str内多少个,判断是否空格,方法删除改变本身,赋值给其他本身不改变,斐波那契时间n空间1,简单递归

# def countnum(a):
# di_num = 0
# ph_num =0
# sp_num = 0
# el_num = 0
# for i in a:
# if i.isdigit():
# di_num +=1
# elif i.isalpha():
# ph_num +=1
# elif i.isspace():
# sp_num +=1
# else:
# el_num +=1
# return (di_num,ph_num,sp_num,el_num)
# ret = countnum("456465hhhgggg $%^&")
# print(ret)
# def islensize(args):
# print(args)
# if len(args) > 5:
# return True
# else:
# return False
# print(islensize(("456122",45,45,12,12,12)))
# def isnone(args):
# for i in args:
# if i is "":
# return True
# print("ok")
# print(isnone([11,22,"",444,"",4445,]))
# def has_space(arg):
# for i in arg:
# if i.isspace():
# return True
# print("ok")
# return False
"""
ret = False
for i in arg:
if i.isspace():
ret = Ture
break#注意提前终止不必要的循环
return ret
"""
# print(has_space("456"))
# print(isnone([11,22,"",6]))
# def fun5(args):
# if len(args) >2 :
# result = args[0:2]
# return result
# else:
# return args
# print(fun5([22,44,55,11,66]))
# def ff(args):#删除偶数会改变初始列表不可用
# result = []
# for i in range(0,len(args)):
# if i %2 ==1:
# result.append(args[i])
# return result
# print(ff([11,22,44,66,77,77]))
# def ff_dict(dic):
# for k,v in dic.items():
# if len(v) > 2:
# dic[k] = v[0:2]
# return dic#可以不return,因为dic自身已经改变,省内存
# print(ff_dict({"k1":"fdgfdfg","k2":[11,22,33],"k3":[11,33]}))
#1 1 2 3 5 8 13
def fib(n):
first = 0
second = 1
i=1
while i<n:
third = first+second
first = second
second = third
i +=1
return third
print(fib(10))
#递归,斐波那契数列
def fib(n,depth,a1,a2):
if depth == n:
return a1
a3 = a1 + a2
r = fib(n,depth+1,a2,a3)#注意需要接收返回值
return r
ret = fib(10,1,0,1)
print(ret)

def q(arg):
#isinstance 判断是否是字符串,列表,元组...
if isinstance(arg,str) or isinstance(arg,list) or isinstance(arg,tuple):
print(arg)
return None
#函数传参传的是引用
def w(a):#例,删除,的是li自己变了.赋值不变是自己生成一个
del a[2:]
a = 123
li = [11,22,33,44]
w(li)
print(li)
原文地址:https://www.cnblogs.com/currynashinians000/p/8605517.html