tags:

views:

193

answers:

5

Hi guys, in c++ I can wrote:

int someArray[8][8];
for (int i=0; i < 7; i++)
   for (int j=0; j < 7; j++)
      someArray[i][j] = 0;

And how can I initialize multi-line arrays in python? I tried:

array = [[],[]]
for i in xrange(8):
   for j in xrange(8):
        array[i][j] = 0
+3  A: 

Here is a shorter way:

array = []
for i in xrange(8):
    array.append( [0] * 8 )
Justin Ethier
+1 for looping way :-)
S.Mark
+7  A: 
>>> [[0]*8 for x in xrange(8)]
[[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, 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]]
>>>
S.Mark
+1 for being the first to use a list comprehension...
Justin Ethier
+3  A: 
array = [[0]*8 for i in xrange(8)]
Phong
+2  A: 
[[0]*8 for x in range(8)]
maker
+6  A: 

You asked about initializing a list of lists. Its a very useful data structure, but it has an important difference from the 2D array in C++: There are no guarantees that all lines have the same length (i.e, that len(a[0])==len(a[1]) (while in C++ you do have that guarantee).

So another solution that might be handy, is using NumPy's array datatype, like this:

import numpy as np
array = np.zeros((8,8))
Ofri Raviv
Nice to know numpy way! +1
S.Mark