Replacing elements between lists in python

Question

I am currently trying to replace elements in a 2D list with elements in another list, so as to implement a game I am making in python. Here is what I have so far:

listA = [1, 3, 5]
listB = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

for a in range(len(listA)):
    alpha = (a/3) #number of rows in listB
    beta = (a/3) #number of columns in listB
    listB[alpha][beta] = 123

When I do this, I get

[[123, 123, 123], [0, 0, 0], [0, 0, 0]]

instead of what I want given the parameters,

[[0, 123, 0], [123, 0, 123], [0, 0, 0]]

Am I doing something wrong?

What do you expect (a/3) to be? It'll never be anything else but 0.. — Martijn Pieters
– Martijn Pieters, Commented Mar 5, 2013 at 15:33
Looping by index is always a bad idea in Python, instead, loop through and build up a new list - using a list comprehension would be a good idea. — Gareth Latty
– Gareth Latty, Commented Mar 5, 2013 at 15:34
Actually, it'll only ever be zero...so your output is not possible! — nneonneo
– nneonneo, Commented Mar 5, 2013 at 15:34
so: 1,3,5 are the indexes where you want that 123 to be put in, and then you want the resulting listB to be split a list of lists? is that it? — Samuele Mattiuzzo
– Samuele Mattiuzzo, Commented Mar 5, 2013 at 15:37

Kevin · Accepted Answer · 2013-03-05 16:03:35Z

3

Instead of iterating through the indices of listA using for a in range(len(listA)):, you should iterate through the elements of listA using for a in listA:

Assuming that the indices in A translate into coordinates in B like so:

0 1 2
3 4 5
6 7 8

Then beta, AKA the column of B corresponding to a, should be calculated as a%3, rather than a/3.

listA = [1, 3, 5]
listB = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

for a in listA:
    #two slashes is integer division, not a comment, 
    #despite markup's color scheme
    alpha = a//3 
    beta = a%3
    listB[alpha][beta] = 123

print listB

Output:

[[0, 123, 0], [123, 0, 123], [0, 0, 0]]

edited Mar 5, 2013 at 16:03

answered Mar 5, 2013 at 15:42

Kevin

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1 Comment

Usman Ali Tahir Over a year ago

I tried @Kevin 's suggestion in idle and it worked, thank you Kevin and everyone else who gave input!

askewchan · Accepted Answer · 2013-03-05 18:06:35Z

3

If you use numpy this is pretty easy:

import numpy as np

A = np.array([1,3,5])
B = np.zeros((3,3))

B.flat[A] = 123

print B

out:

[[   0.  123.    0.]
 [ 123.    0.  123.]
 [   0.    0.    0.]]

Note that what .flat does is return a "flattened" version of your list:

[   0.  123.    0.  123.    0.  123.    0.    0.    0.]

edited Mar 5, 2013 at 18:06

answered Mar 5, 2013 at 15:43

askewchan

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1 Comment

Eryk Sun Over a year ago

Simpler even: B.flat[A] = 123.

Samuele Mattiuzzo · Accepted Answer · 2013-03-05 15:44:30Z

1

>>> for a in range(len(listA)):
...     alpha = (listA[a]/3) #number of rows in listB
...     beta = (listA[a]%3) #number of columns in listB
...     listB[alpha][beta] = 123
... 
>>> listB
[[0, 123, 0], [123, 0, 123], [0, 0, 0]]

you must use the elements inside your listA, or it's pointles to use the indexes generated by range. Also, you should do a bit of math to properly fetch the row and column index

edit: I suggest you take a look at Kevin's answer and explanation, mine is just a quick correction of your code.

answered Mar 5, 2013 at 15:44

Samuele Mattiuzzo

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2 Comments

askewchan Over a year ago

Better to iterate through listA than using range and then indexing listA, as in Kevin's answer.

Samuele Mattiuzzo Over a year ago

yeah, was just correcting his initial code, Kevin's more pythonic and clear indeed

Abhijit · Accepted Answer · 2013-03-05 15:46:12Z

0

>>> listA = [1, 3, 5]
>>> listB = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> listB_unwrapped = list(chain(*listB))
>>> for i in listA:
    listB_unwrapped[i] = 123


>>> listB = zip(*[iter(listB_unwrapped)]*3)
>>> listB
[(0, 123, 0), (123, 0, 123), (0, 0, 0)]
>>>

answered Mar 5, 2013 at 15:46

Abhijit

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