Sort list of lists with groupby and count

I am looking to do a sort of a list of lists. My function needs to return the day that that fewest activity of a certain type and if there is a tie, return the day with the fewest overall activities. Below is a working solution but I feel like it's fairly unpythonic as it needs to convert to a dictionary and back to a list and am looking for a faster way to write this.

print get_day(mylist, 'Activity C') should yield Day 1

print get_day(mylist, 'Activity C')

Day 1

print get_day(mylist, 'Activity A') should yield Day 2

print get_day(mylist, 'Activity A')

Day 2

def get_day(l, activity):
d = {}

for x in l:
if x[0] not in d.keys():
d[x[0]] =
d[x[0]].append(x[1])

d = {k: [v.count(activity), len(v)] for k, v in d.items()}

l = [[k, v[0], v[1]] for k, v in d.items()]

l = sorted(l, key=lambda x: (x[1], x[2]))
return l[0][0]


mylist = [['Day 1', 'Activity A'], ['Day 2', 'Activity A'], ['Day 1', 'Activity A'], ['Day 2', 'Activity C'],
['Day 2', 'Activity D']]


get_day(mylist, 'Activity C')

Day 1

2 Answers
2

First we can write utility for collecting pairs by first coordinate:

from collections import defaultdict


def collect(items):
result = defaultdict(list)
for key, value in items:
result[key].append(value)
return result


After that our get_day function can be written like

get_day

from collections import Counter
from itertools import imap


def get_day(days_activities, target_activity):
activities_by_days = collect(days_activities)
days_by_activities = collect(imap(reversed, days_activities))
days_target_activity_counter = Counter(days_by_activities[target_activity])

def to_target_and_overall_activities_counts(day):
return (days_target_activity_counter[day],
# if there is a tie
len(activities_by_days[day]))

return min(activities_by_days,
key=to_target_and_overall_activities_counts)


Test

# 'Day 1' has fewest overall activities (3 < 4)
>>> mylist = [['Day 1', 'Activity A'],
['Day 1', 'Activity A'],
['Day 2', 'Activity A'],
['Day 2', 'Activity C'],
['Day 1', 'Activity D'],
['Day 2', 'Activity D'],
['Day 2', 'Activity E']]
>>> get_day(mylist, 'Activity C')
'Day 1'
>>> get_day(mylist, 'Activity A')
'Day 2'
>>> get_day(mylist, 'Activity D')
'Day 1'


Can't guarantee speed here without knowing more about the expected input dimensions and use case, but I think this code is more pythonic.

from collections import defaultdict, Counter

def get_day_pythonic(lst, activity):
if not lst:
return
# Count of activities by day
day_act_counts = Counter([d for (d, a) in lst])
# Activity counts per day
act_counter = defaultdict(Counter)
for (d, a) in lst:
act_counter[a][d] += 1
# NOTE: if planning to call this multiple times, should precompute day_act_counts and act_counter.
# Here we sort first by lowest count of activity, then total activity counts, and then day name.
return sorted([(act_counter[activity][d], day_act_counts[d], d) for d in day_act_counts])[0][-1]


EDIT: Faster implementation

def get_day(lst, activity):
if not lst:
return
# Count of all activities by day
day_act_counts = {}
# Count of interested activity by day
act_counter = {}
for (d, a) in lst:
day_act_counts[d] = day_act_counts.get(d, 0) + 1
if a != activity: # don't need exact count for other activities
continue
act_counter[d] = act_counter.get(d, 0) + 1
# Here we take the min first by lowest count of activity, then total activity counts, and then day name.
return min((act_counter.get(d, 0), day_act_counts[d], d) for d in day_act_counts)[-1]


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