FAQ

[Python] efficient updating of nested dictionaries

Omission9
Jan 26, 2004 at 2:33 am
I have a dictionary that looks like this
MY_DICT[KEY_X][KEY_Y][KEY_Z]=FOO

I am having a problem updating this with a simple
MY_DICT.update(NEW_DICT) as update doesn't seem to care about getting
into the inner dicts.
Getting the keys of each and iterating through and updating each one is
terribly slow as the number of keys gets bigger and bigger.
What is the bst way to update my nested dicts?
reply

Search Discussions

13 responses

  • Edward C. Jones at Jan 26, 2004 at 3:33 am

    omission9 wrote:
    I have a dictionary that looks like this
    MY_DICT[KEY_X][KEY_Y][KEY_Z]=FOO

    I am having a problem updating this with a simple
    MY_DICT.update(NEW_DICT) as update doesn't seem to care about getting
    into the inner dicts.
    Getting the keys of each and iterating through and updating each one is
    terribly slow as the number of keys gets bigger and bigger.
    What is the bst way to update my nested dicts?
    Make a table whose rows are (KEY_X, KEY_Y, KEY_Z, FOO). If the table is
    large use MySQL or some other database. For small or medium sized tables
    try "http://members.tripod.com/~edcjones/MultiDict.py".
  • Rich Krauter at Jan 26, 2004 at 3:37 am
    The following is probably too dependent on the data type of the keys,
    but it may be suitable in some programs. It's certainly not a general
    solution for all cases. Others will have much better ideas, but here
    goes anyway ...

    You may want to use a non-nested dict with a 'superkey' composed of the
    concatenation of the three keys, seperated by some delimiter.
    use MY_DICT[KEY_X+'_'+KEY_Y+'_'+KEY_Z]=FOO

    Then you could use update().You would just have to do some pre- and
    post-processing of the keys. i.e. splitting or joining the 'superkey' by
    the delimiter you choose.

    Although, that's probably kind of lame - I bet others will have much
    better suggestions. I'm interested in how other people do this too.
    Rich

    On Sun, 2004-01-25 at 21:33, omission9 wrote:

    I have a dictionary that looks like this
    MY_DICT[KEY_X][KEY_Y][KEY_Z]=FOO

    I am having a problem updating this with a simple
    MY_DICT.update(NEW_DICT) as update doesn't seem to care about getting
    into the inner dicts.
    Getting the keys of each and iterating through and updating each one is
    terribly slow as the number of keys gets bigger and bigger.
    What is the bst way to update my nested dicts?
    -------------- next part --------------
    An HTML attachment was scrubbed...
    URL: http://mail.python.org/pipermail/python-list/attachments/20040125/60ffed61/attachment.html
  • Omission9 at Jan 26, 2004 at 5:03 am

    omission9 wrote:

    I have a dictionary that looks like this
    MY_DICT[KEY_X][KEY_Y][KEY_Z]=FOO

    I am having a problem updating this with a simple
    MY_DICT.update(NEW_DICT) as update doesn't seem to care about getting
    into the inner dicts.
    Getting the keys of each and iterating through and updating each one is
    terribly slow as the number of keys gets bigger and bigger.
    What is the bst way to update my nested dicts?

    So far I have found this on the internet:
    def rUpdate(self,targetDict,itemDict):
    valtab=[]
    for key,val in itemDict.items():
    if type(val)==type({}):
    newTarget=targetDict.setdefault(key,{})
    self.rUpdate(newTarget,val)
    else:
    targetDict[key]=val

    However, this does not seem to handle the fact that each dict has
    multiple keys. :( So far the modification I have made to make it work
    right have failed. Any ideas?
  • Sidharthk at Jan 26, 2004 at 10:44 am
    This is untested code but i think it should work.(fingers crossed)
    Btw i doubt this will be fast though.

    def rec_update(mydict, newdict):
    presentKeysPairs = [(key,value)
    for (key, value) in newdict.items()
    if mydict.has_key(key)]
    newKeysPairs = [(key,value)
    for (key, value) in newdict,items()
    if not mydict.has_key(key)]
    for key, newValue in presentKeysPairs:
    currentValue = mydict[key]
    if isisntance(newValue, dict):
    mydict[key] = rec_update(newValue)
    else:
    mydict[key] = newValue
    mydict.update(dict(newKeysPairs))
    return mydict

    regards

    ps. why can't you simply use tuples to represent the different
    dimensions, even if the number of dimensions vary.
    is there any particular reason why you are using these nested
    dictionaries?
  • Josiah Carlson at Jan 26, 2004 at 7:00 am

    Although, that's probably kind of lame - I bet others will have much
    better suggestions. I'm interested in how other people do this too.
    Rich
    String concatenation is not that lame, but I'd use tuples:
    MY_DICT[(KEY_X, KEY_Y, KEY_Z)] = FOO

    Tuples save on string operations.

    - Josiah
  • Duncan Booth at Jan 26, 2004 at 9:39 am
    Josiah Carlson <jcarlson at nospam.uci.edu> wrote in
    news:bv2e21$j5e$2 at news.service.uci.edu:
    Although, that's probably kind of lame - I bet others will have much
    better suggestions. I'm interested in how other people do this too.
    Rich
    String concatenation is not that lame, but I'd use tuples:
    MY_DICT[(KEY_X, KEY_Y, KEY_Z)] = FOO

    Tuples save on string operations.
    I would omit the extra parentheses here, but its a style thing.

    MY_DICT[KEY_X, KEY_Y, KEY_Z] = FOO

    (Note to original poster: I'd also turn off caps-lock)
  • Rich Krauter at Jan 27, 2004 at 1:11 am

    String concatenation is not that lame, but I'd use tuples:
    MY_DICT[(KEY_X, KEY_Y, KEY_Z)] = FOO
    Tuples save on string operations.

    What a nice way to simplify this common task. That's great. Thanks for
    the advice.
    Rich
    -------------- next part --------------
    An HTML attachment was scrubbed...
    URL: http://mail.python.org/pipermail/python-list/attachments/20040126/2eba9fc8/attachment.htm
  • Ben Finney at Jan 27, 2004 at 1:49 am

    On Mon, 26 Jan 2004 20:11:39 -0500, Rich Krauter wrote:
    What a nice way to simplify this common task. That's great. Thanks for
    the advice.

    [HTML garbage repeating the same content]
    What a hideous way to complicate this simple medium. That sucks.
    Thanks for turning it off in future.

    --
    \ "My roommate got a pet elephant. Then it got lost. It's in the |
    `\ apartment somewhere." -- Steven Wright |
    _o__) |
    Ben Finney <http://bignose.squidly.org/>
  • Rich Krauter at Jan 27, 2004 at 2:32 am
    Oh crap. Sorry about the html emails. I've been meaning to turn that
    off. Thanks for reminding me.
    Rich
  • Ben Finney at Jan 27, 2004 at 2:45 am

    On Mon, 26 Jan 2004 21:32:28 -0500, Rich Krauter wrote:
    Oh crap. Sorry about the html emails. I've been meaning to turn that
    off. Thanks for reminding me.
    Much better! Thanks for being considerate.

    --
    \ "When I turned two I was really anxious, because I'd doubled my |
    `\ age in a year. I thought, if this keeps up, by the time I'm six |
    _o__) I'll be ninety." -- Steven Wright |
    Ben Finney <http://bignose.squidly.org/>
  • Sidharthk at Jan 26, 2004 at 8:04 am
    omission9 <omission9 at invalid.email.info> wrote in message news:<H%_Qb.2609$925.1917 at nwrddc02.gnilink.net>...
    I have a dictionary that looks like this
    MY_DICT[KEY_X][KEY_Y][KEY_Z]=FOO

    I am having a problem updating this with a simple
    MY_DICT.update(NEW_DICT) as update doesn't seem to care about getting
    into the inner dicts.
    Getting the keys of each and iterating through and updating each one is
    terribly slow as the number of keys gets bigger and bigger.
    What is the bst way to update my nested dicts?
    Use a tuple
    MY_DICT[(KEY_X,KEY_Y,KEY_Z)]=FOO

    unless you have a particular reason to use these nested dicts :)
  • Sidharthk at Jan 26, 2004 at 9:01 am
    omission9 <omission9 at invalid.email.info> wrote in message news:<H%_Qb.2609$925.1917 at nwrddc02.gnilink.net>...
    I have a dictionary that looks like this
    MY_DICT[KEY_X][KEY_Y][KEY_Z]=FOO

    I am having a problem updating this with a simple
    MY_DICT.update(NEW_DICT) as update doesn't seem to care about getting
    into the inner dicts.
    Getting the keys of each and iterating through and updating each one is
    terribly slow as the number of keys gets bigger and bigger.
    What is the bst way to update my nested dicts?
    Use Tuples

    MY_DICT[(KEY_X,KEY_Y,KEY_Z)]=FOO

    Unless for some you need to use nested dicts :)
  • Sidharthk at Jan 26, 2004 at 9:01 am
    omission9 <omission9 at invalid.email.info> wrote in message news:<H%_Qb.2609$925.1917 at nwrddc02.gnilink.net>...
    I have a dictionary that looks like this
    MY_DICT[KEY_X][KEY_Y][KEY_Z]=FOO

    I am having a problem updating this with a simple
    MY_DICT.update(NEW_DICT) as update doesn't seem to care about getting
    into the inner dicts.
    Getting the keys of each and iterating through and updating each one is
    terribly slow as the number of keys gets bigger and bigger.
    What is the bst way to update my nested dicts?
    Use Tuples

    MY_DICT[(KEY_X,KEY_Y,KEY_Z)]=FOO

    Unless for some you need to use nested dicts :)

Related Discussions