d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
if 'Alice' in d:
score = d['Alice']
print(score)
d['Alice'] = 60
print(d)
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
if 'Alice' in d:
score = d['Alice']
print(score)
d['Alice'] = 60
print(d)
2023-06-24
ju = [[1,2,3],[5,3,2],[7,3,2]]
print(ju)
a = 0
for a in range(len(ju)):
print(ju[a][0]*ju[a][1]*ju[a][2])
a += 1
print(ju)
a = 0
for a in range(len(ju)):
print(ju[a][0]*ju[a][1]*ju[a][2])
a += 1
2023-06-23
最赞回答 / weixin_慕工程7111902
Computer regarded your『('17')』 of 『age=('17')』 as a 'string',not a 'number'.So it's working like down hereelse: print('adult')If you change your code from____________________________________age=('17') &------------------------...
2023-06-13
set提供isdisjoint()方法，可以快速判断两个集合是否有重合，如果有重合，返回False。##竟然记反了
2023-06-12
有一种方法可以通过key来获取对应的value，这种方法不会引起错误，dict本身提供get方法，把key当作参数传递给get方法，就可以获取对应的value，当key不存在时，也不会报错，而是返回None。
2023-06-11
已采纳回答 / 慕工程9338430
这是个死循环当num为奇数时，跳过了，下面的都不会执行，直接到下次而下一次还是执行奇数，就会这样一直循环，所以运行不出来试试这个num = 1result = 0while num <= 1000: if num % 2 == 1: num += 1 continue result += num num += 1print(result)
2023-06-06
# Enter a code
L = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']
n1=0
for name in L:
if L[n1]=='Candy':
L.pop(n1)
else:
n1+=1
n2=0
for name in L:
if L[n2]=='David':
L.pop(n2)
else:
n2+=1
print(L)
L = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']
n1=0
for name in L:
if L[n1]=='Candy':
L.pop(n1)
else:
n1+=1
n2=0
for name in L:
if L[n2]=='David':
L.pop(n2)
else:
n2+=1
print(L)
2023-06-05
num = 0
sum = 0
while True:
num += 1
if num % 2 != 0:
continue
if num > 1000:
break
sum = sum + num
print(sum)
sum = 0
while True:
num += 1
if num % 2 != 0:
continue
if num > 1000:
break
sum = sum + num
print(sum)
2023-05-24
L = ['Alice', 66, 'Bob', True, 'False', 100]
for n in range(len(L)):
if (n+1) % 2 == 0:
print (L[n])
for n in range(len(L)):
if (n+1) % 2 == 0:
print (L[n])
2023-05-24