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Програмиране с Python

Курс във Факултета по Математика и Информатика към СУ

Решение на Пет малки функции от Николай Велков

Обратно към всички решения

Към профила на Николай Велков

Резултати

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

Код

def extract_type(pairs, actual_type):
return ''.join([''.join([str(pair[0]) for _ in range(0, pair[1])])
for pair in pairs if type(pair[0]) is actual_type])
def flatten_element(element, pairs):
if type(pairs[element]) is not dict:
return [(element, pairs[element])]
else:
result = []
for elem in pairs[element]:
flat_child = flatten_element(elem, pairs[element])
for flat_child in flat_child:
result.append(("{}.{}".format(element,
flat_child[0]), flat_child[1]))
return result
def flatten_dict(non_flat):
flat_elements = [pair for pair in
map(lambda x: flatten_element(x, non_flat), non_flat)]
flat = []
for x in flat_elements:
flat.extend(x)
return {elem[0]: elem[1] for elem in flat}
def unflatten_dict(flat):
result = {}
for elem in flat:
if elem.find('.') == -1:
result[elem] = flat[elem]
else:
sub_dict = filter(
lambda x: x.partition('.')[0] == elem.partition('.')[0]
and x.find('.') != -1, flat
)
dot_erased = {
x: flat["{}.{}".format(elem.partition('.')[0], x)]
for x in map(lambda x: x.partition('.')[2], sub_dict)
}
result[elem.partition('.')[0]] = unflatten_dict(dot_erased)
return result
def reversed_dict(original):
return {original[pair]: pair for pair in original}
def reps(elements):
return tuple([x for x in
filter(lambda x: elements.count(x) > 1, elements)])

Лог от изпълнението

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

OK

История (2 версии и 0 коментара)

Николай обнови решението на 20.03.2015 22:45 (преди над 9 години)

+
+def extract_type(pairs, actual_type):
+ return ''.join([''.join([str(pair[0]) for _ in range(0, pair[1])])
+ for pair in pairs if type(pair[0]) is actual_type])
+
+
+def flatten_element(element, pairs):
+ if type(pairs[element]) is not dict:
+ return [(element, pairs[element])]
+ else:
+ result = []
+
+ for elem in pairs[element]:
+ flat_child = flatten_element(elem, pairs[element])
+ for flat_child in flat_child:
+ result.append(("{}.{}".format(element,
+ flat_child[0]), flat_child[1]))
+
+ return result
+
+
+def flatten_dict(non_flat):
+ flat_elements = [pair for pair in
+ map(lambda x: flatten_element(x, non_flat), non_flat)]
+
+ flat = []
+
+ for x in flat_elements:
+ flat.extend(x)
+
+ return {elem[0]: elem[1] for elem in flat}
+
+
+def unflatten_dict(flat):
+ result = {}
+
+ for elem in flat:
+ if elem.find('.') == -1:
+ result[elem] = flat[elem]
+ else:
+ sub_dict = filter(lambda x: x[0] == elem[0]
+ and x.find('.') != -1, flat)
+ dot_erased = {x: flat["{}.{}".format(elem[0], x)] for
+ x in map(lambda x: x.partition('.')[2], sub_dict)}
+ result[elem[0]] = unflatten_dict(dot_erased)
+
+ return result
+
+
+def reversed_dict(original):
+ return {original[pair]: pair for pair in original}
+
+
+def reps(elements):
+ return tuple([x for x in
+ filter(lambda x: elements.count(x) > 1, elements)])

Николай обнови решението на 20.03.2015 23:52 (преди над 9 години)

-
def extract_type(pairs, actual_type):
return ''.join([''.join([str(pair[0]) for _ in range(0, pair[1])])
for pair in pairs if type(pair[0]) is actual_type])
def flatten_element(element, pairs):
if type(pairs[element]) is not dict:
return [(element, pairs[element])]
else:
result = []
for elem in pairs[element]:
flat_child = flatten_element(elem, pairs[element])
for flat_child in flat_child:
result.append(("{}.{}".format(element,
flat_child[0]), flat_child[1]))
return result
def flatten_dict(non_flat):
flat_elements = [pair for pair in
map(lambda x: flatten_element(x, non_flat), non_flat)]
flat = []
for x in flat_elements:
flat.extend(x)
return {elem[0]: elem[1] for elem in flat}
def unflatten_dict(flat):
result = {}
for elem in flat:
if elem.find('.') == -1:
result[elem] = flat[elem]
else:
- sub_dict = filter(lambda x: x[0] == elem[0]
- and x.find('.') != -1, flat)
- dot_erased = {x: flat["{}.{}".format(elem[0], x)] for
- x in map(lambda x: x.partition('.')[2], sub_dict)}
- result[elem[0]] = unflatten_dict(dot_erased)
+ sub_dict = filter(
+ lambda x: x.partition('.')[0] == elem.partition('.')[0]
+ and x.find('.') != -1, flat
+ )
+ dot_erased = {
+ x: flat["{}.{}".format(elem.partition('.')[0], x)]
+ for x in map(lambda x: x.partition('.')[2], sub_dict)
+ }
+ result[elem.partition('.')[0]] = unflatten_dict(dot_erased)
return result
def reversed_dict(original):
return {original[pair]: pair for pair in original}
def reps(elements):
return tuple([x for x in
filter(lambda x: elements.count(x) > 1, elements)])