Question

我想在功能末尾打印一个短语，但我想要的输出不是打印。 python中没有弹出错误，它只是不打印而无视它的行为。 wordlist是用户输入的单词列表，以查找每个单词出现在他们输入的网站中的次数。 sitewordlist是网站中单词的完整列表。

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Answer 1

您正在使用while循环来充当for循环，但是两者都使用了相同的迭代器x，并且没有在两者之间将其值重置为0。因此，第二个while循环会发现x已经等于len(wordlist)，因此它不执行循环主体。

Answer 2

问题：在您的第一个while中增加x直到等于len(wordlist)-仅在while小于x-有点矛盾。

您可以使用collections.Counter轻松地计算事物并从中获得命令：

len(wordlist)

输出：

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)

全计数器：

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

虽然这里并不十分合适。您可以使用普通的字典来模拟Counter，就像这样：

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})

输出：

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)

或将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.与常规/ defaultdict结合使用：

for ...

具有相同的输出。

Python打印语句在函数末尾不打印任何内容

2 个答案: