最赞回答 / 慕神7009732
s1 = set([1, 2, 3, 4, 5])s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])s3=set()s4=set()print(s1.isdisjoint(s2))a=Falseif s1.isdisjoint(s2)==a: s3=s2-s1 L=list(s3) T=tuple(s3) print(L,T)else: print(s4)a=False 不要带引号，带了引号是表示字符串，不是布尔类型，所以不成立
2022-02-10
最赞回答 / 慕村3371467
答案说的3个是“((1+2),)，(1, )，（1,2,3,4,5）”第一跟第三个是因为没有在tuple里面加逗号（,），所以不算是tuple。
2022-02-09
def square_of_sum(L):
L1=[]
for i in L:
L1.extend([i*i])
return sum(L1)
print(square_of_sum([1, 2, 3, 4, 5]))
L1=[]
for i in L:
L1.extend([i*i])
return sum(L1)
print(square_of_sum([1, 2, 3, 4, 5]))
2022-02-09
names = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']
new_names=[]
for i in names:
name_lower=i.lower()
new_names.append(name_lower)
print(new_names)
new_names=[]
for i in names:
name_lower=i.lower()
new_names.append(name_lower)
print(new_names)
2022-02-09
d = dict()
d['Alice'] = [50, 61, 66]
d['Bob'] = [80, 61, 66]
d['Candy'] = [88, 75, 90]
d['Alice'] = [50, 61, 66]
d['Bob'] = [80, 61, 66]
d['Candy'] = [88, 75, 90]
2022-02-09
num = 2
sum = 0
while True:
if num > 1000:
break
sum = sum + num
num = num + 2
print(sum) # ==> 250500
sum = 0
while True:
if num > 1000:
break
sum = sum + num
num = num + 2
print(sum) # ==> 250500
2022-02-08
# coding: utf-8
s1 = '这是一句中英文混合的{}字符串：{}'
python = 'Python'
hello = 'Hello World!'
result = s1.format(python , hello)
print(result)
s1 = '这是一句中英文混合的{}字符串：{}'
python = 'Python'
hello = 'Hello World!'
result = s1.format(python , hello)
print(result)
2022-02-07
t = (100, 69, 29, 100, 72, 99, 98, 100, 75, 100, 100,42, 88, 100)
print(t.count(100))
print(t.count(100))
2022-02-07
已采纳回答 / 慕UI1212408
你这个是缩进的问题还有continue下面是不运行代码的num=0L = ['Alice', 66, 'Bob', True, 'False', 100]for a in L: num=num+1 if num%2!=0: continue print(a)下载视频
2022-02-04