d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
for key in d:
print(d.get(key))
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
for key in d:
print(d.get(key))
2022-09-14
L=[[1,2,3],[5,3,2],[7,3,2]]
for c in L:
l=c[0]
w=c[1]
h=c[2]
V=l*w*h
print(V)
for c in L:
l=c[0]
w=c[1]
h=c[2]
V=l*w*h
print(V)
2022-09-12
L=['Alice', 'Bob', 'Candy', 'David', 'Ellena']
L[0]='Ellena'
L[1]='Alice'
L[2]='Candy'
L[3]='David'
L[4]='Bob'
print(L)
L[0]='Ellena'
L[1]='Alice'
L[2]='Candy'
L[3]='David'
L[4]='Bob'
print(L)
2022-09-12
L = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']
L.pop(2)
L.pop(2)
print(L)
L.pop(2)
L.pop(2)
print(L)
2022-09-12
names=['Alice', 'Bob', 'Candy', 'David', 'Ellena']
names.insert(5,'Gen')
names.append('Phonebe')
names.append('Zero')
print(names)
names.insert(5,'Gen')
names.append('Phonebe')
names.append('Zero')
print(names)
2022-09-12
L = ['Alice', 66, 'Bob', True, 'False', 100]
n=0
for a in L:
n=n+1
if n%2!=0:
print(a)
n=0
for a in L:
n=n+1
if n%2!=0:
print(a)
2022-09-11
s1='ABC'
s2='123'
s3='xyz'
for a in s1:
for b in s2:
for c in s3:
print(a+b+c)
s2='123'
s3='xyz'
for a in s1:
for b in s2:
for c in s3:
print(a+b+c)
2022-09-11
num=2
sum=0
while num<=1000:
sum=sum+num
num=num+2
continue
print(sum)
sum=0
while num<=1000:
sum=sum+num
num=num+2
continue
print(sum)
2022-09-10
>>> 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
2022-09-10
>>> num=1
>>> sum=1
>>> while num<=10:
... sum=sum*num
... num=num+1
...
>>> print(sum)
3628800
>>> sum=1
>>> while num<=10:
... sum=sum*num
... num=num+1
...
>>> print(sum)
3628800
2022-09-10
最新回答 / qq_慕仙1244835
d.pop('Alice')是已经对dict中的 'Alice'进行删除，而将删除的结果赋值到alice_score，只是为了让人清楚了解删除key后所返回对应的值
2022-09-07
最新回答 / weixin_宝慕林2426525
num = int(input())意思是变量num赋值，输入的信息转化为int数据类型num = input(int())意思是变量num赋值，输入的数据限定为int数据类型，但输入数据不可控所以错误
2022-09-06
最新回答 / 精慕门0234217
T = ((1+2), ((1+2),), ('a'+'b'), (1, ), (1,2,3,4,5))eprint T[2][0]print T[2]<...图片...>当你以T[2][0]的二维方式输出时是将('a'+'b')中的每个部分看作独立的个体但是以T[2]的整体方式输出时就会直接进行运算后输出
2022-09-06