Python练习题答案: 简单有趣＃337：原来的号码【难度：3级】--景越Python编程实例训练营,1000道上机题等你来挑战

简单有趣#337:原来的号码【难度:3级】:

答案1:

from collections import Counter 

EXECUTIONS_ORDER = [('Z', Counter("ZERO"),  '0'),
                    ('W', Counter("TWO"),   '2'),
                    ('U', Counter("FOUR"),  '4'),
                    ('X', Counter("SIX"),   '6'),
                    ('G', Counter("EIGHT"), '8'),
                    ('O', Counter("ONE"),   '1'),
                    ('H', Counter("THREE"), '3'),
                    ('F', Counter("FIVE"),  '5'),
                    ('V', Counter("SEVEN"), '7'),
                    ('I', Counter("NINE"),  '9')]

def original_number(s):
    ans, count, executions = [], Counter(s), iter(EXECUTIONS_ORDER)
    while count:
        c, wordCount, value = next(executions)
        ans.extend([value]*count[c])
        for _ in range(count[c]): count -= wordCount
    return ''.join(sorted(ans))​

答案2:

from collections import Counter, defaultdict 

NUMBERS = ["ZERO", "ONE", "TWO", "THREE", "FOUR", "FIVE", "SIX", "SEVEN", "EIGHT", "NINE"]
VALUES  = {num: str(i) for i,num in enumerate(NUMBERS)}
counts  = Counter(''.join(NUMBERS))

wordsContainningLetter = defaultdict(set)
for n in NUMBERS:
    for c in n: wordsContainningLetter[c].add(n)

EXECUTIONS_ORDER, founds = [], set()
while counts:
    for c,v in counts.copy().items():
        if v == 1:
            try: word = (wordsContainningLetter[c] - founds).pop()
            except KeyError: break
            wordCount = Counter(word)
            founds.add(word)
            EXECUTIONS_ORDER.append( (c, wordCount, VALUES[word]) )
            counts -= wordCount


def original_number(s):
    ans, count, executions = [], Counter(s), iter(EXECUTIONS_ORDER)
    while count:
        c, wordCount, value = next(executions)
        ans.extend([value]*count[c])
        for _ in range(count[c]): count -= wordCount
    return ''.join(sorted(ans))​

答案3:

def original_number(s):
    r, s = [], list(s)
    for word, n in [('ZERO', 0), ('WTO',2), ('XSI',6), ('GEIHT',8), ('THREE',3), 
                    ('UFOR',4), ('ONE',1), ('FIVE',5), ('VSEEN',7), ('NINE',9)]: 
        while word[0] in s: 
            for c in word: s.remove(c)
            r.append(n)
    return ''.join(str(e) for e in sorted(r))​

答案4:

def original_number(s):
    a = s[:]
    code=[0 for _ in range(10)]

    book = [[0, 'Z', 'ZERO'],
            [2, 'W', 'TWO'],
            [6, 'X', 'SIX'],
            [8, 'G', 'EIGHT'],
            [7, 'S', 'SEVEN'],
            [5, 'V', 'FIVE'],
            [4, 'F', 'FOUR'],
            [3, 'T', 'THREE'],
            [1, 'O', 'ONE'],
            [9, 'E', 'NINE']]
    for i in book:
        code[i[0]] = a.count(i[1])
        for j in i[2]:
            a = a.replace(j, '', code[i[0]])
    return ''.join('0123456789'[k]*j for k,j in enumerate(code))​

答案5:

from collections import Counter

NUMS = {
        'ZERO':'0','EIGHT':'8','SIX':'6','SEVEN':'7',
        'THREE':'3','FOUR':'4','FIVE':'5','NINE':'9',
        'TWO':'2','ONE':'1'
        }
        
KEYS = [ 'ZERO','EIGHT','SIX','SEVEN',
         'THREE','FOUR','FIVE','NINE',
         'TWO','ONE' ]         


def original_number(s):
    res = ''; counted = Counter(s)
    for key in KEYS:
        while all(counted[char]>0 for char in key):
            for char in key:
                counted[char] -= 1
            res += NUMS[key]
    return ''.join(sorted(res))​

答案6:

def original_number(s):
    dic = {i:0 for i in range(10)}
    dic[0] = s.count('Z')
    dic[2] = s.count('W')
    dic[4] = s.count('U')
    dic[6] = s.count('X')
    dic[8] = s.count('G')
    dic[1] = s.count('O') - dic[0] - dic[2] - dic[4]
    dic[3] = s.count('H') - dic[8]
    dic[5] = s.count('F') - dic[4]
    dic[7] = s.count('S') - dic[6]
    dic[9] = s.count('I') - dic[5] - dic[6] - dic[8]
    result = ''
    for i in range(10):
      result += str(i)*dic[i]
    return result​

答案7:

import scipy as sc
import scipy.optimize as so


def original_number(s):
    return ''.join([str(k) * int(round(n)) for k, n in enumerate( \
        so.nnls(sc.transpose([sc.histogram([ord(c) - 65 for c in d], range(27))[0] for d in \
                              ['ZERO', 'ONE', 'TWO', 'THREE', 'FOUR', 'FIVE', 'SIX', 'SEVEN', 'EIGHT', 'NINE'] \
                              ]), sc.histogram([ord(c) - 65 for c in s], range(27))[0])[0])])

​

答案8:

def original_number(s):
    s_list = list(s)
    numbers = [(0, 'ZERO'), (2, 'TWO'), (6, 'SIX'), (4, 'FOUR'),  (1, 'ONE'), (5, 'FIVE'), (7, 'SEVEN'), (9, 'NINE'), (3, 'THREE'), (8, 'EIGHT')]  
    secret_number = ''
    for i, number in numbers:
        while all([c in s_list for c in number]):            
            [s_list.remove(c) for c in number]  
            secret_number += str(i)     

    return ''.join(sorted(secret_number))​

答案9:

import re;original_number=lambda s:''.join(n*str(i)for i,n in enumerate(eval(re.sub('([A-Z])',r's.count("\1")','[Z,O-W-U-Z,W,H-G,U,F-U,X,S-X,G,I-G-F+U-X]'))))​

答案10:

def original_number(s):
    
    s = list(s)
    output = []
    nums = (('Z','ZERO','0'),
            ('W','TWO','2'),
            ('U','FOUR','4'),
            ('X','SIX','6'),
            ('G','EIGHT','8'),
            ('O','ONE','1'),
            ('H','THREE','3'),
            ('F','FIVE','5'),
            ('V','SEVEN','7'),
            ('I','NINE','9'))

    for n in nums:
        while n[0] in s:
            output.append(n[2])
            for c in n[1]:           
                del s[s.index(c)]
    
    return ''.join(sorted(output))​

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