Groupby
""" itertools.groupby
Makes an iterator that returns consecutive keys and groups from the iterable.
The iterable needs to already be sorted on the same key function.
"""
import itertools
DEBUG = 1
transactions = [
{'date': '2024-08-01', 'amount': 100},
{'date': '2024-08-02', 'amount': 150},
{'date': '2024-08-01', 'amount': 200},
{'date': '2024-08-03', 'amount': 50},
]
# Sort by date
transactions.sort(key=lambda x: x['date'])
# Group by date
grouped = itertools.groupby(transactions, key=lambda x: x['date']) # Look here
# Output
for date, group in grouped: # grouped is an iterator
print(date, list(group)) # group is a generator
if DEBUG:
print("\nDEBUG / Without itertools.groupby:")
def groupby(items, sortby):
grouped = {}
for t in items:
date = t['date']
if date not in grouped:
grouped[date] = []
grouped[date].append(t['amount']) # Look here
return grouped
grouped = groupby(transactions, 'date')
# Output
for date in grouped: # grouped is a dictionary
print(date, grouped[date])
"""
2024-08-01 [{'date': '2024-08-01', 'amount': 100}, {'date': '2024-08-01', 'amount': 200}]
2024-08-02 [{'date': '2024-08-02', 'amount': 150}]
2024-08-03 [{'date': '2024-08-03', 'amount': 50}]
DEBUG / Without itertools.groupby:
2024-08-01 [100, 200]
2024-08-02 [150]
2024-08-03 [50]
"""
Compress
""" itertools.compress
Makes an interator that returns elements where the corresponding item is true
As an example, lets filter based on a Boolean list
"""
import itertools
DEBUG = 1
products = ['apple', 'banana', 'cherry', 'orange']
in_stock = [True, False, True, False]
available_products = itertools.compress(products, in_stock) # iterator
print(list(available_products))
if DEBUG:
print("\nDEBUG / Without itertools.compress:")
available_products = []
for item, available in zip(products, in_stock):
if available:
available_products.append(item)
"""
['apple', 'cherry']
DEBUG / Without itertools.compress:
['apple', 'cherry']
"""
Accumulate
""" Accumulate sums
Make an iterator that returns accumulated sums from other binary functions.
"""
import itertools
N = [1, 2, 3, 4, 5]
T = []
# without itertools
total = 0
for n in N:
total += n
T.append(total)
print(T)
# itertools
T = itertools.accumulate(N)
print(list(T))
"""
[1, 3, 6, 10, 15]
[1, 3, 6, 10, 15]
"""
Count
""" Count
Make an iterator that returns evenly spaced values beginning with start.
"""
import itertools
# without itertools
def custom_count(start=0, step=1):
n = start
while True:
yield n
n += step
counter = custom_count(10, 1)
first_five = [next(counter) for _ in range(5)]
print(first_five)
# itertools
counter = itertools.count(2.5, 0.5)
first_five = [next(counter) for _ in range(5)]
print(first_five)
"""
[10, 11, 12, 13, 14]
[2.5, 3.0, 3.5, 4.0, 4.5]
"""
Permutations
""" Permutations
Return successive r length permutations of elements from the iterable.
permutations('ABCD', 2) → AB AC AD BA BC BD CA CB CD DA DB DC
permutations(range(3)) → 012 021 102 120 201 210
"""
import itertools
iter = itertools.permutations('ABCD', 2)
print([a + b for a, b in list(iter)])
iter = itertools.permutations(range(3))
print([f"{x}{y}{z}" for x, y, z in list(iter)])
"""
['AB', 'AC', 'AD', 'BA', 'BC', 'BD', 'CA', 'CB', 'CD', 'DA', 'DB', 'DC']
['012', '021', '102', '120', '201', '210']
"""
Last update: 65 days ago
