Python 打印语句在我的函数末尾不打印任何内容

Question

I want to print a phrase at the end of my function, but my desired output is not printing. There are no errors popping up in python, it just isn't printing and acting like it is ignoring it. wordlist is the list of words the user entered to find how many times each word appears in the website they entered. sitewordlist is the entire list of words in the website.

def count(wordlist, sitewordlist):
    x = 0
    while x < len(wordlist):
       numblist = []
       wordcount = sitewordlist.count(wordlist[x])
       numblist.append(wordcount)
       x = x + 1
    final(numblist, wordlist)

def final(numblist, wordlist):    
    y = 0
    while y < len(numblist):
    print("The word" + wordlist[y] + "appears" + numblist[y] + "times.")
    y = y + 1
main()

Answer 1

Problem: in your first while you increase x until it is equal to len(wordlist) - your second while is only entered if x is smaller then len(wordlist) - thats kind of contradictionary.

---

You can use collections.Counter to count things easily and get a dict from it:

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)

Output:

The word is appears 0 times.
The word and appears 6 times.
The word not appears 1 times.
The word are appears 0 times.

---

Full Counter:

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})

---

While is not really appropriate here. You can simulate Counter using a normal dict and while like so:

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)

Output:

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.

Or use for ... in conjunction with a normal / 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)

with identical output.

Answer 2

You're using while-loops to act like for-loops, but you're using the same iterator x in both, and you're not resetting its value to 0 in between. So the second while-loop sees that x is already equal to len(wordlist) , and so it doesn't execute the body of the loop.