>> ... inner.append("new value")
> >> >>
> >> >> thank you. this will do.
> >> >> Just curious, is the above loop can be done in a one-liner?
> >> >
> >> > Ah, that newbie obsession ;)
> >> >
> >> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> >> [inner + ["new value"] if inner[0] == "blah" else inner for inner in
> >> > outer]
> >> > [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
> >> >
> >> > Note that there is a technical difference to be aware of -- matching
> >> > lists are replaced rather than modified.
> >>
> >> I take it you are too sane, or too kind, to suggest the obvious
> >> solution:
> >>
> >> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> >> >>> [inner.append("new value") for inner in outer if inner[0] == "blah"]
> >> [None]
> >> >>> outer
> >> [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
> >>
> >> [snip]
> >
> > Arh!!! this is it :)
> >
> > I'm sure i'll regret this line of code in 2 weeks - after i
> > successfully forget what i wanted to achieve :)
>
> Jokes aside, you should strive to express your intention in your code.
> Be kind to your future self. Write the three-line loop if you want
> in-place modification.
>
> I use comprehensions a lot, but not to save lines. I might make Peter's
> expression above a four-liner to make its structure more visible:
>
> res = [ ( inner + ["new value"]
> if inner[0] == "blah"
> else inner )
> for inner in outer ]
>
> Maybe.
thank you gentlemen. This is very helpful discussion and i appreciate your
explanations - helped a lot.
--
https://mail.python.org/mailman/listinfo/python-list
Re: list of the lists - append after search
Andrew Zyman writes:
> On Thursday, March 2, 2017 at 2:57:02 PM UTC-5, Jussi Piitulainen wrote:
>> Peter Otten <__pete...@web.de> writes:
>>
>> > Andrew Zyman wrote:
>> >
>> >> On Thursday, March 2, 2017 at 11:27:34 AM UTC-5, Peter Otten wrote:
>> >>> Andrew Zyman wrote:
>> >>> .
>> >>> .
>> >>> > End result:
>> >>> > ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
>> >>>
>> >>> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
>> >>> >>> for inner in outer:
>> >>> ... if inner[0] == "blah":
>> >>> ... inner.append("new value")
>> >>
>> >> thank you. this will do.
>> >> Just curious, is the above loop can be done in a one-liner?
>> >
>> > Ah, that newbie obsession ;)
>> >
>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
>> [inner + ["new value"] if inner[0] == "blah" else inner for inner in
>> > outer]
>> > [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
>> >
>> > Note that there is a technical difference to be aware of -- matching
>> > lists are replaced rather than modified.
>>
>> I take it you are too sane, or too kind, to suggest the obvious
>> solution:
>>
>> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
>> >>> [inner.append("new value") for inner in outer if inner[0] == "blah"]
>> [None]
>> >>> outer
>> [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
>>
>> [snip]
>
> Arh!!! this is it :)
>
> I'm sure i'll regret this line of code in 2 weeks - after i
> successfully forget what i wanted to achieve :)
Jokes aside, you should strive to express your intention in your code.
Be kind to your future self. Write the three-line loop if you want
in-place modification.
I use comprehensions a lot, but not to save lines. I might make Peter's
expression above a four-liner to make its structure more visible:
res = [ ( inner + ["new value"]
if inner[0] == "blah"
else inner )
for inner in outer ]
Maybe.
--
https://mail.python.org/mailman/listinfo/python-list
Re: list of the lists - append after search
On Thursday, March 2, 2017 at 2:57:02 PM UTC-5, Jussi Piitulainen wrote:
> Peter Otten <__pete...@web.de> writes:
>
> > Andrew Zyman wrote:
> >
> >> On Thursday, March 2, 2017 at 11:27:34 AM UTC-5, Peter Otten wrote:
> >>> Andrew Zyman wrote:
> >>> .
> >>> .
> >>> > End result:
> >>> > ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
> >>>
> >>> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> >>> >>> for inner in outer:
> >>> ... if inner[0] == "blah":
> >>> ... inner.append("new value")
> >>
> >> thank you. this will do.
> >> Just curious, is the above loop can be done in a one-liner?
> >
> > Ah, that newbie obsession ;)
> >
> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> [inner + ["new value"] if inner[0] == "blah" else inner for inner in
> > outer]
> > [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
> >
> > Note that there is a technical difference to be aware of -- matching
> > lists are replaced rather than modified.
>
> I take it you are too sane, or too kind, to suggest the obvious
> solution:
>
> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> >>> [inner.append("new value") for inner in outer if inner[0] == "blah"]
> [None]
> >>> outer
> [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
>
> [snip]
Arh!!! this is it :)
I'm sure i'll regret this line of code in 2 weeks - after i successfully forget
what i wanted to achieve :)
--
https://mail.python.org/mailman/listinfo/python-list
Re: list of the lists - append after search
Peter Otten <__pete...@web.de> writes:
> Andrew Zyman wrote:
>
>> On Thursday, March 2, 2017 at 11:27:34 AM UTC-5, Peter Otten wrote:
>>> Andrew Zyman wrote:
>>> .
>>> .
>>> > End result:
>>> > ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
>>>
>>> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
>>> >>> for inner in outer:
>>> ... if inner[0] == "blah":
>>> ... inner.append("new value")
>>
>> thank you. this will do.
>> Just curious, is the above loop can be done in a one-liner?
>
> Ah, that newbie obsession ;)
>
outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
[inner + ["new value"] if inner[0] == "blah" else inner for inner in
> outer]
> [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
>
> Note that there is a technical difference to be aware of -- matching
> lists are replaced rather than modified.
I take it you are too sane, or too kind, to suggest the obvious
solution:
>>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
>>> [inner.append("new value") for inner in outer if inner[0] == "blah"]
[None]
>>> outer
[['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
[snip]
--
https://mail.python.org/mailman/listinfo/python-list
Re: list of the lists - append after search
Andrew Zyman wrote:
> On Thursday, March 2, 2017 at 11:27:34 AM UTC-5, Peter Otten wrote:
>> Andrew Zyman wrote:
>> .
>> .
>> > End result:
>> > ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
>>
>> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
>> >>> for inner in outer:
>> ... if inner[0] == "blah":
>> ... inner.append("new value")
>
> thank you. this will do.
> Just curious, is the above loop can be done in a one-liner?
Ah, that newbie obsession ;)
>>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
>>> [inner + ["new value"] if inner[0] == "blah" else inner for inner in
outer]
[['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
Note that there is a technical difference to be aware of -- matching lists
are replaced rather than modified.
>> While this is what you are asking for it will get slower as the list
>> grows. A better solution uses a dictionary:
>
>> >>> lookup = {inner[0]: inner[1:] for inner in outer}
>
> Yes, understood, i don't expect the list to grow above a few thousand
> entries. But i do agree that utilizing the dict is more efficient.
--
https://mail.python.org/mailman/listinfo/python-list
Re: list of the lists - append after search
On Thursday, March 2, 2017 at 11:27:34 AM UTC-5, Peter Otten wrote:
> Andrew Zyman wrote:
> .
> .
> > End result:
> > ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
>
> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> >>> for inner in outer:
> ... if inner[0] == "blah":
> ... inner.append("new value")
thank you. this will do.
Just curious, is the above loop can be done in a one-liner?
> While this is what you are asking for it will get slower as the list grows.
> A better solution uses a dictionary:
> >>> lookup = {inner[0]: inner[1:] for inner in outer}
Yes, understood, i don't expect the list to grow above a few thousand entries.
But i do agree that utilizing the dict is more efficient.
--
https://mail.python.org/mailman/listinfo/python-list
Re: list of the lists - append after search
Andrew Zyman wrote:
> Hello,
> please advise.
>
> I'd like search and append the internal list in the list-of-the-lists.
>
> Example:
> ll =[ [a,1], [b,2], [c,3], [blah, 1000] ]
>
> i want to search for the internal [] based on the string field and, if
> matches, append that list with a value.
>
> if internal_list[0] == 'blah':
> ll[ internal_list].append = [ 'new value']
>
> End result:
> ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
>>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
>>> for inner in outer:
... if inner[0] == "blah":
... inner.append("new value")
...
>>> outer
[['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
While this is what you are asking for it will get slower as the list grows.
A better solution uses a dictionary:
>>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
>>> lookup = {inner[0]: inner[1:] for inner in outer}
>>> lookup
{'a': [1], 'blah': [1000], 'b': [2], 'c': [3]}
>>> lookup["blah"].append("whatever")
>>> lookup
{'a': [1], 'blah': [1000, 'whatever'], 'b': [2], 'c': [3]}
--
https://mail.python.org/mailman/listinfo/python-list
list of the lists - append after search
Hello, please advise. I'd like search and append the internal list in the list-of-the-lists. Example: ll =[ [a,1], [b,2], [c,3], [blah, 1000] ] i want to search for the internal [] based on the string field and, if matches, append that list with a value. if internal_list[0] == 'blah': ll[ internal_list].append = [ 'new value'] End result: ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ] I came up with the following, but the second stmnt is not correct: print [x for x in ll if x[0]== 'blah'] print ll.index([x for x in ll if x[0]=='blah']) output: ValueError: [['blah', 1]] is not in list thank you AZ -- https://mail.python.org/mailman/listinfo/python-list
