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Parsing Python dictionary with multiple objects

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  Parsing Python dictionary with multiple objects anuragpatibandla7@gmail.com - 2014-10-14 08:57 -0700
    Re:Parsing Python dictionary with multiple objects Dave Angel <davea@davea.name> - 2014-10-14 13:02 -0400
      Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-14 14:15 -0700
        Re: Parsing Python dictionary with multiple objects Skip Montanaro <skip.montanaro@gmail.com> - 2014-10-14 16:32 -0500
          Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-14 14:36 -0700
        Re: Parsing Python dictionary with multiple objects MRAB <python@mrabarnett.plus.com> - 2014-10-14 23:58 +0100
    Re:Parsing Python dictionary with multiple objects Dave Angel <davea@davea.name> - 2014-10-14 21:57 -0400
      Re: Parsing Python dictionary with multiple objects Rustom Mody <rustompmody@gmail.com> - 2014-10-14 19:32 -0700
    Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-14 20:40 -0700
      Re: Parsing Python dictionary with multiple objects Rustom Mody <rustompmody@gmail.com> - 2014-10-14 21:50 -0700
      Re: Parsing Python dictionary with multiple objects Dave Angel <davea@davea.name> - 2014-10-15 13:43 -0400
        Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-15 11:35 -0700
    Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-14 20:41 -0700
    Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-15 08:52 -0700
      Re: Parsing Python dictionary with multiple objects Rustom Mody <rustompmody@gmail.com> - 2014-10-15 09:06 -0700
    Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-15 09:28 -0700
      Re: Parsing Python dictionary with multiple objects Rustom Mody <rustompmody@gmail.com> - 2014-10-15 09:43 -0700
        Re: Parsing Python dictionary with multiple objects Rustom Mody <rustompmody@gmail.com> - 2014-10-15 09:59 -0700
    Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-15 09:57 -0700
      Re: Parsing Python dictionary with multiple objects Dave Angel <davea@davea.name> - 2014-10-15 13:48 -0400
    Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-15 10:00 -0700
      Re: Parsing Python dictionary with multiple objects Rustom Mody <rustompmody@gmail.com> - 2014-10-15 10:10 -0700
        Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-15 10:20 -0700
          Re: Parsing Python dictionary with multiple objects Dave Angel <davea@davea.name> - 2014-10-15 13:50 -0400
    Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-15 10:20 -0700
      Re: Parsing Python dictionary with multiple objects Rustom Mody <rustompmody@gmail.com> - 2014-10-15 10:35 -0700
        Re: Parsing Python dictionary with multiple objects Anurag Patibandla <anuragpatibandla7@gmail.com> - 2014-10-15 11:27 -0700
          Re: Parsing Python dictionary with multiple objects Dave Angel <davea@davea.name> - 2014-10-15 18:25 -0400

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#79107 — Parsing Python dictionary with multiple objects

I have a dictionary that looks like this:
{"1":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
"2":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
"3":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
"4":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}}

Now if I have 100 objects like these and I need to split them into 3 smaller dicts in a ratio 2:3:5, how could I do that?
I tried using numpy.random.choice(), but it says it needs to be a 1-d array.

Can someone please help with this?
Thanks

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#79108

anuragpatibandla7@gmail.com Wrote in message:
> I have a dictionary that looks like this:
> {"1":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> "2":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> "3":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> "4":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}}
> 
> Now if I have 100 objects like these and I need to split them into 3 smaller dicts in a ratio 2:3:5, how could I do that?

I really have no idea what that means.  You have 100 dicts of
 dicts? Are the keys unique? If so, you could combine them with a
 loop of update.

> I tried using numpy.random.choice(), but it says it needs to be a 1-d array.
> 

How about random.choice?  It needs a sequence.

To get anything more concrete, you need to specify Python version,
  and make a clearer problem statement,  perhaps with example of
 what you expect.


-- 
DaveA

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#79117

On Tuesday, October 14, 2014 12:59:27 PM UTC-4, Dave Angel wrote:
> anuragpatibandla7@gmail.com Wrote in message:
> 
> > I have a dictionary that looks like this:
> 
> > {"1":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> 
> > "2":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> 
> > "3":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> 
> > "4":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}}
> 
> > 
> 
> > Now if I have 100 objects like these and I need to split them into 3 smaller dicts in a ratio 2:3:5, how could I do that?
> 
> 
> 
> I really have no idea what that means.  You have 100 dicts of
> 
>  dicts? Are the keys unique? If so, you could combine them with a
> 
>  loop of update.
> 
> 
> 
> > I tried using numpy.random.choice(), but it says it needs to be a 1-d array.
> 
> > 
> 
> 
> 
> How about random.choice?  It needs a sequence.
> 
> 
> 
> To get anything more concrete, you need to specify Python version,
> 
>   and make a clearer problem statement,  perhaps with example of
> 
>  what you expect.
> 
> 
> 
> 
> 
> -- 
> 
> DaveA

Hey DaveA,
I am using Python 2.7.
Yes. I have a dict of dicts with keys ranging from 1..100
And the value of each of these keys is another dict. What I need to do is make 3 smaller dicts and assign the values of keys 1..100 in the ratio 2:3:5.
For example, if my original dict is 

d={"1":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
   "2":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
   "3":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
   "4":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}}
   ...
   ...
  "100":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}

I need to have three dicts d1, d2, d3 with d1 containing the values of first 20 keys, d2 containing the values on next 30 keys and d3 containing the values of the next 50.
Can you please help me with this?

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#79118

[Multipart message — attachments visible in raw view] — view raw

Shuffle the keys, then grab the first 20 for one dictionary, the next 30
for the second, and the last 50 for the third.

Skip

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#79119

On Tuesday, October 14, 2014 5:33:01 PM UTC-4, Skip Montanaro wrote:
> Shuffle the keys, then grab the first 20 for one dictionary, the next 30 for the second, and the last 50 for the third.
> 
> Skip

Could you please be more specific?

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#79121

On 2014-10-14 22:15, Anurag Patibandla wrote:
> On Tuesday, October 14, 2014 12:59:27 PM UTC-4, Dave Angel wrote:
>> anuragpatibandla7@gmail.com Wrote in message:
>>
>> > I have a dictionary that looks like this:
>> > {"1":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"},
>> > "2":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"},
>> > "3":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"},
>> > "4":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}}
>> >
>> > Now if I have 100 objects like these and I need to split them into 3 smaller dicts in a ratio 2:3:5, how could I do that?
>>
>> I really have no idea what that means.  You have 100 dicts of
>>  dicts? Are the keys unique? If so, you could combine them with a
>>  loop of update.
>>
>> > I tried using numpy.random.choice(), but it says it needs to be a 1-d array.
>> >
>>
>> How about random.choice?  It needs a sequence.
>>
>> To get anything more concrete, you need to specify Python version,
>> and make a clearer problem statement,  perhaps with example of
>> what you expect.
>>
>
> Hey DaveA,
> I am using Python 2.7.
> Yes. I have a dict of dicts with keys ranging from 1..100
> And the value of each of these keys is another dict. What I need to do is make 3 smaller dicts and assign the values of keys 1..100 in the ratio 2:3:5.
> For example, if my original dict is
>
> d={"1":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"},
>     "2":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"},
>     "3":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"},
>     "4":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}}
>     ...
>     ...
>    "100":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}
>
> I need to have three dicts d1, d2, d3 with d1 containing the values of first 20 keys, d2 containing the values on next 30 keys and d3 containing the values of the next 50.
> Can you please help me with this?
>
You can get a list of the entries using the dict's .items method. It's
then a simple matter of slicing the list.

Note that dicts aren't ordered, i.e. its keys aren't in a fixed order,
so if you want then in a particular order, you'll need to sort the list.

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#79124

anuragpatibandla7@gmail.com Wrote in message:
> I have a dictionary that looks like this:
> {"1":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> "2":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> "3":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> "4":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}}
> 
> Now if I have 100 objects like these and I need to split them into 3 smaller dicts in a ratio 2:3:5, how could I do that?
> I tried using numpy.random.choice(), but it says it needs to be a 1-d array.

> 

What have you actually tried? You haven't shown any actual code.

Look up the method dict.keys, and see how you might use that. Then
 look up random.shuffle, and see what it would do. Also look up
 dict.sort, since your two messages on this thread imply two
 conflicting goals as to which sub dictionaries should go in which
 of your buckets. 

As for extracting 20% of the keys, slicing is your answer. If
 there are 100 keys, 20, 30, and 50 need to be sliced
 off.

Then you'll need a loop to build each result dictionary from its keys.

There are shortcuts,  but it's best to learn the fundamentals first.

Try writing the code. If it doesn’t all work show us what you've
 tried, and what you think is wrong. And when you get an
 exception, show the whole traceback,  don't just paraphrase one
 of the lines.


-- 
DaveA

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#79128

On Wednesday, October 15, 2014 7:35:18 AM UTC+5:30, Dave Angel wrote:
> anurag Wrote in message:
> > I have a dictionary that looks like this:
> > {"1":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> > "2":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> > "3":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}, 
> > "4":{"Key1":"Value1", "Key2":"Value2", "Key3":"Value3"}}
> > Now if I have 100 objects like these and I need to split them into 3 smaller dicts in a ratio 2:3:5, how could I do that?
> > I tried using numpy.random.choice(), but it says it needs to be a 1-d array.


> What have you actually tried? You haven't shown any actual code.

> Look up the method dict.keys, and see how you might use that. Then
>  look up random.shuffle, and see what it would do. Also look up
>  dict.sort, since your two messages on this thread imply two
>  conflicting goals as to which sub dictionaries should go in which
>  of your buckets. 

> As for extracting 20% of the keys, slicing is your answer. If
>  there are 100 keys, 20, 30, and 50 need to be sliced
>  off.

> Then you'll need a loop to build each result dictionary from its keys.

> There are shortcuts,  but it's best to learn the fundamentals first.

> Try writing the code. If it doesn't all work show us what you've
>  tried, and what you think is wrong. And when you get an
>  exception, show the whole traceback,  don't just paraphrase one
>  of the lines.

Yes that is what is in general expected out here -- code --
maybe working, maybe not, maybe incomplete, maybe 'pseudo' etc
Then others here will improve it

However there is one conceptual thing that perhaps should be mentioned: order.

Is your data *essentially* ordered?

And by 'essentially' I mean you think of it independent of python.
Yeah in python dicts are unordered and lists are ordered and one can fudge
one to behave a bit like the other.  But before you fudge, please ponder which
you really need/want.

Below a bit of going from one to other 


# dict -> list
>>> d = {"a":1,"b":2,"c":3}
>>> d
{'a': 1, 'c': 3, 'b': 2}
>>> list(d)
['a', 'c', 'b']

# alternate
>>> d.keys()
['a', 'c', 'b']

>>> d.items()
[('a', 1), ('c', 3), ('b', 2)]
>>> d.values()
[1, 3, 2]

# list -> dict
>>> dict(d.items())
{'a': 1, 'c': 3, 'b': 2}

# round-tripping
>>> dict(d.items()) == d
True

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#79132

Thanks for the response.
Here is the code that I have tried.

from operator import itemgetter
keys = json.keys()
order = list(keys)
q1 = int(round(len(keys)*0.2))
q2 = int(round(len(keys)*0.3))
q3 = int(round(len(keys)*0.5))
b = [q1,q2,q3]
n=0
for i in b:
    queues = order[n:n+i]
        
    n = n+i
    print queues
    
    for j in range(len(queues)):
        q = (queues[j], json.get(queues[j]))
        print q

By this I am able to get the 3 smaller dicts I want, but can you help me assign them to 3 variables?
The dicts need not be ordered but it would be better if they are ordered.

Thanks

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#79134

On Wednesday, October 15, 2014 9:10:54 AM UTC+5:30, Anurag Patibandla wrote:
> Thanks for the response.
> Here is the code that I have tried.

> from operator import itemgetter
> keys = json.keys()
> order = list(keys)
> q1 = int(round(len(keys)*0.2))
> q2 = int(round(len(keys)*0.3))
> q3 = int(round(len(keys)*0.5))
> b = [q1,q2,q3]
> n=0
> for i in b:
>     queues = order[n:n+i]

>     n = n+i
>     print queues

>     for j in range(len(queues)):
>         q = (queues[j], json.get(queues[j]))
>         print q

Converting the end for loop (last 3 lines) into:


    print [(queues[j], json.get(queues[j])) for j in range(len(queues))]

Does that help?

General advice:
1. Instead of writing 'naked' code as you have done, if you wrap it into
functions (preferably small)
2. Contents similar to the original naked code but with print's replaced by 
return's

you make your as well as those trying to help/collaborate with you
life easier

Also the above is a more or mechanical translation. However

something[j] ... for j in range(len(something))

is usually a sign of a C programmer writing python :-)
Usually better to write

x for x in something

So...
Better to write that comprehension as

print [(q, json.get(q)) for q in queues]

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#79162

Anurag Patibandla <anuragpatibandla7@gmail.com> Wrote in message:
> Thanks for the response.
> Here is the code that I have tried.
> 
> from operator import itemgetter
> keys = json.keys()
> order = list(keys)
> q1 = int(round(len(keys)*0.2))
> q2 = int(round(len(keys)*0.3))
> q3 = int(round(len(keys)*0.5))
> b = [q1,q2,q3]
> n=0

threedicts = []

> for i in b:
>     queues = order[n:n+i]
>         
>     n = n+i
>     print queues
>     
>     for j in range(len(queues)):
>         q = (queues[j], json.get(queues[j]))
>         print q
> 

       onedict = {}
       for q in queues:
           onedict[q] = json[q]
       threedicts.append (onedict)
   dict1, dictw, dict3 = threedicts 
           
> By this I am able to get the 3 smaller dicts I want, but can you help me assign them to 3 variables?
> The dicts need not be ordered but it would be better if they are ordered.
> 

dicts are not ordered. If you want the items in a particular
 order, you have to do that after extraction from the dict. There
 is a related type called collections.OrderedDict, which
 'remembers' the order things were added.

-- 
DaveA

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#79167

On Wednesday, October 15, 2014 1:41:13 PM UTC-4, Dave Angel wrote:
> Anurag Patibandla <anuragpatibandla7@gmail.com> Wrote in message:
> 
> > Thanks for the response.
> 
> > Here is the code that I have tried.
> 
> > 
> 
> > from operator import itemgetter
> 
> > keys = json.keys()
> 
> > order = list(keys)
> 
> > q1 = int(round(len(keys)*0.2))
> 
> > q2 = int(round(len(keys)*0.3))
> 
> > q3 = int(round(len(keys)*0.5))
> 
> > b = [q1,q2,q3]
> 
> > n=0
> 
> 
> 
> threedicts = []
> 
> 
> 
> > for i in b:
> 
> >     queues = order[n:n+i]
> 
> >         
> 
> >     n = n+i
> 
> >     print queues
> 
> >     
> 
> >     for j in range(len(queues)):
> 
> >         q = (queues[j], json.get(queues[j]))
> 
> >         print q
> 
> > 
> 
> 
> 
>        onedict = {}
> 
>        for q in queues:
> 
>            onedict[q] = json[q]
> 
>        threedicts.append (onedict)
> 
>    dict1, dictw, dict3 = threedicts 
> 
>            
> 
> > By this I am able to get the 3 smaller dicts I want, but can you help me assign them to 3 variables?
> 
> > The dicts need not be ordered but it would be better if they are ordered.
> 
> > 
> 
> 
> 
> dicts are not ordered. If you want the items in a particular
> 
>  order, you have to do that after extraction from the dict. There
> 
>  is a related type called collections.OrderedDict, which
> 
>  'remembers' the order things were added.
> 
> 
> 
> -- 
> 
> DaveA

Thanks DaveA!
This works perfectly!

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#79133

'json' has my original larger dict

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#79149

Thanks Rustom for the advice.
I am new to Python and getting struck at some basic things. How do I assign the values that I am printing to 3 variables say dict1, dict2, dict3?
When I try to assign them before the print statement like this:
d1, d2, d3 =[(queues[j], json.get(queues[j])) for j in range(len(queues))]

I get an error saying 'need more than one value to unpack'

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#79151

On Wednesday, October 15, 2014 9:22:48 PM UTC+5:30, Anurag Patibandla wrote:
> Thanks Rustom for the advice.
> I am new to Python and getting struck at some basic things. How do I assign the values that I am printing to 3 variables say dict1, dict2, dict3?
> When I try to assign them before the print statement like this:
> d1, d2, d3 =[(queues[j], json.get(queues[j])) for j in range(len(queues))]

> I get an error saying 'need more than one value to unpack'

Probably means your comprehension

[(queues[j], json.get(queues[j])) for j in range(len(queues))]

is having less than 3 values

>>> lst = [1,2,3]
>>> x,y,z = lst
>>> (x,y,z) # note no need to print
(1, 2, 3)
>>> lst=[1]
>>> x,y,z=lst
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: need more than 1 value to unpack
>>> lst=[1,2,3,4]
>>> x,y,z=lst
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: too many values to unpack
================

I suggest youdont directly start with multiple assignment
Instead do it in two steps

dicts = [(queues[j], json.get(queues[j])) for j in range(len(queues))]

d0 = dicts[0]
d1 = dicts[1]
d2 = dicts[2]

When that works go to the more compact form

Also please get rid of the range(len(queues))
Its unpythonic!

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#79152

First the values printed by
'[(queues[j], json.get(queues[j])) for j in range(len(queues))] '
is a list, so I tried to convert it into a dict using dict().
And then I tried doing dict[0] but there is an error which says:
'type' object has no attribute '__getitem__'

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#79153

On Wednesday, October 15, 2014 9:58:49 PM UTC+5:30, Anurag Patibandla wrote:
> First the values printed by
> '[(queues[j], json.get(queues[j])) for j in range(len(queues))] '
> is a list, so I tried to convert it into a dict using dict().
> And then I tried doing dict[0] but there is an error which says:
> 'type' object has no attribute '__getitem__'

print each step in the process until the error step

Also assuming

[(queues[j], json.get(queues[j])) for j in range(len(queues))]

is the same as

[(q, json.get(q) for q in queues]

do

a = [(q, json.get(q) for q in queues]
print a
b = dict(a)
print b
c = b[0]
print c

or whatever it is you are doing

and PASTE (not NARRATE) the results

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#79155

On Wednesday, October 15, 2014 10:13:18 PM UTC+5:30, Rustom Mody wrote:
> On Wednesday, October 15, 2014 9:58:49 PM UTC+5:30, Anurag Patibandla wrote:
> > First the values printed by
> > '[(queues[j], json.get(queues[j])) for j in range(len(queues))] '
> > is a list, so I tried to convert it into a dict using dict().
> > And then I tried doing dict[0] but there is an error which says:
> > 'type' object has no attribute '__getitem__'

Also there are dictionary comprehensions recently added to python:
http://legacy.python.org/dev/peps/pep-0274/

Can generally be used if you are doing
dict(a list comprehension)

To start with though, I suggest you stay with the longer older form
until you understand how it works

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#79154

keys = json.keys()
order = list(keys)
q1 = int(round(len(keys)*0.2))
q2 = int(round(len(keys)*0.3))
q3 = int(round(len(keys)*0.5))
b = [q1,q2,q3]
n=0
for i in b:
    queues = order[n:n+i]
        
    n = n+i
    #print queues
    #print [(queues[j], json.get(queues[j])) for j in range(len(queues))] 
    lists = [(queues[j], json.get(queues[j])) for j in range(len(queues))] 
    #print lists
    dicts = dict(lists)
    print dicts
    print dict[0]

Print dicts works as expected giving me the combine dictionary values. But when I say dict[0]. I see the error:
TypeError: 'type' object has no attribute '__getitem__'

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#79164

Anurag Patibandla <anuragpatibandla7@gmail.com> Wrote in message:

>     dicts = dict(lists)
>     print dicts
>     print dict[0]
> 
> Print dicts works as expected giving me the combine dictionary values. But when I say dict[0]. I see the error:
> TypeError: 'type' object has no attribute '__getitem__'
> 

Of course. You forgot the s in the name dicts. So you were
 referring to the dict class, not your variable.
-- 
DaveA

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