Решение на Пет функции от Антонио Николов

Резултати

10 точки от тестове
0 бонус точки
10 точки общо

16 успешни тест(а)
0 неуспешни тест(а)

Код

import functools

import string

from collections import defaultdict

def is_pangram(sentence):

    all_chars = {'ы', 'э'}

    for char in sentence.lower():

        if ord(char) in range(ord('а'), ord('я') + 1):

            all_chars.add(char)

    return len(all_chars) == 32

def char_histogram(text):

    count_char = {}

    for char in text:

        if not count_char.get(char):

            count_char[char] = text.count(char)

    return count_char

def sort_by(func, arguments):

    return sorted(arguments, key=functools.cmp_to_key(func))

def group_by_type(dictionary):

    grouped = defaultdict(dict)

    for key, value in dictionary.items():

        items_till_now = list(grouped[type(key)].items())

        grouped[type(key)] = dict(items_till_now + list({key: value}.items()))

    return dict(grouped)

def letters(word):

    ignored = string.punctuation + string.whitespace + '–'

    return sorted(list(filter(lambda x: x not in ignored, word.lower())))

def anagrams(words):

    grouped = []

    for faith in words:

        subgroup = list(filter(lambda x: letters(x) == letters(faith), words))

        if subgroup not in grouped:

            grouped.append(subgroup)

    return grouped

Лог от изпълнението

................
----------------------------------------------------------------------
Ran 16 tests in 0.016s

OK

История (5 версии и 2 коментара)

Антонио обнови решението на 14.03.2014 01:33 (преди около 10 години)

+import functools

+from collections import defaultdict

+def is_pangram(sentence):

+    all_chars = {'ы', 'э'}

+    for char in sentence.lower():

+        if ord(char) in range(ord('а'), ord('я') + 1):

+            all_chars.add(char)

+    return len(all_chars) == 32

+def char_histogram(text):

+    count_char = {}

+    for char in text:

+        if not count_char.get(char):

+            count_char[char] = text.count(char)

+    return count_char

+def sort_by(func, arguments):

+    return sorted(arguments, key=functools.cmp_to_key(func))

+def group_by_type(dictionary):

+    grouped = defaultdict(lambda: {})

+    for key, value in dictionary.items():

+        items_till_now = list(grouped[type(key)].items())

+        grouped[type(key)] = dict(items_till_now + list({key: value}.items()))

+    return dict(grouped)

+def anagrams(words):

+    grouped = []

+    for vocable in words:

+        new = list(filter(lambda x: sorted(list(x)) == sorted(list(vocable)),

+                          words))

+        if not new in grouped:

+            grouped.append(new)

+    return grouped

Това на 44-45 ред проблем ли е?

Публикувано преди около 10 години

Антонио обнови решението на 17.03.2014 13:27 (преди около 10 години)

 import functools

+import string

 from collections import defaultdict

 def is_pangram(sentence):

     all_chars = {'ы', 'э'}

     for char in sentence.lower():

         if ord(char) in range(ord('а'), ord('я') + 1):

             all_chars.add(char)

     return len(all_chars) == 32

 def char_histogram(text):

     count_char = {}

     for char in text:

         if not count_char.get(char):

             count_char[char] = text.count(char)

     return count_char

 def sort_by(func, arguments):

     return sorted(arguments, key=functools.cmp_to_key(func))

 def group_by_type(dictionary):

     grouped = defaultdict(lambda: {})

     for key, value in dictionary.items():

         items_till_now = list(grouped[type(key)].items())

         grouped[type(key)] = dict(items_till_now + list({key: value}.items()))

     return dict(grouped)

+def letters(word):

+    ignored = string.punctuation + string.whitespace

+    return sorted(list(filter(lambda x: x not in ignored, word.lower())))

 def anagrams(words):

     grouped = []

     for vocable in words:

-        new = list(filter(lambda x: sorted(list(x)) == sorted(list(vocable)),

-                          words))

-        if not new in grouped:

-            grouped.append(new)

+        subgroup = filter(lambda x: letters(x) == letters(vocable), words)

-    return grouped

+        if subgroup not in grouped:

+            grouped.append(list(subgroup))

+    return grouped

Антонио обнови решението на 17.03.2014 23:07 (преди около 10 години)

 import functools

 import string

 from collections import defaultdict

 def is_pangram(sentence):

     all_chars = {'ы', 'э'}

     for char in sentence.lower():

         if ord(char) in range(ord('а'), ord('я') + 1):

             all_chars.add(char)

     return len(all_chars) == 32

 def char_histogram(text):

     count_char = {}

     for char in text:

         if not count_char.get(char):

             count_char[char] = text.count(char)

     return count_char

 def sort_by(func, arguments):

     return sorted(arguments, key=functools.cmp_to_key(func))

 def group_by_type(dictionary):

     grouped = defaultdict(lambda: {})

     for key, value in dictionary.items():

         items_till_now = list(grouped[type(key)].items())

         grouped[type(key)] = dict(items_till_now + list({key: value}.items()))

     return dict(grouped)

 def letters(word):

     ignored = string.punctuation + string.whitespace

     return sorted(list(filter(lambda x: x not in ignored, word.lower())))

 def anagrams(words):

     grouped = []

-    for vocable in words:

-        subgroup = filter(lambda x: letters(x) == letters(vocable), words)

+    for faith in words:

+        subgroup = list(filter(lambda x: letters(x) == letters(faith), words))

         if subgroup not in grouped:

-            grouped.append(list(subgroup))

+            grouped.append(subgroup)

     return grouped

Ами проблем беше, но гледам си го оправил беше :) Решението ти ми харесва, но няма нужда даваш lambda на defaultdict. То просто взема тип, а не инстанция.

defaultdict(dict) е това, което ти трябва :)