Великденски Яйца

Програмиране с Python

Курс във Факултета по Математика и Информатика към СУ

Решение на Пет малки функции от Димитър Влаховски

Обратно към всички решения

Към профила на Димитър Влаховски

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(text, cls):
return "".join([str(symbol) * times for (symbol, times) in text
if type(symbol) == cls])
def reversed_dict(input_dict):
return {input_dict[key]: key for key in input_dict}
def flatten_dict(input_dict, parent_key='', sep='.'):
result = {}
for key, value in input_dict.items():
new_key = parent_key + sep + key if parent_key != '' else key
if isinstance(value, dict):
result.update(flatten_dict(value, new_key).items())
else:
result[new_key] = value
return dict(result)
def recursive_update(dict1, dict2):
if all([key not in dict1 for key in dict2]):
dict1.update(dict2)
return dict1
if (all([not isinstance(value, dict) for value in dict1.values()]) and
all([not isinstance(value, dict) for value in dict2.values()])):
dict1.update(dict2)
return dict1
for key, value in dict2.items():
if key in dict1:
if isinstance(dict1[key], dict) and isinstance(dict2[key], dict):
dict1[key] = recursive_update(dict1[key], dict2[key])
else:
dict1[key] = dict2[key]
return dict1
def unflatten_dict(input_dict, sep='.'):
result = {}
for key, value in input_dict.items():
dicts = key.split(sep)
dicts.append(value)
recursive_update(result, reduce(lambda x, y: {y: x}, dicts[::-1]))
return result
def reps(input_collection):
return tuple([elem for elem in input_collection
if input_collection.count(elem) > 1])

Лог от изпълнението

...................
----------------------------------------------------------------------
Ran 19 tests in 0.132s

OK

История (2 версии и 0 коментара)

Димитър обнови решението на 23.03.2015 12:59 (преди над 9 години)

+from collections import defaultdict
+from functools import reduce
+
+
+def extract_type(text, cls):
+ return "".join([str(symbol) * times for (symbol, times) in text
+ if type(symbol) == cls])
+
+
+def reversed_dict(input_dict):
+ return {input_dict[key]: key for key in input_dict}
+
+
+def flatten_dict(input_dict, parent_key='', sep='.'):
+ result = {}
+ for key, value in input_dict.items():
+ new_key = parent_key + sep + key if parent_key != '' else key
+ if isinstance(value, dict):
+ result.update(flatten_dict(value, new_key).items())
+ else:
+ result[new_key] = value
+
+ return dict(result)
+
+
+def recursive_update(dict1, dict2):
+ if all([key not in dict1 for key in dict2]):
+ dict1.update(dict2)
+ return dict1
+
+ if (all([not isinstance(value, dict) for value in dict1.values()]) and
+ all([not isinstance(value, dict) for value in dict2.values()])):
+ dict1.update(dict2)
+ return dict1
+
+ for key, value in dict2.items():
+ if key in dict1:
+ if isinstance(dict1[key], dict) and isinstance(dict2[key], dict):
+ dict1[key] = recursive_update(dict1[key], dict2[key])
+ else:
+ dict1[key] = dict2[key]
+
+ return dict1
+
+
+def unflatten_dict(input_dict, sep='.'):
+ result = {}
+ for key, value in input_dict.items():
+ dicts = key.split(sep)
+ dicts.append(value)
+ recursive_update(result, reduce(lambda x, y: {y: x}, dicts[::-1]))
+ return result
+
+
+def reps(input_collection):
+ return tuple([elem for elem in input_collection
+ if input_collection.count(elem) > 1])

Димитър обнови решението на 23.03.2015 13:00 (преди над 9 години)

-from collections import defaultdict
from functools import reduce
def extract_type(text, cls):
return "".join([str(symbol) * times for (symbol, times) in text
if type(symbol) == cls])
def reversed_dict(input_dict):
return {input_dict[key]: key for key in input_dict}
def flatten_dict(input_dict, parent_key='', sep='.'):
result = {}
for key, value in input_dict.items():
new_key = parent_key + sep + key if parent_key != '' else key
if isinstance(value, dict):
result.update(flatten_dict(value, new_key).items())
else:
result[new_key] = value
return dict(result)
def recursive_update(dict1, dict2):
if all([key not in dict1 for key in dict2]):
dict1.update(dict2)
return dict1
if (all([not isinstance(value, dict) for value in dict1.values()]) and
all([not isinstance(value, dict) for value in dict2.values()])):
dict1.update(dict2)
return dict1
for key, value in dict2.items():
if key in dict1:
if isinstance(dict1[key], dict) and isinstance(dict2[key], dict):
dict1[key] = recursive_update(dict1[key], dict2[key])
else:
dict1[key] = dict2[key]
return dict1
def unflatten_dict(input_dict, sep='.'):
result = {}
for key, value in input_dict.items():
dicts = key.split(sep)
dicts.append(value)
recursive_update(result, reduce(lambda x, y: {y: x}, dicts[::-1]))
return result
def reps(input_collection):
return tuple([elem for elem in input_collection
if input_collection.count(elem) > 1])