Re: [Tutor] counting a list of elements
On 04/02/2011 07:00 AM, Knacktus wrote:
Am 01.04.2011 21:31, schrieb Karim:
On 04/01/2011 08:41 PM, Knacktus wrote:
Am 01.04.2011 19:16, schrieb Karim:
Hello,
I would like to get advice on the best practice to count elements in a
list (built from scractch).
The number of elements is in the range 1e3 and 1e6.
1) I could create a generator and set a counter (i +=1) in the loop.
2) or simply len(mylist).
I don't need the content of the list, indeed, in term of memory I
don't
want to wast it. But I suppose len() is optimized too (C
impementation).
If you have some thought to share don't hesitate.
Just a general suggestion: Provide code examples. I know most of the
times you don't have code examples yet as you're thinking of how to
solve your problems. But if you post one of the possible solutions the
experienced guys here will very likely direct you in the proper
direction. But without code it's hard to understand what you're after.
Cheers,
Jan
Thank you all for you answers to clarified I built a collection of
dictionnaries which represent database query on a bug tracking system:
backlog_tables , csv_backlog_table = _backlog_database(table=table,
periods=intervals_list)
backlog_tables is a dictionnary of bug info dictionnaries. The keys of
backlog_tables is a time intervall (YEAR-MONTH) as shown below:
backlog_tables= {'2011-01-01': [{'Assigned Date': datetime.date(2010,
10, 25),
'Category': 'Customer_claim',
'Date': datetime.date(2010, 10, 22),
'Duplicate Date': None,
'Fixed Reference': None,
'Headline': 'Impovement for all test',
'Identifier': '23269',
'Last Modified': datetime.date(2010, 10, 25),
'Priority': 'Low',
'Project': 'MY_PROJECT',
'Reference': 'MY_PROJECT@1.7beta2@20101006.0',
'Resolved Date': None,
'Severity': 'improvement',
'State': 'A',
'Submitter': 'Somebody'},
.
}
_backlog_database() compute the tuple backlog_tables ,
csv_backlog_table:
In fact csv_backlog_table is the same as backlog_tables but instead of
having
the query dictionnaries it holds only the number of query which I use to
create
a CSV file and a graph over time range.
_backlog_database() is as follow:
def _backlog_database(table=None, periods=None):
"""Internal function. Re-arrange database table
according to a time period. Only monthly management
is computed in this version.
@param table the database of the list of defects. Each defect is a
dictionnary with fixed keys.
@param periods the intervals list of months and the first element is the
starting date and the
the last element is the ending date in string format.
@return (periods_tables, csv_table), a tuple of periodic dictionnary
table and
the same keys dictionnary with defect numbers associated values.
"""
if periods is None:
raise ValueError('Time interval could not be empty!')
periods_tables = {}
csv_table = {}
interval_table = []
for interval in periods:
split_date = interval.split('-')
for row in table:
if not len(split_date) == 3:
limit_date = _first_next_month_day(year=int(split_date[0]),
month=int(split_date[1]), day=1)
if row['Date'] < limit_date:
if not row['Resolved Date']:
if row['State'] == 'K':
if row['Last Modified'] >= limit_date:
interval_table.append(row)
elif row['State'] == 'D':
if row['Duplicate Date'] >= limit_date:
interval_table.append(row)
# New, Assigned, Opened, Postponed, Forwarded, cases.
else:
interval_table.append(row)
else:
if row['Resolved Date'] >= limit_date:
interval_table.append(row)
periods_tables[interval] = interval_table
csv_table[interval] = str(len(interval_table))
interval_table = []
return periods_tables, csv_table
This is not the whole function I reduce it on normal case but it shows
what I am doing.
In fact I choose to have both dictionnaries to debug my function and
analyse what's going
on. When everything will be fine I will need only the csv table (with
number per period) to create the graphs.
That's why I was asking for length computing. Honnestly, the actual
queries number is 500 (bug id) but It could be more
in other project. I was ambitious when I sais 1000 to 10
dictionnaries elements but for the whole
list of products we have internally It could be 5.
I see some similarity with my coding style (doing things "by the
way"), which might not be so good ;-).
With this background information I would keep the responsibilities
seperated. Your _backlog_database() function is supposed to do one
thing: Return a dictionary which holds the interval and a list of
result dicts. You could call this dict interval_to_result_tables (to
indicate that the values are lists). That's all your function should do.
Then you want to print a report. This piece of functionality needs to
know how long the lists for each dictionary entry are. Then this
print_report function should be responsible to get the information it
needs by creating it itself or calling another function, which has the
purpose to create the information. Latter would be a bit too much, as
the length would be simply b
Re: [Tutor] counting a list of elements
Am 01.04.2011 21:31, schrieb Karim:
On 04/01/2011 08:41 PM, Knacktus wrote:
Am 01.04.2011 19:16, schrieb Karim:
Hello,
I would like to get advice on the best practice to count elements in a
list (built from scractch).
The number of elements is in the range 1e3 and 1e6.
1) I could create a generator and set a counter (i +=1) in the loop.
2) or simply len(mylist).
I don't need the content of the list, indeed, in term of memory I don't
want to wast it. But I suppose len() is optimized too (C impementation).
If you have some thought to share don't hesitate.
Just a general suggestion: Provide code examples. I know most of the
times you don't have code examples yet as you're thinking of how to
solve your problems. But if you post one of the possible solutions the
experienced guys here will very likely direct you in the proper
direction. But without code it's hard to understand what you're after.
Cheers,
Jan
Thank you all for you answers to clarified I built a collection of
dictionnaries which represent database query on a bug tracking system:
backlog_tables , csv_backlog_table = _backlog_database(table=table,
periods=intervals_list)
backlog_tables is a dictionnary of bug info dictionnaries. The keys of
backlog_tables is a time intervall (YEAR-MONTH) as shown below:
backlog_tables= {'2011-01-01': [{'Assigned Date': datetime.date(2010,
10, 25),
'Category': 'Customer_claim',
'Date': datetime.date(2010, 10, 22),
'Duplicate Date': None,
'Fixed Reference': None,
'Headline': 'Impovement for all test',
'Identifier': '23269',
'Last Modified': datetime.date(2010, 10, 25),
'Priority': 'Low',
'Project': 'MY_PROJECT',
'Reference': 'MY_PROJECT@1.7beta2@20101006.0',
'Resolved Date': None,
'Severity': 'improvement',
'State': 'A',
'Submitter': 'Somebody'},
.
}
_backlog_database() compute the tuple backlog_tables , csv_backlog_table:
In fact csv_backlog_table is the same as backlog_tables but instead of
having
the query dictionnaries it holds only the number of query which I use to
create
a CSV file and a graph over time range.
_backlog_database() is as follow:
def _backlog_database(table=None, periods=None):
"""Internal function. Re-arrange database table
according to a time period. Only monthly management
is computed in this version.
@param table the database of the list of defects. Each defect is a
dictionnary with fixed keys.
@param periods the intervals list of months and the first element is the
starting date and the
the last element is the ending date in string format.
@return (periods_tables, csv_table), a tuple of periodic dictionnary
table and
the same keys dictionnary with defect numbers associated values.
"""
if periods is None:
raise ValueError('Time interval could not be empty!')
periods_tables = {}
csv_table = {}
interval_table = []
for interval in periods:
split_date = interval.split('-')
for row in table:
if not len(split_date) == 3:
limit_date = _first_next_month_day(year=int(split_date[0]),
month=int(split_date[1]), day=1)
if row['Date'] < limit_date:
if not row['Resolved Date']:
if row['State'] == 'K':
if row['Last Modified'] >= limit_date:
interval_table.append(row)
elif row['State'] == 'D':
if row['Duplicate Date'] >= limit_date:
interval_table.append(row)
# New, Assigned, Opened, Postponed, Forwarded, cases.
else:
interval_table.append(row)
else:
if row['Resolved Date'] >= limit_date:
interval_table.append(row)
periods_tables[interval] = interval_table
csv_table[interval] = str(len(interval_table))
interval_table = []
return periods_tables, csv_table
This is not the whole function I reduce it on normal case but it shows
what I am doing.
In fact I choose to have both dictionnaries to debug my function and
analyse what's going
on. When everything will be fine I will need only the csv table (with
number per period) to create the graphs.
That's why I was asking for length computing. Honnestly, the actual
queries number is 500 (bug id) but It could be more
in other project. I was ambitious when I sais 1000 to 10
dictionnaries elements but for the whole
list of products we have internally It could be 5.
I see some similarity with my coding style (doing things "by the way"),
which might not be so good ;-).
With this background information I would keep the responsibilities
seperated. Your _backlog_database() function is supposed to do one
thing: Return a dictionary which holds the interval and a list of result
dicts. You could call this dict interval_to_result_tables (to indicate
that the values are lists). That's all your function should do.
Then you want to print a report. This piece of functionality needs to
know how long the lists for each dictionary entry are. Then this
print_report function should be responsible to get the information it
needs by creating it itself or calling another function, which has the
purpose to create the information. Latter would be a bit too much, as
the length would be simply be:
number_of_tables = len(interval_to_result
Re: [Tutor] counting a list of elements
On 04/01/2011 08:41 PM, Knacktus wrote:
Am 01.04.2011 19:16, schrieb Karim:
Hello,
I would like to get advice on the best practice to count elements in a
list (built from scractch).
The number of elements is in the range 1e3 and 1e6.
1) I could create a generator and set a counter (i +=1) in the loop.
2) or simply len(mylist).
I don't need the content of the list, indeed, in term of memory I don't
want to wast it. But I suppose len() is optimized too (C impementation).
If you have some thought to share don't hesitate.
Just a general suggestion: Provide code examples. I know most of the
times you don't have code examples yet as you're thinking of how to
solve your problems. But if you post one of the possible solutions the
experienced guys here will very likely direct you in the proper
direction. But without code it's hard to understand what you're after.
Cheers,
Jan
Thank you all for you answers to clarified I built a collection of
dictionnaries which represent database query on a bug tracking system:
backlog_tables , csv_backlog_table = _backlog_database(table=table,
periods=intervals_list)
backlog_tables is a dictionnary of bug info dictionnaries. The keys of
backlog_tables is a time intervall (YEAR-MONTH) as shown below:
backlog_tables= {'2011-01-01': [{'Assigned Date': datetime.date(2010,
10, 25),
'Category': 'Customer_claim',
'Date': datetime.date(2010, 10, 22),
'Duplicate Date': None,
'Fixed Reference': None,
'Headline': 'Impovement for all test',
'Identifier': '23269',
'Last Modified': datetime.date(2010, 10, 25),
'Priority': 'Low',
'Project': 'MY_PROJECT',
'Reference': 'MY_PROJECT@1.7beta2@20101006.0',
'Resolved Date': None,
'Severity': 'improvement',
'State': 'A',
'Submitter': 'Somebody'},
.
}
_backlog_database() compute the tuple backlog_tables , csv_backlog_table:
In fact csv_backlog_table is the same as backlog_tables but instead of
having
the query dictionnaries it holds only the number of query which I use to
create
a CSV file and a graph over time range.
_backlog_database() is as follow:
def _backlog_database(table=None, periods=None):
"""Internal function. Re-arrange database table
according to a time period. Only monthly management
is computed in this version.
@param table the database of the list of defects. Each defect is
a dictionnary with fixed keys.
@param periods the intervals list of months and the first element
is the starting date and the
the last element is the ending date in string format.
@return (periods_tables, csv_table), a tuple of periodic
dictionnary table and
the same keys dictionnary with defect numbers associated
values.
"""
if periods is None:
raise ValueError('Time interval could not be empty!')
periods_tables = {}
csv_table = {}
interval_table = []
for interval in periods:
split_date = interval.split('-')
for row in table:
if not len(split_date) == 3:
limit_date =
_first_next_month_day(year=int(split_date[0]), month=int(split_date[1]),
day=1)
if row['Date'] < limit_date:
if not row['Resolved Date']:
if row['State'] == 'K':
if row['Last Modified'] >= limit_date:
interval_table.append(row)
elif row['State'] == 'D':
if row['Duplicate Date'] >= limit_date:
interval_table.append(row)
# New, Assigned, Opened, Postponed, Forwarded,
cases.
else:
interval_table.append(row)
else:
if row['Resolved Date'] >= limit_date:
interval_table.append(row)
periods_tables[interval] = interval_table
csv_table[interval] = str(len(interval_table))
interval_table = []
return periods_tables, csv_table
This is not the whole function I reduce it on normal case but it shows
what I am doing.
In fact I choose to have both dictionnaries to debug my function and
analyse what's going
on. When everything will be fine I will need only the csv table (with
number per period) to create the graphs.
That's why I was asking for length computing. Honnestly, the actual
queries number is 500 (bug id) but It could be more
in other project. I was ambitious when I sais 1000 to 10
dictionnaries elements but for the whole
list of products we have internally It could be 5.
Regards
Karim
Karim
___
Tutor maillist - Tuto
Re: [Tutor] counting a list of elements
Am 01.04.2011 19:16, schrieb Karim: Hello, I would like to get advice on the best practice to count elements in a list (built from scractch). The number of elements is in the range 1e3 and 1e6. 1) I could create a generator and set a counter (i +=1) in the loop. 2) or simply len(mylist). I don't need the content of the list, indeed, in term of memory I don't want to wast it. But I suppose len() is optimized too (C impementation). If you have some thought to share don't hesitate. Just a general suggestion: Provide code examples. I know most of the times you don't have code examples yet as you're thinking of how to solve your problems. But if you post one of the possible solutions the experienced guys here will very likely direct you in the proper direction. But without code it's hard to understand what you're after. Cheers, Jan Karim ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] counting a list of elements
The nice thing about Python is you don't have to build things from scratch,
and I imagine most people would discourage on account of it being a waste of
time, other then education value.
But the "best practice" in my humble opinion would be to use the built in
len function. If what you are after is to see how often 1e3 comes up in your
list, then you would use list.count('1e3').
In general, its best to use the python built ins.
On Fri, Apr 1, 2011 at 1:16 PM, Karim wrote:
>
>
> Hello,
>
> I would like to get advice on the best practice to count elements in a list
> (built from scractch).
> The number of elements is in the range 1e3 and 1e6.
>
> 1) I could create a generator and set a counter (i +=1) in the loop.
>
> 2) or simply len(mylist).
>
> I don't need the content of the list, indeed, in term of memory I don't
> want to wast it. But I suppose len() is optimized too (C impementation).
>
> If you have some thought to share don't hesitate.
>
> Karim
> ___
> Tutor maillist - Tutor@python.org
> To unsubscribe or change subscription options:
> http://mail.python.org/mailman/listinfo/tutor
>
___
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] counting a list of elements
Am 01.04.2011 19:16, schrieb Karim: Hello, I would like to get advice on the best practice to count elements in a list (built from scractch). It's not clear to me what your starting point is. If you don't have a list and don't need a list, but have a large number of objects you create in your code sequentially or a large number of other events and you want to know how many of those objects exist / events have occured, then simply use a counter while creating. If you have a list (for whatever reason), then use len(). HTH, Jan The number of elements is in the range 1e3 and 1e6. 1) I could create a generator and set a counter (i +=1) in the loop. 2) or simply len(mylist). I don't need the content of the list, indeed, in term of memory I don't want to wast it. But I suppose len() is optimized too (C impementation). If you have some thought to share don't hesitate. Karim ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] counting a list of elements
On 4/1/2011 10:16 AM Karim said... I would like to get advice on the best practice to count elements Well, I suspect you're more interested in knowing the total count of how many as opposed to counting how many. To get the total count of how many use len(mylist). Emile ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] counting a list of elements
Hello, I would like to get advice on the best practice to count elements in a list (built from scractch). The number of elements is in the range 1e3 and 1e6. 1) I could create a generator and set a counter (i +=1) in the loop. 2) or simply len(mylist). I don't need the content of the list, indeed, in term of memory I don't want to wast it. But I suppose len() is optimized too (C impementation). If you have some thought to share don't hesitate. Karim ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
