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