Великденски Яйца

Програмиране с Python

Курс във Факултета по Математика и Информатика към СУ

Решение на Пет малки функции от Виктор Драганов

Обратно към всички решения

Към профила на Виктор Драганов

Резултати

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

Код

__author__ = 'Veke'
def extract_type(input, filter_type):
filtered = ''
for tup in input:
if type(tup[0]) is filter_type:
filtered += tup[0] * tup[1]
return filtered
def reversed_dict(input):
output = {}
for key in input.keys():
if not (key is output.values()):
output[input[key]] = key
return output
def flatten_dict(d):
# If the value type is <class 'dict'> for every value
# we add to key '.' + value's key.
def expand_sub_dict(key, value):
print('1')
if isinstance(value, dict):
return [(key + '.' + k, v) for k, v in flatten_dict(value).items()]
else:
# Bottom of the recursion
return [(key, value)]
# We expand every item of the dictionary
items = [i for k, v in d.items() for i in expand_sub_dict(k, v)]
return dict(items)
def unflatten_dict(d):
items = dict()
for key, value in d.items():
# We split the key to parts which are parted '.'
parts = key.split('.')
it = items
for part in parts[:-1]:
if part not in it:
it[part] = dict()
it = it[part]
it[parts[-1]] = value
return items
def reps(collection):
output = [i for i in collection if collection.count(i) > 1]
return output

Лог от изпълнението

...................
----------------------------------------------------------------------
Ran 19 tests in 0.140s

OK

История (2 версии и 0 коментара)

Виктор обнови решението на 23.03.2015 16:40 (преди около 9 години)

+_author__ = 'Veke'
+
+
+def extract_type(input, filter_type):
+ filtered = ''
+ for tup in input:
+ if type(tup[0]) is filter_type:
+ filtered += tup[0] * tup[1]
+ return filtered
+
+
+def reversed_dict(input):
+ output = {}
+ for key in input.keys():
+ if not (key is output.values()):
+ output[input[key]] = key
+ return output
+
+
+def flatten_dict(d):
+ # If the value type is <class 'dict'> for every value
+ # we add to key '.' + value's key
+ def expand_sub_dict(key, value):
+ print('1')
+ if isinstance(value, dict):
+ return [(key + '.' + k, v) for k, v in flatten_dict(value).items()]
+ else:
+ # Bottom of the recursion
+ return [(key, value)]
+
+ # We expand every item of the dictionary
+ items = [i for k, v in d.items() for i in expand_sub_dict(k, v)]
+
+ return dict(items)
+
+
+def reps(collection):
+ output = [i for i in collection if collection.count(i) > 1]
+ return output

Виктор обнови решението на 23.03.2015 16:56 (преди около 9 години)

-_author__ = 'Veke'
+__author__ = 'Veke'
def extract_type(input, filter_type):
filtered = ''
for tup in input:
if type(tup[0]) is filter_type:
filtered += tup[0] * tup[1]
return filtered
def reversed_dict(input):
output = {}
for key in input.keys():
if not (key is output.values()):
output[input[key]] = key
return output
def flatten_dict(d):
# If the value type is <class 'dict'> for every value
- # we add to key '.' + value's key
+ # we add to key '.' + value's key.
def expand_sub_dict(key, value):
print('1')
if isinstance(value, dict):
return [(key + '.' + k, v) for k, v in flatten_dict(value).items()]
else:
# Bottom of the recursion
return [(key, value)]
# We expand every item of the dictionary
items = [i for k, v in d.items() for i in expand_sub_dict(k, v)]
return dict(items)
+
+
+def unflatten_dict(d):
+ items = dict()
+ for key, value in d.items():
+ # We split the key to parts which are parted '.'
+ parts = key.split('.')
+ it = items
+ for part in parts[:-1]:
+ if part not in it:
+ it[part] = dict()
+ it = it[part]
+ it[parts[-1]] = value
+ return items
def reps(collection):
output = [i for i in collection if collection.count(i) > 1]
return output