删除列表［5,7,21,3,6,0,11,26,2,1,33,9］中第二个大于15的数。谢谢！
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- 兰振lanzhen 2021-06-19 17:39已采纳
1、先对列表排序；
2、循环列表，判断数值大小后删除；
ls =[5,7,21,3,6,0,11,26,2,1,33,9] ls.sort() n = 0 for i in ls: if i>15 : n+=1 if n==2: print(i) ls.remove(i)点赞 打赏 评论分享 - Roc-xb 2021-06-19 17:36
#!/usr/bin/python # -*- coding: UTF-8 -*- """ @author: YangPC @time:2021/06/19 @QQ:327844761 @微信公众号：ewbang """ if __name__ == '__main__': ls = [5, 7, 21, 3, 6, 0, 11, 26, 2, 1, 33, 9] # 先排序 ls.sort() print(ls) for index, item in enumerate(ls): if item > 15 and item != ls[-1]: ls.remove(ls[index + 1]) break print(ls)程序输出截图：如有帮助，望采纳。点击我回答右上角【采纳】按钮。
点赞 打赏 评论分享 - Keep Doing this 2021-06-19 17:43
a=[5,7,21,3,6,0,11,26,2,1,33,9]
m=0
t=0
for i in range(0,len(a)):
j=i
for j in range(0,len(a)):
if a[i]<a[j]:
m=a[i];
a[i]=a[j]
a[j]=m
for i in range(0,len(a)):
if a[i]>15:
t+=1
if t==2:
print("大于15的第二个数为:{}".format(a[i]))点赞 打赏 评论分享 - Keep Doing this 2021-06-19 17:51
a=[5,7,21,3,6,0,11,26,2,1,33,9] t=0 for i in range(0,len(a)): j=i for j in range(0,len(a)): if a[i]<a[j]: m=a[i]; a[i]=a[j] a[j]=m for i in range(0,len(a)): if a[i]>15: t+=1 if t==2: print("大于15的第二个数为:{}".format(a[i])) k=i a.remove(a[k]) print(a)点赞 打赏 评论分享