已采纳回答 / 慕码人0617
布尔值是逻辑判断时用的and：且的关系，a and b，及a和b的都为真时，才为真or：或的关系，a or b，只要有一个为真就返回真，都假才为假not：相当于取反， not a，若a为真，则 not a就是假，反之亦然
2021-05-26
d = {
'Alice': [45],
'Bob': [60],
'Candy': [75],
}
d['Alice']=[50, 61, 66]
d['Bob']=[80, 61, 66]
d['Candy']=[88, 75, 90]
for d1 in d:
name=d1
score=d.get(d1)
print('name={},score={}'.format(name,score))
'Alice': [45],
'Bob': [60],
'Candy': [75],
}
d['Alice']=[50, 61, 66]
d['Bob']=[80, 61, 66]
d['Candy']=[88, 75, 90]
for d1 in d:
name=d1
score=d.get(d1)
print('name={},score={}'.format(name,score))
2021-05-25
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
for d1 in d:
print(d.get(d1))
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
for d1 in d:
print(d.get(d1))
2021-05-25
a =
if a <= 3:
print("baby")
else:
if a <=6:
print("kid")
else:
if a <=18:
print("teenager")
elif a >18:
print("adult")
if a <= 3:
print("baby")
else:
if a <=6:
print("kid")
else:
if a <=18:
print("teenager")
elif a >18:
print("adult")
2021-05-20
答案是错的
函数定义的正确计算方式：
def sub_sum(L):
sum1 = 0
sum2 = 0
for item in L:
if item % 2 == 0:
sum1 += item
else:
sum2 += item
return sum1, sum2
函数定义的正确计算方式：
def sub_sum(L):
sum1 = 0
sum2 = 0
for item in L:
if item % 2 == 0:
sum1 += item
else:
sum2 += item
return sum1, sum2
2021-05-20
