Python 列表 删除元素

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!

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)

 

#!/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。

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]))

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)