python学习(13)
random.uniform(a,b) 随机生成a,b之间的一个浮点数
random.uniform(1,20) 1.0130916166719703
习题1：生成[“z1”,”y2”,”x3”,”w4”,”v5”]
#coding=utf-8
result = []
for i in range(1,6):
result.append(chr(122-i+1)+str(i))
print(result) #coding=utf-8 result = [] for i in range(1,6): result.append(chr(97+i-1)+str(i)) print(result)
习题2：拼接一个字符串的首字母、末尾字母、中间字母为一个字符串 要考虑奇数偶数长度，如果只有1个2个字母的字符串
方式1：直接拼字符串
s = "abcdefghig"
result_str = ""
result_str += s[0]
if len(s)%2 == 1:
middle_letter = s[len(s)//2]
else:
middle_letter = s[len(s)//2-1]
middle_letter += s[len(s)//2]
print(middle_letter)
result_str += middle_letter
result_str += s[-1]
print("拼接后的字符串: ",result_str)方式2：利用列表
def join_str(s):
result = []
result.append(s[0])
if len(s)%2==1:
result.append(s[len(s)//2])
else:
result.append(s[len(s)//2-1])
result.append(s[len(s)//2])
result.append(s[-1])
return "".join(result)
s = "abcdefghig"
s2 = "abcdefghi"
print("拼接后的字符串: ",join_str(s))
print("拼接后的字符串: ",join_st切片没有越界
s = "a" s[5:] #没有越界 '' s[5] Traceback (most recent call last): File "<stdin>", line 1, in <module> IndexError: string index out of range a = [1] a[0:1] = [1,2,3,4] a [1, 2, 3, 4] a =[1] a[0:2] = [1,2,3,4] a [1, 2, 3, 4] a = [1] a[0] 1 a[0] = [1,2,3,4] a [[1, 2, 3, 4]] a = [1] a[0:7] [1] a[0:7] = [1,2,3,4] a [1, 2, 3, 4] a [1, 2, 3, 4] a[0:2] = [7,8,9,10] a [7, 8, 9, 10, 3, 4] a[0] = [55,66] a [[55, 66], 8, 9, 10, 3, 4] a[0:5] = [1] a [1, 4]
习题3：S = “i am, a boy ”把boy替换为m 利用切片赋值特性
算法：先把字符串转换为列表 遍历列表，找到boy字符串 替换boy字符串为m
#coding=utf-8
s = "i am, a boy!!"
list_s = list(s)
sub_length = len("boy")
for i in range(len(list_s)):
if "".join(list_s[i:i+sub_length]) == "boy":
list_s[i:i+sub_length] = "m"#可以用切片直接替换
print("".join(list_s)) 习题4：非递归实现生成斐波那契数列
#coding=utf-8
def fib(n):
result = []
if n <=0 or not isinstance(n,int):
return result
for i in range(n):
if i < 2:
result.append(1)
else:
result.append(result[i-1] + result[i-2])
return result
print("斐波那契数列: ",fib(5))
print("斐波那契数列: ",fib(-2))习题5：非递归实现求第n项斐波那契数列值 算法：遍历n，如果n等于0或1直接返回1 如果n >=2,记录前面两个项的和，并重新记录新的第1个，第2个数；
#coding=utf-8
def fib(n):
a,b=1,1
sum = 0
if n <=0 or not isinstance(n,int):
return -1
for i in range(n):
if i < 2:
sum = 1
else:
sum = a + b
a = b
b = sum
return sum
print(fib(5))习题6：递归求两个数的最大公约数 方式1：
算法：
找出两个数的较小数（第一次） 如果两个数除以较小数余数都等于0的话，添加到一个列表中 不然的话继续调用函数自身，并且num-1传入到num参数中。
def max_common_divisor(a,b,num=0,result=[]):
if num == 0:
if a < b:
num = a
else:
num = b
if a%num == 0 and b%num == 0:
print(result)
result.append(num)
else:
max_common_divisor(a,b,num-1)
return result#函数递归调用结束后才会执行此代码
print(max_common_divisor(24,120)) 方式2：更相减损法
def get(small,big):
if small > big:
small,big = big,small
if small == big:
return small
return get(small,big-small)
print(get(30,24))方式3:辗转相除法
def gain(small, big):
if small > big:
small, big = big, small
if small == 0:
return big
return gain(small, big % small)习题7：利用递归，处理嵌套列表，生成列表[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] 定义全局变量：（尽量不使用全局变量）
l = [1,2,[3,4,[5,6,7,[8,9,[10,11]]]]]
result = []
def func(p):
global result
for v in p:
if not isinstance(v,list):
result.append(v)
else:
func(v)
return result
print(func(l))引用传参：
def get_list(l,result=[]):
for v in l:
if not isinstance(v,list):
result.append(v)
else:
get_list(v)
return result
l = [1,2,[3,4,[5,6,7,[8,9,[10,11]]]]]
print(get_list(l))