Matrix (Jagged Array) for in loop initialization

Question

I'm trying to do an inline uniform distribution initialization, but don't seems to find how to do so, if possible.

My working code is:

test = [[1,2,3],[4,5,6],[7,8,9]]
p = []
for row in test:
    prow = [1.0/(len(row)*len(test)) for x in row]
    p.append(prow)
print p
[[0.1111111111111111, 0.1111111111111111, 0.1111111111111111],
 [0.1111111111111111, 0.1111111111111111, 0.1111111111111111],
 [0.1111111111111111, 0.1111111111111111, 0.1111111111111111]]

My question is, is there a way to narrow the "for row in test" to the same form of the second loop and end up with only 1 line? Something like:

p = [1.0/(len(row)*len(test)) for x in row for row in test]
print p
[0.1111111111111111,
 0.1111111111111111,
 0.1111111111111111,
 0.1111111111111111,
 0.1111111111111111,
 0.1111111111111111,
 0.1111111111111111,
 0.1111111111111111,
 0.1111111111111111]

But obviously the right result. Clues?

You guys are fast! Thanks! Can't even accept the answer yet. — mimoralea, Jan 21 '14 at 05:15

score 4 · Answer 1 · answered Jan 21 '14 at 05:12

4

I think something like:

[[1.0/(len(row)*len(test))]*len(row) for row in test]

if I'm reading your code correctly. The magic here is because every element in the row gets the same (immutable) value, you can just create a list of the correct length with the correct value ala [value]*length.

answered Jan 21 '14 at 05:12

mgilson

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Magic indeed. Hadn't even thought about this approach. – mimoralea Jan 21 '14 at 05:25

score 3 · Accepted Answer · edited May 23 '17 at 10:31

Yeah, what you're looking for is this:

In [13]: test = [[1,2,3],[4,5,6],[7,8,9]]

In [14]: p = [[1.0/(len(row)*len(test)) for x in row] for row in test]

In [15]: p
Out[15]: 
[[0.1111111111111111, 0.1111111111111111, 0.1111111111111111],
 [0.1111111111111111, 0.1111111111111111, 0.1111111111111111],
 [0.1111111111111111, 0.1111111111111111, 0.1111111111111111]]

PS: You might be interested in list comprehension syntax

Matrix (Jagged Array) for in loop initialization

2 Answers2