Setting values in matrices - Python

问题

I made i matrix 5x3

field = []
fields = []
    for i in range(0,5):
        for j in range(0,3):
            x = 1
            field.append(x)
    fields.append(field)

When i want to change one field

fields[2][2] = 0

i get this:

fields[0][0] = 1
fields[0][1] = 1
fields[0][2] = **0**
fields[1][0] = 1
fields[1][1] = 1
fields[1][2] = **0**
fields[2][0] = 1
fields[2][1] = 1
fields[2][2] = **0**
fields[3][0] = 1
fields[3][1] = 1
fields[3][2] = **0**
fields[4][0] = 1
fields[4][1] = 1
fields[4][2] = **0**

Instead one change i am getting five

回答1:

Its because you have the reference to the same field across all rows.

You want this:

for i in range(0,5):
    field = []
    for j in range(0,3):
        x = 1
        field.append(x)
    fields.append(field)

field should get reset for every row. That's why you should have it inside the first loop. Now your fields[2][2] = 0 would work.

>>> fields
[[1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1]]
>>> fields[2][2] = 0
>>> fields
[[1, 1, 1], [1, 1, 1], [1, 1, 0], [1, 1, 1], [1, 1, 1]]

回答2:

The reason this happens is that each row of your list refers to the same object, the list named field. You can see this by looking at the id of each row. You will find that:

id(fields[0]) == id(fields[1])
id(fields[0]) == id(fields[2])

and so on. Each row is in fact the same object.

You need to create a separate list for each row:

fields = []
for i in range(0,5):
    field = []
    for j in range(0,3):
        x = 1
        field.append(x)
    fields.append(field)

And now you will see that:

id(fields[0]) != id(fields[1])

and so on. And your attempts to modify individual elements will behave as you intend.

But really, if you want to work with matrices, then you should use numpy.

来源：https://stackoverflow.com/questions/15988855/setting-values-in-matrices-python

标签

python

arrays

multidimensional-array

matrix

setting