请用英文写下逻辑清晰的一段话包含以下短语：free of charge免费 book the ticket订票 the ultimate goal最终目的 constant exposure经常接触 add fuel to the fire火上浇油 run out of patience忍无可忍 in case of emergency万一情况紧急 without thinking twice不假思索 be meant as a nod to意为认可 be caught playing the football被逮到踢足球 apply to online learning适用于网上学习 有效管理实时间manage your time effectively 经常反思 regularly reflect on

Free of charge bookings for the ultimate goal of constant exposure can add fuel to the fire, but when you run out of patience in case of emergency, it’s better to act without thinking twice. This can be meant as a nod to the effort of being caught playing football and can also apply to online learning, so it's important to ensure you manage your time effectively and regularly reflect on your progress.

相关问题

请写出 “Education is of criticalimportance to the future of our country and our party.”这个句子的语法树

这句话 "Education is of critical importance to the future of our country and our party." 可以分析出一个典型的简单句结构，其中包含了三个核心部分：

1. 主语 (Subject)：Education
2. 谓语 (Predicate)：is of critical importance
3. 宾语 (Object)：to the future of our country and our party

具体到语法树上，可以分为以下几个层级：

- 最高层：主句 (S)
- 第一层：谓词中心 (VP) - 由系动词 "is" 构成
- 第二层：定语 (AdjP) - 描述 "critical importance"
- 子定语 (AdvP)：of critical importance
- 第三层：介词短语 (PP) - 表示方向或关联，"to the future of our country and our party"
- 名词短语 (NP)：the future of our country and our party

所以整个句子的语法树看起来像这样：

   S
  / \
 VP   PP
  |     |
 AdjP   NP
      /  \
 AdvP  Noun phrase
     |      |
    critical  future of our country and our party

用python写一个小程序,任意在这个小程序中输入一段英文,统计出现频率最高的三组短语

在Python中，我们可以使用`collections`模块中的`Counter`类来统计英文单词的频率，并利用`nlTK`库对文本进行预处理，包括分词。下面是一个简单的示例程序：

import re
from collections import Counter

# 导入nltk库并下载停用词列表
import nltk
nltk.download('stopwords')

from nltk.corpus import stopwords
from nltk.tokenize import word_tokenize, sent_tokenize

def get_top_phrases(text):
    # 预处理步骤
    stop_words = set(stopwords.words('english'))
    text = text.lower()
    words = word_tokenize(text)
    filtered_words = [word for word in words if word.isalnum() and word not in stop_words]
    
    # 统计词频
    word_counts = Counter(filtered_words)

    # 分割短语（这里简单地假设每个空格后的连续词是一组短语）
    phrases = [' '.join(word) for word in filtered_words]
    phrase_counts = Counter(phrases)

    # 找到频率最高的三个短语
    top_phrases = phrase_counts.most_common(3)

    return top_phrases

if __name__ == "__main__":
    input_text = input("请输入一段英文: ")
    top_phrases_result = get_top_phrases(input_text)
    print(f"出现频率最高的三个短语及频率分别是:")
    for phrase, freq in top_phrases_result:
        print(f"{phrase}: {freq} times")