Search for mapping solution

[Markus Joschko]

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

 

[Achim Domma]

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

 

[Max M]

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

regards Max M

 

[John J. Lee]

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

 

[Markus Joschko]

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.

 

[Markus Joschko]

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

 

[Mike C. Fletcher]

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

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/

 

[Bengt Richter]

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

 

[Alan Kennedy]

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.

 

[Markus Joschko]

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

 

[Martin Maney]

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

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. :-/