def all_sub(r, c, mat): # returns all sub matrices of order r * c in mat
arr_of_subs = []
if (r == len(mat)) and (c == len(mat[0])):
arr_of_subs.append(mat)
return arr_of_subs
for i in range(len(mat) - r + 1):
for j in range(len(mat[0]) - c + 1):
temp_mat = []
for ki in range(i, r + i):
temp_row = []
for kj in range(j, c + j):
temp_row.append(mat[ki][kj])
temp_mat.append(temp_row)
arr_of_subs.append(temp_mat)
return arr_of_subs
def get_all_sub_mat(mat):
rows = len(mat)
cols = len(mat[0])
def ContinSubSeq(lst):
size=len(lst)
for start in range(size):
for end in range(start+1,size+1):
yield (start,end)
for start_row,end_row in ContinSubSeq(list(range(rows))):
for start_col,end_col in ContinSubSeq(list(range(cols))):
yield [i[start_col:end_col] for i in mat[start_row:end_row] ]
def extract(mat, n, n1, m, m1):
l=[]
for i in range(n-1, n1):
r=[]
for j in range(m-1, m1):
if mat[i][j] != []:
r.append(mat[i][j])
l.append(r)
return l
# set 1 in i1 and j1
# set dimension+1 in i2 and j2
res = []
for i1 in range(1, 3):
for i2 in range(1,3):
for j1 in range(1, 3):
for j2 in range(1, 3):
li= extract(mat, i1,i2,j1,j2)
if li !=[] and i2 >= i1 and j2>=j1 :
res.append(li)
print res
假设矩阵
分为3个功能:
^{pr2}$运行这个
或者更简单,放在一个函数中:
我做了一个函数,允许从矩阵中提取矩阵,我用它来提取所有可能的组合,你会找到脚本, 这个脚本解决了你的问题
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