Write through the wall cracks, Transformation array and list of encountered at least through the wall

Entitled:

Problem-solving ideas:

　　　The new array is composed of: a new array for each column and the columns are the original array corresponding to the front row and the sum [1,2,3] ---> [1,3,6].

　　　Each column is a crack in the wall, the more repeats of each column of numbers, the fewer wall wear!

Code:

Import numpy AS NP 
B = np.zeros (Shape = (l, 5), DTYPE = int) 
NP = np.random.randint (1,10, size = (5, 5 ))
 for I in np.T: 
    B = + I 
    m = len (SET (List (b.reshape (-1)))) - ##. 1 there is a problem here, because b is an Array, must reshape (-1) can look, see the following reasons Code!
     Print (m)

 

# Input: 
Import numpy AS NP 
X = np.random.randint (l, 5, size = (l, 5), DTYPE = int) 
X 


# Output: 
Array ([[. 1,. 3,. 1,. 3,. 1]] )

# Input: 
List (X) 

# Output: 

[Array ([ . 1,. 3,. 1,. 3,. 1 ])] 


# Input: 

x.reshape ( -1 ) 

# Output: 

Array ([ . 1,. 3,. 1,. 3, 1 ]) 

# input: 
list (x.reshape (-1 )) single-line array ## must reshape it, can list (), to give a regular list! 

# output: 

[ 1, 3, 1, 3, 1 ]
 
# Therefore there can be re-set to: 
# input: 
sET (List (x.reshape ( -1 ))) 

# output: { 1, 3}

 Topic algorithm: feeling very low

lst = [[1,2,2,1],[3,1,2],[1,3,2],[1,3,2],[2,4],[3,1,2],[1,3,1,1]]
m = len(lst)
lt = []
for i in lst:
    n=0
    for j in i[:-1]:
       n+=j
       lt.append(n)
lt1=[]
for i in lt:
   lt1.append(lt.count(i))
y=m-max(lt1)
print(y)