think python 第8章字符串

8.1字符串是一个序列

8.2len

8.3traveral with a for loop

1.通过while循环遍历字符串

index = 0 f = 'fruit' while index < len(f): letter = f[index] print(letter) index += 1

2.通过for循环遍历

f = 'fruit' for i in f: print(i)

8.4string slices

数组[a:b] 左闭右开

8.5 strings are immutable

字符串是不可变的，即不可以修改一个已经存在的字符。如果我们想把‘hello，world’改变为‘hello，python’

string1 = 'hello,world' string2 = string1[:6] + 'python' print(string2)

8.6searching

def find(word,letter): index = 0 　　# 读取字符串单字符的下标 while index < len(word) if word[index] == letter return index index += 1 return -1

8.7looping and counting

word = 'banana' count = 0 for letter in word: if letter == "a": count += 1 print(count)

封装起来：

def count(word,leter): number = 0 for i in word: if i == letter: number += 1 print(number)

也可以这样来封装（其实是一样的）

word = input('the word is:') letter = input("the letter is:") def count(word,letter): number = 0 for i in word: if i == letter: number += 1 return number print(count(word,letter))

8.8string methods

语法(函数) == 方法.语法()

8.9in运算符

8.10string comparison

字符串的‘大小’比较：

（1）大写祖母小于小写字母；

（2）按顺序每个字母进行比较，一旦不一致，只需比较这个即可，后面的不用理会

（3）banana < bananas

8.11debugging

修改后的代码如下：

def is_reverse(word1,word2): if len(word1) != len(word2): return False else: i = 0 j = len(word2) -1 while j >= 0: if word1[i] != word2[j-1]: return False i += 1 j -= 1 return True

8.13exercises

例题8.10 首先用第6章的方法实现回文单词

def first(word): return word[0] def last(word): return word[-1] def middle(word): return word[1:-1] def is_palindrome(word): if len(word) <= 1: return True if first(word) != last(word): return False return is_palindrome(middle(word)) print is_palindrome('allen')#语法错误？ print is_palindrome('bob') print is_palindrome('otto') print is_palindrome('a') print is_palindrome('')

还可以这样来实现回文单词的判定：

def is_palindrome(word): if word[::-1] == word: return True else: return False

 

转载于:https://www.cnblogs.com/Kingwjk/p/7886647.html