names = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']
name_set = set(names)
for x in name_set:
if 'bob' in x.lower():
print('True')
name_set = set(names)
for x in name_set:
if 'bob' in x.lower():
print('True')
2023-01-03
# Enter a code
d = {'Alice': [50, 61, 66], 'Bob': [80, 61, 66], 'Candy': [88, 75, 90]}
for key,value in d.items():
for x in value:
print(key,x)
d = {'Alice': [50, 61, 66], 'Bob': [80, 61, 66], 'Candy': [88, 75, 90]}
for key,value in d.items():
for x in value:
print(key,x)
2023-01-03
# Enter a code
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
names = d.keys()
for x in names:
if 'Alice' in d:
d.pop('Alice')
print(d)
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
names = d.keys()
for x in names:
if 'Alice' in d:
d.pop('Alice')
print(d)
2023-01-03
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
print('old-->{}'.format(d))
for x in d:
if 'Alice' in d:
d[x]=60
print('new-->{}'.format(d))
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
print('old-->{}'.format(d))
for x in d:
if 'Alice' in d:
d[x]=60
print('new-->{}'.format(d))
2023-01-03
# Enter a code
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
L = ['Alice', 'Bob', 'Candy', 'Mimi', 'David']
for name in L:
if name in d:
print(name,d.get(name))
else:
print('None')
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
L = ['Alice', 'Bob', 'Candy', 'Mimi', 'David']
for name in L:
if name in d:
print(name,d.get(name))
else:
print('None')
2023-01-03
T = ((1+2), ((1+2),), ('a'+'b'), (1, ), (1,2,3,4,5))
s = 0
for i in T:
if(type(i) == tuple):
s += 1
print(s)
s = 0
for i in T:
if(type(i) == tuple):
s += 1
print(s)
2023-01-01
最新回答 / weixin_慕村9391395
a = 'python'print('hello,', a or 'world')因为Python把0、空字符串和None看成False。所以a 是true。在计算a or b时，如果 a 是 True，则根据或运算法则，整个计算结果必定为 True，因此返回 a结果：
hello, python...
2022-12-31
最新回答 / 慕用4035794
and 表示与逻辑运算，这种情况下，第一个操作数是True的话，那就看输出结果就取决于第二个数了。这是为什么呢？因为第一个数已经是True了，and 与的运算结果取决于第二个数的真假。
2022-12-28
S=(100, 69, 29, 100, 72, 99, 98, 100, 75, 100, 100, 42, 88, 100)
print(S.count(100))
print(S.count(100))
2022-12-26
L=[[1,2,3],[5,3,2],[7,3,2]]
for l in L:
a=l[0]
b=l[1]
c=l[2]
S=(a*b+a*c+b*c)*2
print(S)
for l in L:
a=l[0]
b=l[1]
c=l[2]
S=(a*b+a*c+b*c)*2
print(S)
2022-12-23