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)
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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)