請(qǐng)問(wèn)這個(gè)運(yùn)行結(jié)果是對(duì)的嗎
請(qǐng)問(wèn)這個(gè)運(yùn)行結(jié)果是對(duì)的嗎_
請(qǐng)問(wèn)這個(gè)運(yùn)行結(jié)果是對(duì)的嗎_
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
for l in L:
? ? for s in S.copy():
? ? ? ? if l==s:
? ? ? ? ? ? S.remove(l)
? ? ? ? else:
? ? ? ? ? ? S.add(l)
print(S)
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set([1,?2,?3,?4,?5,?6,?7,?8,?9,?10,?11])
2021-07-23
不對(duì),嵌套for循環(huán),走外面的for循環(huán)取L值,里面的for循環(huán)取S值,值會(huì)重復(fù),例如外面取1,里面取3,1==3,就會(huì)執(zhí)行add方法,沒(méi)有實(shí)現(xiàn)你想要的。只要遍歷一遍就可以,判斷從L中取得值與S中的值一不一樣就可以了
2021-12-29
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
print(S)
for x in L:
? ? if(x in S):
? ? ? ? S.remove(x)
? ? else:
? ? ? ? S.add(x)
print(S)
? ? ? ??
2021-09-13
其實(shí)還可以這樣寫
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
s=list(S)
for a in L:
??? num=0
??? for b in s:
??????? if a==b:
??????????? num+=1
??? if num>=1:
??????? S.remove(a)
??? else:
??????? S.add(a)
print(S)
2021-07-29
l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
s = set([1, 3, 5, 7, 9, 11])
for ss in l:
??? if ss not in s:
??????? s.add(ss)
print s