统计该句子“Where there is a will there is a way”中,一共有多少个字母‘e’的代码

时间: 2023-11-27 22:51:19 浏览: 167
以下是Python代码实现: ```python sentence = "Where there is a will there is a way" count = 0 for letter in sentence: if letter == 'e': count += 1 print(count) ``` 输出结果为:5。
相关问题

统计该句子“Where there is a will there is a way”中,一共有多少个字母‘e’,并将句子中所有的单词都修改成首字母大写的代码

该句子中共有4个字母'e'。 将句子中所有的单词都修改成首字母大写的代码如下: ```python sentence = "Where there is a will there is a way" title_sentence = sentence.title() print(title_sentence) ``` 输出结果为: ``` Where There Is A Will There Is A Way ```

no pain, no gain, where there is a will, there is a way 如何用python算词的频次

要用 Python 算出一段文本中每个单词出现的频次,可以按照如下步骤进行: 1. 将文本中的所有单词提取出来,并且将它们全部转换成小写字母,方便后续统计。 可以使用正则表达式对文本进行分词,代码如下: ``` import re text = "No pain, no gain. Where there is a will, there is a way." words = re.findall(r'\b\w+\b', text.lower()) ``` 2. 遍历单词列表,统计每个单词出现的次数。 可以使用一个字典来存储每个单词出现的次数,代码如下: ``` word_freq = {} for word in words: if word in word_freq: word_freq[word] += 1 else: word_freq[word] = 1 ``` 3. 输出每个单词和它的频次。 可以使用一个循环遍历字典中的每一项,依次输出单词和它的频次,代码如下: ``` for word, freq in word_freq.items(): print(f"{word}: {freq}") ``` 完整代码如下: ``` import re text = "No pain, no gain. Where there is a will, there is a way." words = re.findall(r'\b\w+\b', text.lower()) word_freq = {} for word in words: if word in word_freq: word_freq[word] += 1 else: word_freq[word] = 1 for word, freq in word_freq.items(): print(f"{word}: {freq}") ```

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1444. Elephpotamus Time limit: 0.5 second Memory limit: 64 MB Harry Potter is taking an examination in Care for Magical Creatures. His task is to feed a dwarf elephpotamus. Harry remembers that elephpotamuses are very straightforward and imperturbable. In fact, they are so straightforward that always move along a straight line and they are so imperturbable that only move when attracted by something really tasty. In addition, if an elephpotamus stumbles into a chain of its own footprints, it falls into a stupor and refuses to go anywhere. According to Hagrid, elephpotamuses usually get back home moving along their footprints. This is why they never cross them, otherwise they may get lost. When an elephpotamus sees its footprints, it tries to remember in detail all its movements since leaving home (this is also the reason why they move along straight lines only, this way it is easier to memorize). Basing on this information, the animal calculates in which direction its burrow is situated, then turns and goes straight to it. It takes some (rather large) time for an elephpotamus to perform these calculations. And what some ignoramuses recognize as a stupor is in fact a demonstration of outstanding calculating abilities of this wonderful, though a bit slow-witted creature. Elephpotamuses' favorite dainty is elephant pumpkins, and some of such pumpkins grow on the lawn where Harry is to take his exam. At the start of the exam, Hagrid will drag the elephpotamus to one of the pumpkins. Having fed the animal with a pumpkin, Harry can direct it to any of the remaining pumpkins. In order to pass the exam, Harry must lead the elephpotamus so that it eats as many pumpkins as possible before it comes across its footprints. Input The first input line contains the number of pumpkins on the lawn N (3 ≤ N ≤ 30000). The pumpkins are numbered from 1 to N, the number one being assigned to the pumpkin to which the animal is brought at the start of the trial. In the next N lines, the coordinates of the pumpkins are given in the order corresponding to their numbers. All the coordinates are integers in the range from −1000 to 1000. It is guaranteed that there are no two pumpkins at the same location and there is no straight line passing through all the pumpkins. Output In the first line write the maximal number K of pumpkins that can be fed to the elephpotamus. In the next K lines, output the order in which the animal will eat them, giving one number in a line. The first number in this sequence must always be 1.写一段Java完成此目的

7-3 Score Processing 分数 10 作者 翁恺 单位 浙江大学 Write a program to process students score data. The input of your program has lines of text, in one of the two formats: Student's name and student id, as <student id>, <name>, and Score for one student of one course, as <student id>, <course name>, <marks>. Example of the two formats are: 3190101234, Zhang San 3190101111, Linear Algebra, 89.5 Comma is used as the seperator of each field, and will never be in any of the fields. Notice that there are more than one word for name of the person and name of the course. To make your code easier, the score can be treated as double. The number of the students and the number of the courses are not known at the beginning. The number of lines are not known at the beginning either. The lines of different format appear in no order. One student may not get enrolled in every course. Your program should read every line in and print out a table of summary in .csv format. The first line of the output is the table head, consists fields like this: student id, name, <course name 1>, <course name 2>, ..., average where the course names are all the courses read, in alphabet order. There should be one space after each comma. Then each line of the output is data for one student, in the ascended order of their student id, with score of each course, like: 3190101234, Zhang San, 85.0, , 89.5, , , 87.3 For the course that hasn't been enrolled, leave a blank before the comma, and should not get included in the average. The average has one decimal place. There should be one space after each comma. And the last line of the output is a summary line for average score of every course, like: , , 76.2, 87.4, , , 76.8 All the number output, including the averages have one decimal place. Input Format As described in the text above. Output Format As described in the text above. The standard output is generated by a program compiled by gcc, that the round of the first decimal place is in the "gcc way". Sample Input 3180111435, Operating System, 34.5 3180111430, Linear Algebra, 80 3180111435, Jessie Zhao 3180111430, Zhiwen Yang 3180111430, Computer Architecture, 46.5 3180111434, Linear Algebra, 61.5 3180111434, Anna Teng Sample Output student id, name, Computer Architecture, Linear Algebra, Operating System, average 3180111430, Zhiwen Yang, 46.5, 80.0, , 63.2 3180111434, Anna Teng, , 61.5, , 61.5 3180111435, Jessie Zhao, , , 34.5, 34.5 , , 46.5, 70.8, 34.

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