python – Find all indexes for the matching value in a list of ordered dict

Question

I’m having a list of ordered dict which looks similarly as below

[OrderedDict([('a', 1), ('b', 2)]), OrderedDict([('a', 1), ('b', 3)]), OrderedDict([('a', 2), ('b', 2)]), OrderedDict([('a', 3), ('b', 2)]), OrderedDict([('a', 1), ('b', 3)])]

I wanted to store indexes of list in an array which are the a value as 1
So, my list would contain below elements

[0,1,4]

I’m having a traditional script to get these values but since my original list is holding more than a million ordered dicts, it’s taking a longer time to fetch the elements.

for ele in range(len(liso)):
 if(liso[ele]['a'] ==1):
  giso.add(ele)

Can someone help me to rewrite the above script using map or filter to optimize the query?

Asked By: N Raghu

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Source

Answer 1

List Comprehension: [i for i, x in enumerate(liso) if x['a'] == 1]

Filter:

If you’re using Python 2: filter(lambda i: liso[i]['a'] == 1, xrange(len(liso)))

If you’re using Python 3: list(filter(lambda i: liso[i]['a'] == 1, range(len(liso))))

Answered By: Jacob G.

Answer 2

Ducks would work well here. It builds a database-style index on Python objects.

pip install ducks

from ducks import Dex

objects = [OrderedDict([('a', 1), ('b', 2)]), OrderedDict([('a', 1), ('b', 3)]), OrderedDict([('a', 2), ('b', 2)]), OrderedDict([('a', 3), ('b', 2)]), OrderedDict([('a', 1), ('b', 3)])]

# build index on objects
dex = Dex(objects, ['a', 'b'])

# get matching objects
dex[{'a': 1}]  # gets objects where a == 1
dex[{'b': 2}]  # gets objects where b == 2
dex[{'a': 1, 'b': 2}]  # gets objects where a == 1 and b == 2
dex[{'a': {'<': 3}}] # gets objects where a < 3

On a million-item dataset, those queries will run 10 to 100x faster than filter / list comprehensions, which must scan through every object every time.

Answered By: manimino

python – Find all indexes for the matching value in a list of ordered dict

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