CODEWITHSUNDEEP
pythonlistdictionary
adohertyd
adohertyd

Reputation: 2689

Accessing values in a list of nested dictionaries

My dictionary looks like this:

docScores = {0:[{u'word':2.3},{u'the':8.7},{u'if':4.1},{u'Car':1.7}],
             1:[{u'friend':1.2},{u'a':5.2},{u'you':3.8},{u'person':0.8}],
             ...
             29:[{u'yard':1.5},{u'gardening':2.8},{u'paint':3.7},{u'brush':1.6}]
            }

I want to sum the values of each inner dict for each list and store it in a new dict, with the new dict having key values of {0:2.3+8.7+4.1+1.7, 1:1.2+5.2+3.8+0.8, ... etc} i.e. 

for x in docScores[0]: #{0:
    for x in docScores[0][0].values(): #{,2.3}.
        sum = sum+x #where sum = 0 before loop
        docSum[0] = sum
    repeat this loop for every document

Any variation that I have tried is giving me unexpected outputs. Can anyone give me the correct syntax for this?

Upvotes: 2

Views: 6087

Answers (5)

dawg
dawg

Reputation: 103744

This dict comprehension works:

docScores = {0:[{u'word':2.3},{u'the':8.7},{u'if':4.1},{u'Car':1.7}],
             1:[{u'friend':1.2},{u'a':5.2},{u'you':3.8},{u'person':0.8}],
             29:[{u'yard':1.5},{u'gardening':2.8},{u'paint':3.7},{u'brush':1.6}]
            }

sum_d={k:sum(d.values()[0] for d in v) for k,v in docScores.items()}

print sum_d

Prints:

{0: 16.8, 1: 11.0, 29: 9.6}

However, changing your data structure may be easier. You could have a dict of dicts:

>>> NdocScores = {0:{u'word':2.3,u'the':8.7,u'if':4.1,u'Car':1.7},
...              1:{u'friend':1.2,u'a':5.2,u'you':3.8,u'person':0.8},
...              29:{u'yard':1.5,u'gardening':2.8,u'paint':3.7,u'brush':1.6}
...             }

Which allows each docs data to be directly accessed:

>>> NdocScores[0]
{u'Car': 1.7, u'the': 8.7, u'word': 2.3, u'if': 4.1}
>>> NdocScores[0][u'Car']
1.7
>>> sum(NdocScores[1].values())
11.0

>>> NdocScores[29]
{u'gardening': 2.8, u'yard': 1.5, u'brush': 1.6, u'paint': 3.7}

Or, just have a list of dicts with the position in the list corresponding to the doc index:

>>> lofdicts=[v for k,v in NdocScores.items()]
>>> lofdicts
[{u'Car': 1.7, u'the': 8.7, u'word': 2.3, u'if': 4.1}, {u'a': 5.2, u'person': 0.8, u'you': 3.8, u'friend': 1.2}, {u'gardening': 2.8, u'yard': 1.5, u'brush': 1.6, u'paint': 3.7}]
>>> lofdicts[0]
{u'Car': 1.7, u'the': 8.7, u'word': 2.3, u'if': 4.1}
>>> sum(lofdicts[1].values())
11.0

Upvotes: 3

mgilson
mgilson

Reputation: 309841

new_dict={}

docScores = {0:[{u'word':2.3},{u'the':8.7},{u'if':4.1},{u'Car':1.7}],
             1:[{u'friend':1.2},{u'a':5.2},{u'you':3.8},{u'person':0.8}],
             29:[{u'yard':1.5},{u'gardening':2.8},{u'paint':3.7},{u'brush':1.6}]
            }

for k,v in docScores.items():
    new_dict[k]=sum( sum(d.values()) for d in v )

print (new_dict) #{0: 16.8, 1: 11.0, 29: 9.6}

As others have mentioned, you could make this into a dictionary comprehension (python 2.7+):

new_dict = {k : sum( sum(d.values()) for d in v ) for k,v in docScores.items() }

But at this point I think that the comprehension is getting very difficult to comprehend (and therefore I wouldn't do it).

Also, someone should probably point out that if all your dictionary keys are sequential integers starting from 0 and going to 29, You probably shouldn't be using a dictionary to store this data -- maybe a list would be more appropriate ...

EDIT

using a list:

new_list = [sum( sum(d.values()) for d in v ) for _,v in sorted(docScores.items()) ]

Upvotes: 2

jsbueno
jsbueno

Reputation: 110208

docScores = {0:[{u'word':2.3},{u'the':8.7},{u'if':4.1},{u'Car':1.7}],
             1:[{u'friend':1.2},{u'a':5.2},{u'you':3.8},{u'person':0.8}],
             2:[{u'yard':1.5},{u'gardening':2.8},{u'paint':3.7},{u'brush':1.6}]
            }


result = dict(enumerate(sum (sum(word.values()) for word in  word_list[1]) for word_list in sorted(docScores.items())  ) )

Upvotes: 0

jamylak
jamylak

Reputation: 133514

>>> doc_scores = {
        0: [{u'word': 2.3}, {u'the': 8.7}, {u'if': 4.1}, {u'Car': 1.7}],
        1: [{u'friend': 1.2}, {u'a': 5.2}, {u'you': 3.8}, {u'person': 0.8}],
        29: [{u'yard': 1.5}, {u'gardening': 2.8}, {u'paint': 3.7}, {u'brush': 1.6}]
}
>>> dict((k, sum(n for d in v for n in d.itervalues())) 
         for k, v in doc_scores.iteritems())
{0: 16.8, 1: 11.0, 29: 9.6}

If you only have one value in each of the dicts in the lists you can shorten this:

>>> dict((k, sum(d.values()[0] for d in v)) for k, v in doc_scores.iteritems())
{0: 16.8, 1: 11.0, 29: 9.6}

Upvotes: 1

Denis
Denis

Reputation: 7343

And more oneline solve )

sum(reduce(lambda x, y: x+y, [d.values() for d in v for _,v in docScores.iteritems()]))

Upvotes: 1

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