Home | Trees | Indices | Help |
|
---|
|
Dictionary that remembers insertion order
|
|||
|
|||
|
|||
new empty dictionary |
|
||
|
|||
|
|||
|
|||
|
|||
|
|||
|
|||
None |
|
||
None |
|
||
a shallow copy of od |
|
||
list of (key, value) pairs in od |
|
||
an iterator over the (key, value) items of D |
|
||
an iterator over the keys in od |
|
||
an iterator over the values of D |
|
||
list of keys in od |
|
||
v, remove specified key and return the corresponding |
|
||
(k, v), return and remove a (key, value) pair |
|
||
od.get(k,d), also set od[k]=d if k not in od |
|
||
None |
|
||
list of values in od |
|
||
a set-like object providing a view on od's items |
|
||
a set-like object providing a view on od's keys |
|
||
an object providing a view on od's values |
|
||
Inherited from Inherited from |
|
|||
New ordered dictionary with keys from S |
|
|
|||
__marker = <__builtin__.object object>
|
|||
Inherited from |
|
|||
Inherited from |
|
del od[y]
|
od==y. Comparison to another OD is order-sensitive while comparison to a regular mapping is order-insensitive.
|
Initialize an ordered dictionary. The signature is the same as regular dictionaries, but keyword arguments are not recommended because their insertion order is arbitrary.
|
iter(od)
|
od!=y
|
Return state information for pickling
|
repr(od)
|
reversed(od) |
od[i]=y
|
Update D from mapping/iterable E and F. If E present and has a .keys() method, does: for k in E: D[k] = E[k] If E present and lacks .keys() method, does: for (k, v) in E: D[k] = v In either case, this is followed by: for k, v in F.items(): D[k] = v
|
Remove all items from od.
|
|
If not specified, the value defaults to None.
|
|
od.iteritems -> an iterator over the (key, value) pairs in od
|
|
od.itervalues -> an iterator over the values in od
|
|
value. If key is not found, d is returned if given, otherwise KeyError is raised.
|
Pairs are returned in LIFO order if last is true or FIFO order if false.
|
|
Update D from mapping/iterable E and F. If E present and has a .keys() method, does: for k in E: D[k] = E[k] If E present and lacks .keys() method, does: for (k, v) in E: D[k] = v In either case, this is followed by: for k, v in F.items(): D[k] = v
|
|
|
|
|
|
__marker
|
Home | Trees | Indices | Help |
|
---|
Generated by Epydoc 3.0.1 on Tue Mar 8 12:39:48 2016 | http://epydoc.sourceforge.net |