Великденски Яйца

Програмиране с Python

Курс във Факултета по Математика и Информатика към СУ

Решение на Пет малки функции от Диана Генева

Обратно към всички решения

Към профила на Диана Генева

Резултати

  • 10 точки от тестове
  • 0 бонус точки
  • 10 точки общо
  • 19 успешни тест(а)
  • 0 неуспешни тест(а)

Код

from functools import reduce
def extract_type(pairs, wanted_type):
pairs_of_type = (list(filter(lambda x: type(x[0]) is wanted_type, pairs)))
return reduce(lambda x, y: x + str(y[0])*y[1], pairs_of_type, "")
def reversed_dict(regular_order):
return {value: key for key, value in regular_order.items()}
def reps(collection):
return tuple(filter(lambda x: collection.count(x) > 1, collection))
def flatten_dict(unflattened):
pair_queue = list(unflattened.items())
flattened = {}
while pair_queue:
key, value = pair_queue[0]
if(type(value) is dict):
for internal_key, internal_value in value.items():
pair_queue.append((key + '.' + internal_key, internal_value))
else:
flattened[key] = value
pair_queue.remove((key, value))
return flattened
def unflatten_dict(flattened):
if type(flattened) is not dict:
return flattened
unflattened = {}
for key, value in flattened.items():
if '.' not in key:
unflattened[key] = value
else:
first = key.split('.', 1)[0]
rest = key.split('.', 1)[1]
if first not in unflattened:
unflattened[first] = {rest: value}
else:
unflattened[first][rest] = value
return {key: unflatten_dict(value) for key, value in unflattened.items()}

Лог от изпълнението

...................
----------------------------------------------------------------------
Ran 19 tests in 0.136s

OK

История (1 версия и 0 коментара)

Диана обнови решението на 19.03.2015 20:35 (преди над 9 години)

+from functools import reduce
+
+
+def extract_type(pairs, wanted_type):
+ pairs_of_type = (list(filter(lambda x: type(x[0]) is wanted_type, pairs)))
+ return reduce(lambda x, y: x + str(y[0])*y[1], pairs_of_type, "")
+
+
+def reversed_dict(regular_order):
+ return {value: key for key, value in regular_order.items()}
+
+
+def reps(collection):
+ return tuple(filter(lambda x: collection.count(x) > 1, collection))
+
+
+def flatten_dict(unflattened):
+ pair_queue = list(unflattened.items())
+ flattened = {}
+ while pair_queue:
+ key, value = pair_queue[0]
+ if(type(value) is dict):
+ for internal_key, internal_value in value.items():
+ pair_queue.append((key + '.' + internal_key, internal_value))
+ else:
+ flattened[key] = value
+ pair_queue.remove((key, value))
+ return flattened
+
+
+def unflatten_dict(flattened):
+ if type(flattened) is not dict:
+ return flattened
+ unflattened = {}
+ for key, value in flattened.items():
+ if '.' not in key:
+ unflattened[key] = value
+ else:
+ first = key.split('.', 1)[0]
+ rest = key.split('.', 1)[1]
+ if first not in unflattened:
+ unflattened[first] = {rest: value}
+ else:
+ unflattened[first][rest] = value
+ return {key: unflatten_dict(value) for key, value in unflattened.items()}