CODEWITHSUNDEEP
pythonperformancedictionarynestedlist-comprehension
Bob Lafleur
Bob Lafleur

Reputation: 3

How can I improve performance of nested comprehension?

I am trying to use python 3.x comprehension to create a nested dictionary structure. My comprehension syntax works, but it is very slow, especially with a large data set. I have also created my desired data structure using loops and it runs much faster, but I would like to know if there is a way to improve this comprehension to make it more efficient and potentially run as fast as, or faster than my loop code.

My input data is a list of dictionaries, each dictionary outlining the specifics of an amateur radio contact (log entry). Here is a random subset of my data (limited to 20 entries, and non-essential keys in the dictionary removed to make this more clear)

[{'BAND': '20M',
  'CALL': 'AA9GL',
  'COUNTRY': 'UNITED STATES OF AMERICA',
  'QSO_DATE': '20170528',
  'TIME_ON': '132100'},
 {'BAND': '20M',
  'CALL': 'KE4BFI',
  'COUNTRY': 'UNITED STATES OF AMERICA',
  'QSO_DATE': '20150704',
  'TIME_ON': '034600'},
 {'BAND': '20M',
  'CALL': 'W8OTR',
  'COUNTRY': 'UNITED STATES OF AMERICA',
  'QSO_DATE': '20190119',
  'TIME_ON': '194645'},
 {'BAND': '10M',
  'CALL': 'FY5FY',
  'COUNTRY': 'FRENCH GUIANA',
  'QSO_DATE': '20150328',
  'TIME_ON': '161953'},
 {'BAND': '17M',
  'CALL': 'KD5FOY',
  'COUNTRY': 'UNITED STATES OF AMERICA',
  'QSO_DATE': '20190121',
  'TIME_ON': '145630'},
 {'BAND': '10M',
  'CALL': 'K5GQ',
  'COUNTRY': 'UNITED STATES OF AMERICA',
  'QSO_DATE': '20150110',
  'TIME_ON': '195326'},
 {'BAND': '10M',
  'CALL': 'CR5L',
  'COUNTRY': 'PORTUGAL',
  'QSO_DATE': '20151025',
  'TIME_ON': '182351'},
 {'BAND': '20M',
  'CALL': 'AD4TR',
  'COUNTRY': 'UNITED STATES OF AMERICA',
  'QSO_DATE': '20170325',
  'TIME_ON': '144606'},
 {'BAND': '40M',
  'CALL': 'EA8FJ',
  'COUNTRY': 'CANARY ISLANDS',
  'QSO_DATE': '20170618',
  'TIME_ON': '020300'},
 {'BAND': '10M',
  'CALL': 'PY2DPM',
  'COUNTRY': 'BRAZIL',
  'QSO_DATE': '20150104',
  'TIME_ON': '205900'},
 {'BAND': '17M',
  'CALL': 'MM0HVU',
  'COUNTRY': 'SCOTLAND',
  'QSO_DATE': '20170416',
  'TIME_ON': '130200'},
 {'BAND': '10M',
  'CALL': 'LW3DG',
  'COUNTRY': 'ARGENTINA',
  'QSO_DATE': '20161029',
  'TIME_ON': '210629'},
 {'BAND': '10M',
  'CALL': 'LW3DG',
  'COUNTRY': 'ARGENTINA',
  'QSO_DATE': '20151025',
  'TIME_ON': '210714'},
 {'BAND': '20M',
  'CALL': 'EI7HDB',
  'COUNTRY': 'IRELAND',
  'QSO_DATE': '20170423',
  'TIME_ON': '184000'},
 {'BAND': '20M',
  'CALL': 'KM0NAS',
  'COUNTRY': 'UNITED STATES OF AMERICA',
  'QSO_DATE': '20180102',
  'TIME_ON': '142151'},
 {'BAND': '10M',
  'CALL': 'PY2TKB',
  'COUNTRY': 'BRAZIL',
  'QSO_DATE': '20150328',
  'TIME_ON': '223535'},
 {'BAND': '40M',
  'CALL': 'EB1DJ',
  'COUNTRY': 'SPAIN',
  'QSO_DATE': '20170326',
  'TIME_ON': '232430'},
 {'BAND': '40M',
  'CALL': 'LU6PCK',
  'COUNTRY': 'ARGENTINA',
  'QSO_DATE': '20150615',
  'TIME_ON': '000200'},
 {'BAND': '17M',
  'CALL': 'G3RKF',
  'COUNTRY': 'ENGLAND',
  'QSO_DATE': '20190121',
  'TIME_ON': '144315'},
 {'BAND': '20M',
  'CALL': 'UA1ZKI',
  'COUNTRY': 'EUROPEAN RUSSIA',
  'QSO_DATE': '20170508',
  'TIME_ON': '141400'}]

I want to create a dictionary where each key is a band (10M, 20M, etc) and the value will be a dictionary listing the counties contacted on that band as keys and a count of contacts for each country on that band as the values. Here is what my output looks like:

{'10M': {'ARGENTINA': 2,
         'BRAZIL': 2,
         'FRENCH GUIANA': 1,
         'PORTUGAL': 1,
         'UNITED STATES OF AMERICA': 1},
 '17M': {'ENGLAND': 1, 'SCOTLAND': 1, 'UNITED STATES OF AMERICA': 1},
 '20M': {'EUROPEAN RUSSIA': 1, 'IRELAND': 1, 'UNITED STATES OF AMERICA': 5},
 '40M': {'ARGENTINA': 1, 'CANARY ISLANDS': 1, 'SPAIN': 1}}

This is the comprehension that I came up with to create the output. It works, and with the limited data set shown here, it runs quickly, but with an input list of a couple thousand entries, it takes quite a long time to run.

worked_dxcc_by_band = {
    z["BAND"]: {
        x["COUNTRY"]: len([y["COUNTRY"]
                           for y in log_entries
                           if y["COUNTRY"] == x["COUNTRY"] and y["BAND"] == z["BAND"]])
        for x in log_entries
        if x["BAND"] == z["BAND"]
    }
    for z in log_entries
}

Because this is a triple-nested comprehension, and all 3 loops run through the entire log_entries list, I am assuming that is why it gets very slow.

Is there a more efficient way to accomplish this with comprehension? I am fine using my loop to process the data but I am trying to enhance my skills regarding comprehensions so I thought this would be a good exercise!

This is what I am doing without using comprehension: I have a function analyize_log_entry which I call as I load each log entry in from a file.

from collections import Counter

worked_dxcc_by_band = {}


def analyze_log_entry(entry):
    if "BAND" in entry:
        if "COUNTRY" in entry:
            if entry["BAND"] in worked_dxcc_by_band:
                worked_dxcc_by_band[entry["BAND"]][entry["COUNTRY"]] += 1
            else:
                worked_dxcc_by_band[entry["BAND"]] = Counter()
                worked_dxcc_by_band[entry["BAND"]][entry["COUNTRY"]] = 1

This in itself may not be that efficient but my full code has many similar blocks within the analyze_log_entry function that build multiple dictionaries. Because I am only going through all of my data once, and building the dictionaries where appropriate, it is probably much more efficient than using comprehension, which is essentially multiple loops. As I said, this is more of an exercise to learn how to accomplish the same task with different methods.

Upvotes: 0

Views: 132

Answers (1)

Andrej Kesely
Andrej Kesely

Reputation: 195408

EDIT: Dictionary comprehension version:

out = {band: dict(Counter(v['COUNTRY'] for v in g)) for band, g in groupby(sorted(data, key=lambda k: k['BAND']), lambda k: k['BAND'])}

You can combine itertools.groupby and collections.Counter:

from itertools import groupby
from collections import Counter

s = sorted(data, key=lambda k: k['BAND'])

out = {}
for band, g in groupby(s, lambda k: k['BAND']):
    c = Counter(v['COUNTRY'] for v in g)
    out[band] = dict(c)

from pprint import pprint
pprint(out)

Prints:

{'10M': {'ARGENTINA': 2,
         'BRAZIL': 2,
         'FRENCH GUIANA': 1,
         'PORTUGAL': 1,
         'UNITED STATES OF AMERICA': 1},
 '17M': {'ENGLAND': 1, 'SCOTLAND': 1, 'UNITED STATES OF AMERICA': 1},
 '20M': {'EUROPEAN RUSSIA': 1, 'IRELAND': 1, 'UNITED STATES OF AMERICA': 5},
 '40M': {'ARGENTINA': 1, 'CANARY ISLANDS': 1, 'SPAIN': 1}}

EDIT: Without modules:

out = {}
for i in data:
    out.setdefault(i['BAND'], {}).setdefault(i['COUNTRY'], 0)
    out[i['BAND']][i['COUNTRY']] += 1

from pprint import pprint
pprint(out)

Benchmark:

from timeit import timeit

from itertools import groupby
from collections import Counter

def sol_orig():
    worked_dxcc_by_band = {z["BAND"]: {x["COUNTRY"] : len([y["COUNTRY"] for y in data if y["COUNTRY"] == x["COUNTRY"] and y["BAND"] == z["BAND"]]) for x in data if x["BAND"] == z["BAND"]} for z in data}
    return worked_dxcc_by_band

def solution():
    out = {band: dict(Counter(v['COUNTRY'] for v in g)) for band, g in groupby(sorted(data, key=lambda k: k['BAND']), lambda k: k['BAND'])}
    return out

def solution_2():
    out = {}
    for i in data:
        out.setdefault(i['BAND'], {}).setdefault(i['COUNTRY'], 0)
        out[i['BAND']][i['COUNTRY']] += 1
    return out

t1 = timeit(lambda: solution(), number=10000)
t2 = timeit(lambda: solution_2(), number=10000)
t3 = timeit(lambda: sol_orig(), number=10000)

print(t1)
print(t2)
print(t3)

Prints:

0.18113317096140236
0.08159565401729196
3.5367472909856588

Upvotes: 3

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