Проблема: в вашем первом while вы увеличиваете x до тех пор, пока оно не станет равным len(wordlist) - ваш второй while вводится только в том случае, если x меньше len(wordlist) - это противоречиво.
Вы можете использовать collections.Counter, чтобы легко считать вещи и получать от них подсказку:
from collections import Counter
def count(wordlist, sitewordlist):
data = Counter(sitewordlist)
for w in wordlist:
print(f"The word {w} appears {data.get(w,0)} times.")
text = """n 1066, William of Normandy introduced what, in later centuries, became referred
to as a feudal system, by which he sought the advice of a council of tenants-in-chief (a
person who held land) and ecclesiastics before making laws. In 1215, the tenants-in-chief
secured Magna Carta from King John, which established that the king may not levy or collect
any taxes (except the feudal taxes to which they were hitherto accustomed), save with the
consent of his royal council, which gradually developed into a parliament. Over the
centuries, the English Parliament progressively limited the power of the English monarchy
which arguably culminated in the English Civil War and the trial and execution of Charles
I in 1649. After the restoration of the monarchy under Charles II, and the subsequent
Glorious Revolution of 1688, the supremacy of Parliament was a settled principle and all
future English and later British sovereigns were restricted to the role of constitutional
monarchs with limited executive authority. The Act of Union 1707 merged the English
Parliament with the Parliament of Scotland to form the Parliament of Great Britain.
When the Parliament of Ireland was abolished in 1801, its former members were merged
into what was now called the Parliament of the United Kingdom.
(quote from: https://en.wikipedia.org/wiki/Parliament_of_England)""".split()
# some cleanup
text[:] = [t.strip(".,-!?1234567890)([]{}\n") for t in text]
words = ["is","and","not","are"]
count(words,text)
Выход:
The word is appears 0 times.
The word and appears 6 times.
The word not appears 1 times.
The word are appears 0 times.
Полный счетчик:
Counter({'the': 22, 'of': 15, 'Parliament': 7, '': 6, 'and': 6, 'a': 5, 'which': 5,
'English': 5, 'in': 4, 'to': 4, 'were': 3, 'with': 3, 'was': 3, 'what': 2, 'later': 2,
'centuries': 2, 'feudal': 2, 'council': 2, 'tenants-in-chief': 2, 'taxes': 2, 'into': 2,
'limited': 2,'monarchy': 2, 'Charles': 2, 'merged': 2, 'n': 1, 'William': 1, 'Normandy': 1,
'introduced': 1, 'became': 1, 'referred': 1, 'as': 1, 'system': 1, 'by': 1, 'he': 1,
'sought': 1, 'advice': 1, 'person': 1, 'who': 1, 'held': 1, 'land': 1, 'ecclesiastics': 1,
'before': 1, 'making': 1, 'laws': 1, 'In': 1, 'secured': 1, 'Magna': 1, 'Carta': 1,
'from': 1, 'King': 1, 'John': 1, 'established': 1, 'that': 1, 'king': 1, 'may': 1,
'not': 1, 'levy': 1, 'or': 1, 'collect': 1, 'any': 1, 'except': 1, 'they': 1,
'hitherto': 1, 'accustomed': 1, 'save': 1, 'consent': 1, 'his': 1, 'royal': 1,
'gradually': 1, 'developed': 1, 'parliament': 1, 'Over': 1, 'progressively': 1, 'power': 1,
'arguably': 1, 'culminated': 1, 'Civil': 1, 'War': 1, 'trial': 1, 'execution': 1,
'I': 1, 'After': 1, 'restoration': 1, 'under': 1, 'II': 1, 'subsequent': 1, 'Glorious': 1,
'Revolution': 1, 'supremacy': 1, 'settled': 1, 'principle': 1, 'all': 1, 'future': 1,
'British': 1, 'sovereigns': 1, 'restricted': 1, 'role': 1, 'constitutional': 1,
'monarchs': 1, 'executive': 1, 'authority': 1, 'The': 1, 'Act': 1, 'Union': 1,
'Scotland': 1, 'form': 1, 'Great': 1, 'Britain': 1, 'When': 1, 'Ireland': 1,
'abolished': 1, 'its': 1, 'former': 1, 'members': 1, 'now': 1, 'called': 1, 'United': 1,
'Kingdom': 1, 'quote': 1, 'from:': 1,
'https://en.wikipedia.org/wiki/Parliament_of_England': 1})
Пока здесь не совсем уместно. Вы можете смоделировать Counter используя обычный dict и, в то же время, сделайте так:
def count_me_other(words,text):
wordlist = words.split()
splitted = (x.strip(".,!?") for x in text.split())
d = {}
it = iter(splitted)
try:
while it:
c = next(it)
if c not in d:
d[c]=1
else:
d[c]+=1
except StopIteration:
for w in wordlist:
print(f"The word {w} appears {d.get(w,0)} times.")
wordlist = "A C E G I K M"
text = "A B C D E F G A B C D E F A B C D E A B C D A B C A B A"
count_me_other(wordlist,text)
Выход:
The word A appears 7 times.
The word C appears 5 times.
The word E appears 3 times.
The word G appears 1 times.
The word I appears 0 times.
The word K appears 0 times.
The word M appears 0 times.
Или используйте for ... в сочетании с нормальным / defaultdict:
def count_me_other_2(words,text):
wordlist = words.split()
splitted = (x.strip(".,!?") for x in text.split())
d = {}
for w in splitted:
if w not in d:
d[w]=1
else:
d[w]+=1
for w in wordlist:
print(f"The word {w} appears {d.get(w,0)} times.")
def count_me_other_3(words,text):
from collections import defaultdict
wordlist = words.split()
splitted = (x.strip(".,!?") for x in text.split())
d = defaultdict(int)
for w in splitted:
d[w] += 1
for w in wordlist:
print(f"The word {w} appears {d.get(w,0)} times.")
count_me_other_2(wordlist,text)
count_me_other_3(wordlist,text)
с одинаковым выходом.