基于NLTK的NLP计时器多时间单位组合识别开发咨询

原代码无法识别组合时长的核心原因是，小时、分钟、秒的判断逻辑完全独立，每次匹配到单位就把句子里所有剩余数字全塞给该单位，没有建立数字和相邻单位的对应关系，只要句子里同时出现两个时间单位就会出现数值匹配错误。

改造核心逻辑 #

顺着分词后的文本顺序逐词扫描即可，逻辑非常适合初学者理解：
核心匹配规则：数字永远绑定紧跟在它后面的第一个时间单位，符合日常自然语言表达习惯，不会出现匹配错位。

• 提前定义好所有需要过滤的无意义停用词，以及时间单位的匹配规则
• 初始化小时、分钟、秒的默认值为0，再加一个临时变量暂存当前扫描到的数字
• 逐词遍历处理好的文本：
  • 碰到数字就先存到临时变量里
  • 碰到时间单位，就把之前暂存的数字赋值给对应单位，赋值完清空临时变量
  • 碰到停用词直接跳过
• 最后把匹配到的三个时间值传给原有的倒计时函数即可，不需要改动倒计时逻辑

改造后完整代码 #

import time
import datetime
from nltk_utils import bag_of_words, tokenize
import multiprocessing
from playsound import playsound


def is_number(x):
    if type(x) == str:
        x = x.replace(',', '')
    try:
        float(x)
    except:
        return False
    return True

def text2int (textnum, numwords={}):
    units = [
        'zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight',
        'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen',
        'sixteen', 'seventeen', 'eighteen', 'nineteen',
    ]
    tens = ['', '', 'twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety']
    scales = ['hundred', 'thousand', 'million', 'billion', 'trillion']
    ordinal_words = {'first':1, 'second':2, 'third':3, 'fifth':5, 'eighth':8, 'ninth':9, 'twelfth':12}
    ordinal_endings = [('ieth', 'y'), ('th', '')]

    if not numwords:
        numwords['and'] = (1, 0)
        for idx, word in enumerate(units): numwords[word] = (1, idx)
        for idx, word in enumerate(tens): numwords[word] = (1, idx * 10)
        for idx, word in enumerate(scales): numwords[word] = (10 ** (idx * 3 or 2), 0)

    textnum = textnum.replace('-', ' ')

    current = result = 0
    curstring = ''
    onnumber = False
    lastunit = False
    lastscale = False

    def is_numword(x):
        if is_number(x):
            return True
        if word in numwords:
            return True
        return False

    def from_numword(x):
        if is_number(x):
            scale = 0
            increment = int(x.replace(',', ''))
            return scale, increment
        return numwords[x]

    for word in textnum.split():
        if word in ordinal_words:
            scale, increment = (1, ordinal_words[word])
            current = current * scale + increment
            if scale > 100:
                result += current
                current = 0
            onnumber = True
            lastunit = False
            lastscale = False
        else:
            for ending, replacement in ordinal_endings:
                if word.endswith(ending):
                    word = "%s%s" % (word[:-len(ending)], replacement)

            if (not is_numword(word)) or (word == 'and' and not lastscale):
                if onnumber:
                    curstring += repr(result + current) + " "
                curstring += word + " "
                result = current = 0
                onnumber = False
                lastunit = False
                lastscale = False
            else:
                scale, increment = from_numword(word)
                onnumber = True

                if lastunit and (word not in scales):
                    curstring += repr(result + current)
                    result = current = 0

                if scale > 1:
                    current = max(1, current)

                current = current * scale + increment
                if scale > 100:
                    result += current
                    current = 0

                lastscale = False
                lastunit = False
                if word in scales:
                    lastscale = True
                elif word in units:
                    lastunit = True

    if onnumber:
        curstring += repr(result + current)

    return curstring

# 统一抽离停用词表，避免重复编写相同内容
STOP_WORDS = {'second', 'seconds', 'minute', 'minutes', 'hour', 'hours', 'please',
              'set', 's', 'timer', '``', '{', '}', 'text', ':', "''", 'a', 'for',
              'and', 'if', 'privacy', 'time', 'but', 'end', 'put', 'me', 'my', 'will',
              'you', 'now', 'right', 'privacy', 'rite', 'wright', 'write', 'your', 'go', 'ahead', 't'}

def countdown(h, m, s):
    total_seconds = h * 3600 + m * 60 + s
    while total_seconds > 0:
        timer = datetime.timedelta(seconds = total_seconds)
        print(timer, end="\r")
        time.sleep(1)
        total_seconds -= 1
    print('timer ended')

if __name__ == "__main__":
    # 测试输入，支持多单位组合、英文数词、阿拉伯数字混合输入
    input_text = "please will you set a 10 minutes 45 seconds timer"
    sentence_to_num = text2int(input_text)
    tokens = tokenize(sentence_to_num)
    print("分词结果：", tokens)

    # 初始化时间值和临时数字存储
    h = m = s = 0
    current_num = None

    for word in tokens:
        # 碰到数字先暂存
        if is_number(word):
            current_num = int(float(word.replace(',', '')))
            continue
        # 碰到时间单位就绑定前面暂存的数字
        if word in ["hour", "hours"] and current_num is not None:
            h = current_num
            current_num = None
            print(f"{h} hours starting now")
        elif word in ["minute", "minutes"] and current_num is not None:
            m = current_num
            current_num = None
            print(f"{m} minutes starting now")
        elif word in ["second", "seconds"] and current_num is not None:
            s = current_num
            current_num = None
            print(f"{s} seconds starting now")
        # 其他停用词直接跳过
        else:
            continue

    # 启动倒计时
    countdown(h, m, s)

效果说明 #

• 支持纯数字输入（如10 minutes 45 seconds）、英文数词输入（如ten minutes forty five seconds）、多单位混合输入（如1 hour 5 minutes 30 seconds）
• 自动跳过所有无意义的停用词，不会出现原代码里数字重复赋值给多个单位的问题
• 代码逻辑线性简单，后续要加天、毫秒等其他时间单位，只需要在单位判断的分支里加对应规则即可

内容的提问来源于stack exchange，提问作者Brandon Jones