Question

我正在编写一个措辞数字的数字测试，以便在学校进行评估，并且需要生成一个随机的数字。我可以这样做，但我需要一个随机数的答案，以检查它是否正确。我使用了四个列表来表示问题，例如数千，数百，数十和单位。然后我使用每个列表中的随机选择打印出措辞号码。我有另一个列表，其中包含实际数字，并且需要知道如何匹配措辞号码与实际号码的匹配程度。因此，如果它生成一千，我如何匹配它1.或九百我如何匹配它9.有更快的方式吗？如果有必要，我可以提供我的代码，但它有点长。我也是初学者。

import random
wordt=['one thousand', 'two thousand', 'three thousand', 'four thousand', 'five thousand', 'six thousand', 'seven thousand', 'eight thousand', 'nine thousand']
wordh=['one hundered', 'two hundered', 'three hundered', 'four hundered', 'five hundered', 'six hundered', 'seven hundered', 'eight hundered', 'nine hundered']
wordte=['twenty', 'thirty', 'fourty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety']
wordu=['one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine']
print(wordt[random.randint(0,8)], end=' ')
print(wordh[random.randint(0,8)], end=' ')
print('and', end=' ')
print(wordte[random.randint(0,6)], end=' ')
print(wordu[random.randint(0,8)])``
number=['1', '2', '3', '4', '5', '6', '7', '8', '9']

Answer 1


import random

def AllSame (t):
    """
    Check if string is composed with only one, repeating character. For instance:
    >>> AllSame("aaaaaaa")
    True
    """
    if not t: return False
    l = t[0]
    c = 1
    for x in t[1:]: c += x==l
    return c==len(t)

# English numbers:
digits = {
    "1": "one", "2": "two", "3": "three", "4": "four", "5": "five", "6": "six", "7": "seven", "8": "eight", "9": "nine", "0": "zero",
    "10": "ten", "11": "eleven", "12": "twelve", "13": "thirteen", "14": "fourteen", "15": "fifteen", "16": "sixteen", "17": "seventeen", "18": "eighteen", "19": "nineteen",
    "20": "twenty", "30": "thirty", "40": "forty", "50": "fifty", "60": "sixty", "70": "seventy", "80": "eighty", "90": "ninety",
    "100": "one hundred", "200": "two hundred", "300": "three hundred", "400": "four hundred", "500": "five hundred", "600": "six hundred", "700": "seven hundred", "800": "eight hundred", "900": "nine hundred", "00": "",
    "1000": "one thousand", "2000": "two thousand", "3000": "three thousand", "4000": "four thousand", "5000": "five thousand", "6000": "six thousand", "7000": "seven thousand", "8000": "eight thousand", "9000": "nine thousand", "000": "thousand",
    "1000000": "million", "000000": "million",
    "1000000000": "billion",
    # For declanative languages. Where thousand changes form.
    # E.g. In Croatian: 1000=Tisucu 21000=Dvadeset i [jedna tisuca]
    "001": "one thousand", "002": "two thousand", "003": "three thousand", "004": "four thousand", "005": "five thousand", "006": "six thousand", "007": "seven thousand", "008": "eight thousand", "009": "nine thousand"}

def NumberToWord (num, digits=digits, lj=" ", kz=0):
    """
    Explodes a number to word(s).
    digits is a dictionary containing all major number-word mappings. See the example.
    lj is the word inserted between tens and ones.
    kz is internal for recursive control, do not use it manually.
    """
    num = num.strip()
    if kz==1:
        t = ""
        for x in num:
            if x=="0" and len(t)==0: continue
            t += x
        num = t
    if not num: return ""
    if kz==-1:
        e = ""
        for x in num: e += digits[x]+" "
        return e[:-1]
    if AllSame(num) and num[0]=="0":
        return (len(num)*(digits[num[0]]+" "))[:-1]
    if num.startswith("0"): return NumberToWord(num, digits, lj, -1)
    if digits.has_key(num): return digits[num]
    l = len(num)
    if l==2: return digits[num[0]+"0"]+lj+digits[num[1]]
    if l==3 or l==4: return NumberToWord(num[0]+((l-1)*"0"), digits, lj, 1)+" "+NumberToWord(num[1:], digits, lj, 1)
    if l==5:
        d1 = num[0]; d2 = num[1]
        if d1=="1": return (NumberToWord(num[:2], digits, lj, 1)+" "+digits["000"]+" "+NumberToWord(num[2:], digits, lj, 1)).strip()
        return (digits[d1+"0"]+" "+digits["00"+d2]+" "+NumberToWord(num[2:], digits, lj, 1)).strip()
    if l==6:
        d1 = num[0]; d2 = num[1]; d3 = num[2]
        if d2=="1": m = digits[d2+d3]+" "+digits["000"]
        else: m = digits[d2+"0"]+" "+digits["00"+d3]
        return (digits[d1+"00"]+" "+m+" "+NumberToWord(num[3:], digits, lj, 1)).strip()
    return NumberToWord(num, digits, lj, -1)

print NumberToWord(str(random.randint(0, 999999)))

列表中的随机输出由单独的列表

1 个答案: