最新回答 / 努力小废尘
肯定不一样啦，下面的程序数值会更大些。他在数值原来的基础上先加一了，在进行sum值的运算；而上面的程序先进行sum值的运算，在进行原先数值的增大，所以两边的数值运算结果不一样
2021-09-16
# Enter a code
t=(100, 69, 29, 100, 72, 99, 98, 100, 75, 100, 100, 42, 88, 100)
a=t.count(100)
print(a)
t=(100, 69, 29, 100, 72, 99, 98, 100, 75, 100, 100, 42, 88, 100)
a=t.count(100)
print(a)
2021-09-16
# Enter a code
s1 = set([1, 2, 3, 4, 6, 8, 10])
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
flag = s1.isdisjoint(s2)
if not flag:
for item in s1:
if item not in s2:
continue
print(item)
s1 = set([1, 2, 3, 4, 6, 8, 10])
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
flag = s1.isdisjoint(s2)
if not flag:
for item in s1:
if item not in s2:
continue
print(item)
2021-09-16
# Enter a code
s1 = set([1, 2, 3, 4, 6, 8, 10])
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
flag = s1.isdisjoint(s2)
if not flag:
for item in s1:
if item not in s2:
continue
print(item)
s1 = set([1, 2, 3, 4, 6, 8, 10])
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
flag = s1.isdisjoint(s2)
if not flag:
for item in s1:
if item not in s2:
continue
print(item)
2021-09-16
dist = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
name = '孙雯'
if name in dist.keys():
dist.pop(name)
else:
print('{} not in dist'.format(name))
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
name = '孙雯'
if name in dist.keys():
dist.pop(name)
else:
print('{} not in dist'.format(name))
2021-09-16
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
for name in d:
print(d.get(name))
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
for name in d:
print(d.get(name))
2021-09-16
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49,
'Gaven':86
}
print(d)
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49,
'Gaven':86
}
print(d)
2021-09-16
# Enter a code
s1='ABC'
s2='123'
s3='xyz'
for x in s1:
for y in s2:
for z in s3:
print(x+y+z)
s1='ABC'
s2='123'
s3='xyz'
for x in s1:
for y in s2:
for z in s3:
print(x+y+z)
2021-09-16
# Enter a code
T=(100, 69, 29, 100, 72, 99, 98, 100, 75, 100, 100, 42, 88, 100)
print(T.count(100))
T=(100, 69, 29, 100, 72, 99, 98, 100, 75, 100, 100, 42, 88, 100)
print(T.count(100))
2021-09-16
# Enter a code
a=1
b=1
while a>0 and a<=10:
b=b*a
a=a+1
print(b)
a=1
b=1
while a>0 and a<=10:
b=b*a
a=a+1
print(b)
2021-09-15
最新回答 / qq_棠梨煎雪故人来_03907185
你这应该是从c啊之类的语言里面学的习惯吧，当你for i in L的时候，它迭代的不是下标1、2、3，它迭代的是里面的元素，在这句里迭代的是[1,2,3]和后面的两个列表。所以你这句话按机器理解的是area*=L[[1,2,3][5,3,2]]，很明显，这是错的。所以这个错误报告给你说list的下标应该是整数或者是划分，不能是个列表。想达到你想要的效果，就for i in range(len(L)吧，但就算是按下标算，你思路也是错的。
2021-09-15
最新回答 / 鹏鹏1209393
template='{} {} {} ，{} {} {}.'
l='life'
i='is'
s='short'
y='you'
n='need'
p='python'
result=template.format(l,i,s,y,n,p)
2021-09-15
最新回答 / weixin_慕容1556897
有关系！例如： “{0}{2}{1}".format(100,200,300) 输出结果为：100 300 200 “{0}{1}{2}".format(100,200,300) 输出结果为：100 200 300 “{ }{ }{ }".format(100,200,300) 输出结果为：100 200 300 默认顺序输出
2021-09-15
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)
2021-09-15