如何在给定矩阵中找到所有子矩阵值都大于阈值(不使用numpy和Scipy)

Question

我正在研究微控制器，它不支持numpy或scipy。我想提取矩阵中所有大于给定阈值的子矩阵。

myMtarix = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 204, 0, 0, 0, 0, 0, 0, 60, 0, 0, 284, 0],
[0, 100, 0, 74, 421, 157, 0, 0, 0, 0, 317, 364, 736, 245, 1470, 0],
[0, 717, 0, 258, 879, 496, 0, 0, 0, 0, 0, 671, 766, 725, 1429, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

myMatrix=6*16矩阵，我想提取

subMatrix1 = [[100],
[717]]
subMatrix2 = [[0, 204, 0],
[74, 421, 157],
[258, 879, 496]]
subMatrix3 = [[0, 60, 0, 0, 284],
[317, 364, 736, 245, 1470],
[0,671, 766, 725, 1429]]

阈值=10

我试过这种

这里，收集值>阈值及其指标

threshold = 10
pressureIndexes = []
pressurePoints = []
reqValue = []
reqIndex = []

reqValue = [myMatrix[i][j] for i in range(numRows) for j in range(numCols) if(myMatrix[i][j] > threshold)]
reqIndex = [[i,j] for i in range(numRows) for j in range(numCols) if(myMatrix[i][j] > threshold)]

从这里，我试图提取子矩阵的精确边界

Xend = Xstart = reqIndex[0][0]
Yend = Ystart = reqIndex[0][1]
dummy = []  
for i in range(1,len(reqIndex)):
PXstart = Xstart
PXend  = Xend
PYstart = Ystart
PYend  = Yend
Xstart = min(Xstart,reqIndex[i][0])
Xend = max(Xend,reqIndex[i][0])
Ystart = min(PYstart,reqIndex[i][1])
Yend = max(PYend,reqIndex[i][1])
if(abs(PXend-Xend) > 2 or abs(PYend-Yend) > 2):
dummy.append([[PXstart,PXend],[PYstart,PYend]])
Xend = Xstart = reqIndex[i][0]
Yend = Ystart = reqIndex[i][1]
dummy.append([[Xstart,Xend],[Ystart,Yend]])
print()

for i in dummy:
print(i)
print('---------------------------------------------------------')
sleep(1)

输出：

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 204, 0, 0, 0, 0, 0, 0, 60, 0, 0, 284, 0]
[0, 100, 0, 74, 421, 157, 0, 0, 0, 0, 317, 364, 736, 245, 1470, 0]
[0, 717, 0, 258, 879, 496, 0, 0, 0, 0, 0, 671, 766, 725, 1429, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]


[[1, 1], [4, 4]]  # [[rowStarting,rowEnding],[colStarting,colEnding]] for value 204
[[1, 1], [11, 11]]
[[1, 3], [1, 14]]

Answer 1

这里有一种方法，它将返回一个没有0行或列的子矩阵列表：

mm       = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 204, 0, 0, 0, 0, 0, 0, 60, 0, 0, 284, 0],
[0, 100, 0, 74, 421, 157, 0, 0, 0, 0, 317, 364, 736, 245, 1470, 0],
[0, 717, 0, 258, 879, 496, 0, 0, 0, 0, 0, 671, 766, 725, 1429, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]


## values below threshold are set to 0
def threshold(m, th=10):
for l in range(len(m)):
for c in range(len(m[l])):
v=m[l][c]
if v < th:
v = 0
m[l][c] = v
return m

## helper functions to handle matrixes
## work with a colum values
def column(m, c):
return list(map(lambda x: x[c], m))

## cut and drop left of matrix
def cutleftofcolumn(m, c):
return list(map(lambda x: x[c+1:], m))

## cut and drop right of matrix
def cutrightofcolumn(m, c):
return list(map(lambda x: x[:c], m))

## remove 0 borders of the matrix
def removeBorders(m: list) -> list:
test = True
while test:
test = False
#if sum(m[0])==0:
if all(map(lambda x: x==0, m[0])):
m = m[1:]
test = True
#if sum(m[-1])==0:
if all(map(lambda x: x==0, m[-1])):
m = m[:-1]
test = True
#if sum(column(m, 0))==0:
if all(map(lambda x: x==0, column(m, 0))):
m=cutleftofcolumn(m, 0)
test = True
#if sum(column(m, -1))==0:
if all(map(lambda x: x==0, column(m, -1))):
m=cutrightofcolumn(m, -1)
test = True
return m

## split on a 0 column and returns 2 matrices
def splitv(m, c):
a=cutleftofcolumn(m, c)
b=cutrightofcolumn(m, c)
return [a, b]

## split on a 0 line and returns 2 matrices
def splith(m, l):
return [m[0:l], m[l+1:]]

mm = threshold(mm, 10)
listMatrices = [mm]

test = True
while (test):
test=False
index = 0
while (len (listMatrices)>index):
## get first matrix of the list
m = listMatrices.pop(0)
## removes 0 borders of the matrix
m = removeBorders(m)

lll=[]    
## try to find an horizontal split on 0 line
if lll==[]:
for l in range(len(m)):
#if sum(m[l])==0:
if all(map(lambda x: x==0, m[l])):
lll = lll + splith(m, l)
test = True
break

## try to find a vertical split on 0 column
if lll==[]:
for c in range(len(m[0])):
cc = column(m, c)
#if sum(cc)==0:
if all(map(lambda x: x==0, cc)):
lll = lll + splitv(m, c)
#print("Csum=0", c)
test = True
break

if lll==[]:
listMatrices.append(m)
else:
for l in lll:
listMatrices.append(l)
index+=1

print(listMatrices)

输出为：

[[[0, 60, 0, 0, 284], [317, 364, 736, 245, 1470], [0, 671, 766, 725, 1429]],
[[0, 204, 0], [74, 421, 157], [258, 879, 496]],
[[100], [717]]]