A = ["aaa","ddd","ccc"]
B = ["aaa","ddd","eee"]
text = input("输入:")
if A[0] in text and A[1] in text:
print(A,B)
elif A[2] in text:
print(A)
elif B[2] in text:
print(B)
else:
print('not found')
#第二个也是一样的,都是对数组切片,然后与输入字串比较,符合要求就输出
A = ["123","124","125"]
B = ["123","124","126"]
text = input("输入:")
if A[0] in text and A[1] in text:
print(A,B)
elif A[2] in text:
print(A)
elif B[2] in text:
print(B)
else:
print('not found')
虽然实现的方式不优雅,但确实能够解决你的问题,代码如下:
def plastic(l):
l_sort = sorted(l)
result = list(range(len(l)))
for n,i in enumerate(l_sort,1):
result[l.index(i)] = n
return result
S=[[5,4,3,2,0],[6,5,4,0,1],[0,6,5,1,2],[1,7,6,0,3]]
S=list(map(plastic,S))
print(S)
输出:
[[5, 4, 3, 2, 1], [5, 4, 3, 1, 2], [1, 5, 4, 2, 3], [2, 5, 4, 1, 3]]
matrix = [[1, 2, 3, 4],[5, 6, 7, 8],[9, 10, 11, 12]]# 矩阵转置
# 矩阵的列数
colomn = len(matrix[0])
# 转置矩阵的行数,设置空矩阵[[], [], [], []]
transformMatrix = [[] for i in range(colomn)]
for ele in matrix:
for i in range(colomn):
# transformMatrix[i]标识新矩阵的第i行
# ele[i]标识原有矩阵的第i列
transformMatrix[i].append(ele[i])
print transformMatrix