3
mat=[[0,1,5],[1,3,6],[-1,4,4],[1,2,2],[7,3,7],[2,5,3]]

mat matrix shape could be a 10000*5. here just an example

Here I define a function. It tries to find mat[:,0] < be or mat[:,0] > ba or mat[:,1] < bb. If one column match the condition, the element[i,0] < be the element[i,0]=be, then copy the column to another matrix "swape". Also delete this column from matrix "mat". Same as mat[:,0] > ba or mat[:,1] < bb. For mat[:,1] < bb, the colume will copy to "swapt",mat[:,0] > ba don't copy, just delete.

example be=0, ba=6,bb=3

The return list should be :

mat=[[1,3,6],[2,5,3]]
swape=[[-1,4,4]]
swapt=[[1,2,2],[0,1,5]]

The function will return mat, swape, and swapt

def swapco(be,ba,bb,mat):
        swape=np.array()
        swapt=np.array()
        leng=np.shape(mat)[0]
        for i in range(leng):
            if mat[i,2]<bb: 
                mat[i,2]=bb   
                np.append(swapt,i,1)
                np.delete(mat, i, 0)
            else:
                if mat[i,0]>=ba:    
                    mat[i,0]=ba 
                    np.append(swape,i,1)
                    np.delete(mat, i, 0)
                elif mat[i,0]<=be:
                    mat[i,0]=be 
                    np.append(swape,i,1)
                    np.delete(mat, i, 0)
            i+=1

        return swape, swapt

In my code, I found the matrix mat length always reduce once some columns matched the condition. It will append and delete wrong column. Also the append is append a address or a deepcopy?

If using

for col in mat:

Then how to delete itself in mat? Or any efficiency way to write this code?

Question updated...

rfeynman
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  • It's not entirely clear to me what you want to achieve with your code: it looks like you're trying to append to some array the indices of the columns you've checked for a certain condition. Could you update your post and add a numerical example, based on e.g. `mat=np.reshape(np.arange(0,15),(5,3)); ba=11; be = 5; bb=3`? – Oliver W. Mar 02 '15 at 23:44
  • You should not really be deleting rows or columns of a matrix inside a loop. Keep in mind that if I delete row 1 when i = 1, than row 3 becomes row 2 and row 2 becomes row 1. Now when i = 2, you're looking at the _new_ row 2, which is the _old_ row 3. The _old_ row 2 never gets checked in the loop. – Bi Rico Mar 03 '15 at 00:55

1 Answers1

2

It's really unclear from your code what you're trying to do, but let me write a little example for you that might help you get started cleaning up your code.

import numpy as np

def split(mat, a, b):
    assert a < b
    where_less_than_a = mat[:, 0] < a
    where_less_than_b = mat[:, 0] < b

    less_than_a = mat[where_less_than_a, :]
    between_a_b = mat[(~where_less_than_a) & where_less_than_b, :]
    greater_eq_b = mat[(~where_less_than_b), :]
    return less_than_a, between_a_b, greater_eq_b

mat = np.arange(27).reshape((9, 3))
x, y, z = split(mat, 4., 17.)
print(x)
# [[0 1 2]
#  [3 4 5]]
print(y)
# [[ 6  7  8]
#  [ 9 10 11]
#  [12 13 14]
#  [15 16 17]]
print(z)
# [[18 19 20]
#  [21 22 23]
#  [24 25 26]]

I hope the above example helps you get started on your own project. There is one thing that might be a little confusing, the ~ in the code above is used as a "logical not" operator. It can be used in this way with numpy arrays of type bool, but be careful because it has a different meaning for other python objects (it's called the complement operator).

Bi Rico
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    You answer is very helpful. So in where_less_than_b = mat[:, 0] < b, could I use bool operation here. Such as mat[:, 0] < b & mat[:, 0] > c assert b>c? It looks give me wrong notice... – rfeynman Mar 03 '15 at 14:39
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    You can totally combine comparison operations, for example `(mat[:, 0] < a) & (mat[:, 0] > c)`. Make sure to include the parenthesis. The assert isn't really needed, it's just lazy error checking. – Bi Rico Mar 03 '15 at 19:58