最贊回答 / qq_慕函數(shù)1463992
s1 = set([1,10]) 2個(gè)值也就是 for a in s1:??循環(huán)2次了。一次if a in s2:? ? ? ? print(a)一次else:? ? ? ? print('none')
2023-02-06
若
a = 0
b = "0"
c = ""
那我們是不是就可以認(rèn)為a是整數(shù)，b是字符串，c是空值也就是none
a = 0
b = "0"
c = ""
那我們是不是就可以認(rèn)為a是整數(shù)，b是字符串，c是空值也就是none
2023-02-02
最贊回答 / weixin_慕少4178098
因?yàn)槟氵@個(gè)代碼多算了一個(gè)101*101，L=[]a=0while a<=100:? ? a=a+1? ? ?#####最后一次運(yùn)算a=101? ? L.append(a*a)? #####這里最后一次是101*101，再返回到while才停止循環(huán)print(sum(L))正確的是把a(bǔ)=a+1放到L.append(a*a)的下面：L=[]a=0while a<=100:? ? L.append(a*a)? #####這里是100*100????a=a+1? ?#####現(xiàn)在a=101，返回到wh...
2023-01-15
最新回答 / weixin_慕少4178098
我運(yùn)行了一下，沒(méi)啥大問(wèn)題，最后的輸出報(bào)錯(cuò)了，改成none就沒(méi)問(wèn)題了，能正常運(yùn)行
2023-01-14
def func(**kwargs):
s = 0
for i in n:
print('name:{},gander:{},age:{}'.format(n[s],g[s],a[s]))
s += 1
n = ['Alice','Bob','Candy']
g = ['F','M','F']
a = ['18','19','20']
func(names = n,gander = g,age = a)
s = 0
for i in n:
print('name:{},gander:{},age:{}'.format(n[s],g[s],a[s]))
s += 1
n = ['Alice','Bob','Candy']
g = ['F','M','F']
a = ['18','19','20']
func(names = n,gander = g,age = a)
2023-01-10
我的似乎區(qū)分python & python3兩個(gè)命令
2023-01-09
