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An efficient way to convert a list dictionary to a pair list of keys and values

I have a list dictionary as follows (it can be more than 1M elements, it is also assumed that the dictionary is sorted by key)

import scipy.sparse as sp
d = {0: [0,1], 1: [1,2,3], 
     2: [3,4,5], 3: [4,5,6], 
     4: [5,6,7], 5: [7], 
     6: [7,8,9]}

I want to know what is the most efficient way (quick way for a large dictionary) to convert it to a list of row and column indexes, for example:

r_index = [0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 6, 6, 6]
c_index = [0, 1, 1, 2, 3, 3, 4, 5, 4, 5, 6, 5, 6, 7, 7, 7, 8, 9]

Here are some solutions that I still have:

  • Using iteration

    row_ind = [k for k, v in d.iteritems() for _ in range(len(v))] # or d.items() in Python 3
col_ind = [i for ids in d.values() for i in ids]

  • Using pandas library

    import pandas as pd
df = pd.DataFrame.from_dict(d, orient='index')
df = df.stack().reset_index()
row_ind = list(df['level_0'])
col_ind = list(df[0])

  • Using itertools

    import itertools
indices = [(x,y) for x, y in itertools.chain.from_iterable([itertools.product((k,), v) for k, v in d.items()])]
indices = np.array(indices)
row_ind = indices[:, 0]
col_ind = indices[:, 1]

I'm not sure how this is the fastest way to deal with this problem if there are many elements in my dictionary. Thank!

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3 answers

python , , . cpython - pypy - . .

import time
l = range(10000)
x = dict([(k, list(l)) for k in range(1000)])

def org(d):
    row_ind = [k for k, v in d.items() for _ in range(len(v))]
    col_ind = [i for ids in d.values() for i in ids]

def ext(d):
    row_ind = [k for k, v in d.items() for _ in range(len(v))]
    col_ind = []
    for ids in d.values():
        col_ind.extend(ids)

def ext_both(d):
    row_ind = []
    for k, v in d.items():
        row_ind.extend([k] * len(v))
    col_ind = []
    for ids in d.values():
        col_ind.extend(ids)

functions = [org, ext, ext_both]
for func in functions:
    begin = time.time()
    func(x)
    elapsed = time.time() - begin
    print(func.__name__ + ": "  + str(elapsed))

python2:

org: 0.512559890747
ext: 0.340406894684
ext_both: 0.149670124054
+2

:

import time
l = xrange(10000)
x = dict([(k, list(l)) for k in xrange(1000)])


def f(d):
    row_ind = [k for k, v in d.iteritems() for _ in range(len(v))]
    col_ind = [i for ids in d.values() for i in ids]


def ff(d):
    import pandas as pd
    df = pd.DataFrame.from_dict(d, orient='index')
    df = df.stack().reset_index()
    row_ind = list(df['level_0'])
    col_ind = list(df[0])


def fff(d):
    import itertools
    import numpy as np
    indices = [(x, y) for x, y in itertools.chain.from_iterable(
        [itertools.product((k,), v) for k, v in d.items()])]
    indices = np.array(indices)
    row_ind = indices[:, 0]
    col_ind = indices[:, 1]

alternatives = [f, ff, fff]
for func in alternatives:
    begin = time.time()
    func(x)
    print time.time() - begin

:

0.977538108826
5.26920008659
6.98472499847

. , , . , .

0

, . def . import timer @timer.Timer() - . Google. : https://wiki.python.org/moin/PythonDecorators

-2

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