Re: problem with interface of operator.itemgetter
2009/8/17 Simon Forman sajmik...@gmail.com You can use a little helper function to create your itemgetter like this: def makeItemGetter(indexes): I = itemgetter(*indexes) if len(indexes) 1: return I return lambda thing: (I(thing),) If indexes contains only one index the itemgetter is wrapped in a lambda that turns its output into a tuple. Thanks. I just thought if the stdlib could support some function like item*s *getter would be better, it always returns tuple instead of a item or a tuple of items. -- http://mail.python.org/mailman/listinfo/python-list
Re: problem with interface of operator.itemgetter
On Thu, Aug 13, 2009 at 11:16 PM, dou dounirvana...@gmail.com wrote:
I have a function to do some thing like LEFT JOIN in SQL, the function use
the itemgetter to get the ON and SELECT parameters of the two table(list
of list), the problem is that itemgetter may return a value or a tuple of
values, because of the inconsistent return type of itemgetter, I have to do
2^3 IF check,
the function like this:
def left_join(table1, on_index1, table2, on_index2, getter1, getter2):
do thing like:
SELECT on1(table1), g1(table1), g2(table2) FROM table1
LEFT JOIN table2
ON on1(table1) == on2(table2)
on1 = itemgetter(*on_index1)
on2 = itemgetter(*on_index2)
g1 = itemgetter(*getter1)
g2 = itemgetter(*getter2)
d2 = {}
d2 = dict((on2(r2), r2) for r2 in table2)
#if itemgetter always return tuple, below could simple as one line
#return [ list(on1(r1) + g1(r1) + g2(d2.get(on1(r1 for r1 in table1
]
len_on = len(on_index1)
len_g1 = len(getter1)
len_g2 = len(getter2)
if len_on == 1:
if len_g1 == 1 and len_g2 == 1:
return [ [on1(r1), g1(r1), g2(d2.get(on1(r1)))] for r1 in table1
]
elif len_g1 == 1 and len_g2 1:
return [ list((on1(r1),g1(r1))+g2(d2.get(on1(r1 for r1 in
table1 ]
elif len_g1 1 and len_g2 == 1:
return [ list((on1(r1),)+g1(r1)+(g2(d2.get(on1(r1))),)) for r1
in table1 ]
else: #len_g1 1 and len_g2 1:
return [ list((on1(r1),)+g1(r1)+g2(d2.get(on1(r1 for r1 in
table1 ]
else: # len_on 1
if len_g1 == 1 and len_g2 == 1:
return [ list(on1(r1))+[g1(r1),g2(d2.get(on1(r1)))] for r1 in
table1 ]
elif len_g1 == 1 and len_g2 1:
return [ list(on1(r1)+(g1(r1),)+g2(d2.get(on1(r1 for r1 in
table1 ]
elif len_g1 1 and len_g2 == 1:
return [ list(on1(r1)+g1(r1)+(g2(d2.get(on1(r1))),)) for r1 in
table1 ]
else: #len_g1 1 and len_g2 1:
return [ list(on1(r1)+g1(r1)+g2(d2.get(on1(r1 for r1 in
table1 ]
so is there a way to force itemgetter to return tuple even when
itemgetter(only_have_one_argument)? or some other function to do this?
Thanks.
You can use a little helper function to create your itemgetter like this:
def makeItemGetter(indexes):
I = itemgetter(*indexes)
if len(indexes) 1:
return I
return lambda thing: (I(thing),)
If indexes contains only one index the itemgetter is wrapped in a
lambda that turns its output into a tuple.
HTH,
~Simon
--
http://mail.python.org/mailman/listinfo/python-list
problem with interface of operator.itemgetter
I have a function to do some thing like LEFT JOIN in SQL, the function use
the itemgetter to get the ON and SELECT parameters of the two table(list
of list), the problem is that itemgetter may return a value or a tuple of
values, because of the inconsistent return type of itemgetter, I have to do
2^3 IF check,
the function like this:
def left_join(table1, on_index1, table2, on_index2, getter1, getter2):
do thing like:
SELECT on1(table1), g1(table1), g2(table2) FROM table1
LEFT JOIN table2
ON on1(table1) == on2(table2)
on1 = itemgetter(*on_index1)
on2 = itemgetter(*on_index2)
g1 = itemgetter(*getter1)
g2 = itemgetter(*getter2)
d2 = {}
d2 = dict((on2(r2), r2) for r2 in table2)
#if itemgetter always return tuple, below could simple as one line
#return [ list(on1(r1) + g1(r1) + g2(d2.get(on1(r1 for r1 in table1
]
len_on = len(on_index1)
len_g1 = len(getter1)
len_g2 = len(getter2)
if len_on == 1:
if len_g1 == 1 and len_g2 == 1:
return [ [on1(r1), g1(r1), g2(d2.get(on1(r1)))] for r1 in table1
]
elif len_g1 == 1 and len_g2 1:
return [ list((on1(r1),g1(r1))+g2(d2.get(on1(r1 for r1 in
table1 ]
elif len_g1 1 and len_g2 == 1:
return [ list((on1(r1),)+g1(r1)+(g2(d2.get(on1(r1))),)) for r1
in table1 ]
else: #len_g1 1 and len_g2 1:
return [ list((on1(r1),)+g1(r1)+g2(d2.get(on1(r1 for r1 in
table1 ]
else: # len_on 1
if len_g1 == 1 and len_g2 == 1:
return [ list(on1(r1))+[g1(r1),g2(d2.get(on1(r1)))] for r1 in
table1 ]
elif len_g1 == 1 and len_g2 1:
return [ list(on1(r1)+(g1(r1),)+g2(d2.get(on1(r1 for r1 in
table1 ]
elif len_g1 1 and len_g2 == 1:
return [ list(on1(r1)+g1(r1)+(g2(d2.get(on1(r1))),)) for r1 in
table1 ]
else: #len_g1 1 and len_g2 1:
return [ list(on1(r1)+g1(r1)+g2(d2.get(on1(r1 for r1 in
table1 ]
so is there a way to force itemgetter to return tuple even when
itemgetter(only_have_one_argument)? or some other function to do this?
Thanks.
--
http://mail.python.org/mailman/listinfo/python-list
