Search for mapping solution

Discussion in 'Python' started by Markus Joschko, Jul 6, 2003.

  1. Hi,
    stated in a post befor, I'm a java programmer, fascinated about the elegant
    way python solves iterations. Maybe you can show me a solution how to map
    the following

    I have a List:

    Name - Number - Costs

    lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]

    Now I want to have it in a dictionary(name,costs) Should look like
    {'fred':'0,60' , 'sam':'1'}

    What's an elegant way to do it? I can use a lot of loops, but I assume, that
    there is a better way of doing so.

    Thanks,
    Markus
    Markus Joschko, Jul 6, 2003
    #1
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  2. Markus Joschko

    Achim Domma Guest

    "Markus Joschko" <> wrote in message
    news:be9t66$2otne$...
    > Name - Number - Costs
    >
    > lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]
    >
    > Now I want to have it in a dictionary(name,costs) Should look like
    > {'fred':'0,60' , 'sam':'1'}


    I would do it like this:

    lines = [['fred','333','0.10'],['sam','444','1'],['fred','333','0.50']]
    costs = {}
    for name,number,price in lines:
    costs[name] = costs.setdefault(name,0)+float(price)
    print costs

    Achim
    Achim Domma, Jul 6, 2003
    #2
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  3. Markus Joschko

    Max M Guest

    Markus Joschko wrote:

    > Hi,
    > stated in a post befor, I'm a java programmer, fascinated about the elegant
    > way python solves iterations. Maybe you can show me a solution how to map
    > the following
    >
    > I have a List:
    >
    > Name - Number - Costs
    >
    > lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]
    >
    > Now I want to have it in a dictionary(name,costs) Should look like
    > {'fred':'0,60' , 'sam':'1'}
    >
    > What's an elegant way to do it? I can use a lot of loops, but I assume, that
    > there is a better way of doing so.


    lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]
    costs = {}
    for name, items, price in lines:
    costs[name] = costs.setdefault(name, 0.0) +
    float(price.replace(',','.'))

    print costs

    >>> {'fred': 0.59999999999999998, 'sam': 1.0}


    regards Max M
    Max M, Jul 6, 2003
    #3
  4. Markus Joschko

    John J. Lee Guest

    Markus Joschko <> writes:
    [...]
    > I have a List:
    >
    > nName - Number - Costs

    [...]
    > > lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]

    [...]

    Note that tuples were designed for that sort of 'mini-object' use (and
    were not intended primarily as immutable lists).

    lines = [('fred','333','0,10'), ('sam','444','1'), ('fred','333','0,50')]


    though of course it's no disaster if you end up with a list of
    3-element lists instead of 3-tuples, if it's convenient to build the
    list with zip or whatever.


    John
    John J. Lee, Jul 6, 2003
    #4
  5. >
    > Python 2.2.3 (#42, May 30 2003, 18:12:08) [MSC 32 bit (Intel)] on
    > win32
    > Type "help", "copyright", "credits" or "license" for more information.
    >>>> lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]
    >>>> d = dict([(x[0], x[2]) for x in lines])
    >>>> d

    > {'sam': '1', 'fred': '0,50'}


    fred should be 0,60. The 3rd column should be summarized.


    >
    > OK, I'll write and maintain a 1000-line perl program as penance ...
    >
    > sometimes-concise-is-elegant-too-ly yrs.
    >
    Markus Joschko, Jul 6, 2003
    #5
  6. >
    > lines = [['fred','333','0.10'],['sam','444','1'],['fred','333','0.50']]
    > costs = {}
    > for name,number,price in lines:
    > costs[name] = costs.setdefault(name,0)+float(price)
    > print costs
    >

    thanks it works. But maybe I can complicate the example a little bit
    (because in real world it's more complicated):

    What if I every list in lines has 20 or more entries and I have only the
    index number to access the name, e.g.

    lines = [['elem1','elem2','fred','elem3',.......;'elem
    17','333','elem18','0.10'],[...],[...]]


    what I want to say: I can't be sure that the name is always on the third
    position. That's dynamic. I know it before I parse the list, but
    I can't say

    for elem1,elem2,name,.... cause it can also be

    for elem1,name,elem3 ....


    Thanks for the answer,
    Markus
    Markus Joschko, Jul 6, 2003
    #6
  7. result = {}
    for (name, whatever, costs) in lines:
    costs = float(number.replace(',','.'))
    dict[name] = dict.get( name, 0.0) + costs

    (that's untested, but you should get the idea). Note, however, floating
    point is generally a poor choice for accounting applications, so you may
    want to look into the libraries for fixed-point calculations.

    HTH,
    Mike

    Markus Joschko wrote:

    >Hi,
    >stated in a post befor, I'm a java programmer, fascinated about the elegant
    >way python solves iterations. Maybe you can show me a solution how to map
    >the following
    >
    >I have a List:
    >
    >Name - Number - Costs
    >
    >lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]
    >
    >Now I want to have it in a dictionary(name,costs) Should look like
    >{'fred':'0,60' , 'sam':'1'}
    >
    >What's an elegant way to do it? I can use a lot of loops, but I assume, that
    >there is a better way of doing so.
    >
    >Thanks,
    > Markus
    >
    >

    _______________________________________
    Mike C. Fletcher
    Designer, VR Plumber, Coder
    http://members.rogers.com/mcfletch/
    Mike C. Fletcher, Jul 6, 2003
    #7
  8. On Sun, 06 Jul 2003 21:17:36 +0200, Markus Joschko <> wrote:

    >Hi,
    >stated in a post befor, I'm a java programmer, fascinated about the elegant
    >way python solves iterations. Maybe you can show me a solution how to map
    >the following
    >
    >I have a List:
    >
    >Name - Number - Costs
    >
    >lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]


    >
    >Now I want to have it in a dictionary(name,costs) Should look like
    >{'fred':'0,60' , 'sam':'1'}

    Should same-name costs just be added, and the number just be ignored?
    Assuming so, do you actually want the costs in the final dict to be represented
    as localized strings, or should they be floating point numbers -- or, should they
    be fixed point in effect?
    >
    >What's an elegant way to do it? I can use a lot of loops, but I assume, that
    >there is a better way of doing so.
    >

    If the names were all different, it would be a snap

    >>> lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]
    >>> d = dict([(name,cost) for name,num,cost in lines])
    >>> d

    {'sam': '1', 'fred': '0,50'}

    but, your example seems to have further requirements, so maybe:

    ====< jocsch.py >==============================================
    lines = [['fred','333','0,10'],['sam','444','1'],['fred','333','0,50']]

    # might want to use locale-sensitive fixed point for currency, but we'll fake it here ;-)
    def str2num(s): return ',' in s and int(s.replace(',','')) or 100*int(s) # units of 0,01
    def num2str(n): h,u = divmod(abs(n),100); s='-'[:n<0]; return u and '%s%d,%02d'%(s,h,u) or '%s%d'%(s,h)

    d={}
    for name,num,cost in lines:
    cost = str2num(cost) # units of 0,01
    d[name] = d.get(name, 0) + cost # accumulate same-name costs in integral units
    for name in d.keys(): d[name] = num2str(d[name]) # mixed literal string syntax again

    print lines
    print d
    ===============================================================
    Result:

    [14:27] C:\pywk\clp>jocsch.py
    [['fred', '333', '0,10'], ['sam', '444', '1'], ['fred', '333', '0,50']]
    {'sam': '1', 'fred': '0,60'}

    A uniform format (i.e., '1,00' instead of '1') would have simplified conversions a little ;-)

    Regards,
    Bengt Richter
    Bengt Richter, Jul 6, 2003
    #8
  9. Markus Joschko

    Alan Kennedy Guest

    Sean Ross wrote:

    > Hi. You've left out the accumulating part of the OP's requirements:


    I know :-(

    The real temptation I have to resist is deciding to answer someone
    question without reading the whole question properly.

    I tried to cancel the post as soon as I realised, but it was obviously
    too late.

    No more posting for me for a while.

    --
    alan kennedy
    -----------------------------------------------------
    check http headers here: http://xhaus.com/headers
    email alan: http://xhaus.com/mailto/alan
    Alan Kennedy, Jul 6, 2003
    #9
  10. Thanks for all the answers. Nice to have such a community.

    For me it's really interesting to see all the possible solutions. I learned
    some new things in this discussion

    Greetings,
    Markus
    Markus Joschko, Jul 7, 2003
    #10
  11. Markus Joschko

    Martin Maney Guest

    I dedicate this monument to Alan Kennedy, without whom I would probably
    never have tried to compress things so. :)

    Sean Ross <> wrote:
    > lines = [['fred','333','0.10'],['sam','444','1'],['fred','333','0.50']]
    > costs = {}
    > # nearly identical to Achim's solution, but with a list comp.
    > [costs.__setitem__(name, costs.get(name,0)+float(price))
    > for name, number, price in lines]
    > print costs
    > # outputs: {'sam': 1.0, 'fred': 0.60}


    It isn't really one line though, is it? For truly cryptic terseness
    you want to swing functional (I shall adopt your interpretation of the
    third element although seeing it as a list of integers would have
    allowed for additional functional yumminess):

    >>> lines = [['fred','333','0.10'],['sam','444','1'],['fred','333','0.50']]
    >>> reduce(lambda d,x: d.update({x[0]: d.get(x[0],0.0) + float(x[2])}) or d, [{}] + lines)

    {'fred': 0.59999999999999998, 'sam': 1.0}

    Mind you, I normally despise the whole one-liner phenomenon, which
    makes me almost pleased about the inelegant necessity of that 'or d' -
    blame it on either update's lack of a return value or lambda's
    emasculation. But as I've been working my way through SICP as this
    summer's project, it certainly seems odd. :-/

    --
    automation: replacing what works with something that almost works,
    but which is faster and cheaper. - attributed to Roger Needham
    Martin Maney, Jul 10, 2003
    #11
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