# Enter a code
L = []
for n in range(1, 101):
L.append(n*n)
print(L)
print(sum(L))
L = []
for n in range(1, 101):
L.append(n*n)
print(L)
print(sum(L))
2023-12-27
# Enter a code
s1 = set([1, 2, 3, 4, 5])
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
if not s1.isdisjoint(s2):
print(s1.intersection(s2))
s1 = set([1, 2, 3, 4, 5])
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
if not s1.isdisjoint(s2):
print(s1.intersection(s2))
2023-12-27
# Enter a code
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
for n in L:
if n not in S:
S.add(n)
else:
S.remove(n)
print(S)
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
for n in L:
if n not in S:
S.add(n)
else:
S.remove(n)
print(S)
2023-12-27
# Enter a code
L = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']
L.append('Zero')
L.append('Phoebe')
L.append('Gen')
L.sort()
print(L)
L = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']
L.append('Zero')
L.append('Phoebe')
L.append('Gen')
L.sort()
print(L)
2023-12-27
T = ('Alice', 'Bob', 'Candy', 'David', 'Ellena')
# 通過(guò)下標(biāo)的方式訪問(wèn)元素
print(T[0]) # ==> Alice
print(T[4]) # ==> Ellena
# 切片
print(T[1:3]) # ==> ('Bob', 'Candy')
candy錯(cuò)了吧，應(yīng)該是David吧
# 通過(guò)下標(biāo)的方式訪問(wèn)元素
print(T[0]) # ==> Alice
print(T[4]) # ==> Ellena
# 切片
print(T[1:3]) # ==> ('Bob', 'Candy')
candy錯(cuò)了吧，應(yīng)該是David吧
2023-12-25
num = 3.14 * 1.57
print (round (num , 2))
round 需要嵌套在print里面
print (round (num , 2))
round 需要嵌套在print里面
2023-12-19
最贊回答 / 翎棟
L = [75, 92, 59, 68, 99]def avg(l):? ? return sum(l) / len(l)def avg2(l):? ? sum1 = 0? ? for i in? l:? ? ? ? sum1 += i? ? return sum1 / len(l)average = avg2(L)print(average)
2023-12-19
最新回答 / 快樂(lè)雞哥
sum=0是給總成績(jī)一開(kāi)始賦0，根據(jù)循環(huán)，依次把L列表中的分?jǐn)?shù)加到sum中?？梢岳斫鉃?+75、0+75+92、...
2023-12-19
最新回答 / 還能學(xué)
for循環(huán)L1分別會(huì)是 [1,2,3]|[5,3,2]|[7,3,2]? ? ? ? ? ? L1[0]? ? ? ? ? ?0 1 2? 0 1 2? ?0 1 2? ? ? ? ? ?每次取L1的0號(hào)元素
2023-12-15
直接使用 get 方法如果找不到默認(rèn)返回 None 的特性，代碼如下。
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
names = ["Alice", "Bob", "Candy", "Mimi", "David"]
for name in names :
print(d.get(name))
d = {
'Alice': 45,
'Bob': 60,
'Candy': 75,
'David': 86,
'Ellena': 49
}
names = ["Alice", "Bob", "Candy", "Mimi", "David"]
for name in names :
print(d.get(name))
2023-12-13
最贊回答 / weixin_慕田峪3431918
python3之前的版本中整型除整型得到的依然是整型，即省略小數(shù)，之后的版本整型除整型得到的的是浮點(diǎn)數(shù)
2023-12-12
最新回答 / weixin_慕無(wú)忌2350060
?if not isinstance(x,int) or not isinstance(x,float):這里應(yīng)該用and，否則任何類(lèi)型進(jìn)去都是true
2023-12-12