Copying list in python: deep vs shallow copy: gotcha for me in python?

Question

So this is what I tried to do.

vectorized = [0] * length
for i,key in enumerate(foo_dict.keys()):
    vector = vectorized
    vector[i] = 1
    print vector
    vector = vectorized
print vectorized

So what I was hoping was for example the length is 4. So i create a 4 dimension vector:

  vectorized=[0,0,0,0]

now, depending on the index of dictionary (which is also of length 4 in this case) create a vector with value 1 while rest has zero

so vector = [1, 0,0,0]  , [0,1,0,0] and so on..

Now instead what is happening is:

 vector = [1,0,0,0],[1,1,0,0] .. and finally [1,1,1,1]

even vectorized is now

   [1,1,1,1]

Whats wrong I am doing. and how do i achieve what i want to achieve. Basically I am trying to create unit vectors. Thanks

I wish more question were this quality. +1. I would only recommend improving the title. — Trufa
– Trufa, Commented Mar 30, 2012 at 0:57

Ignacio Vazquez-Abrams · Accepted Answer · 2012-03-30 00:14:18Z

9

This line (these lines, really):

vector = vectorized

copies the list reference. You need to do a shallow copy of the sequence contents.

vector = vectorized[:]

answered Mar 30, 2012 at 0:14

Ignacio Vazquez-Abrams

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Comments

Mark Byers · Accepted Answer · 2012-03-30 00:14:40Z

5

You are creating a single list and then giving it several different names. Remember that a = b doesn't create a new object. It just means that a and b are both names for the same thing.

Try this instead:

for ...:
    vector = [0] * length
    ...

answered Mar 30, 2012 at 0:14

Mark Byers

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David Robinson · Accepted Answer · 2012-03-30 00:15:40Z

4

The line

vector = vectorized

is not making a copy of vectorized. Any time you change vector from then on, you are also changing `vectorized.

You can change the first line to:

vector = vectorized[:]

or

import copy
vector = copy.copy(vectorized)

If you want to make a copy.

answered Mar 30, 2012 at 0:15

David Robinson

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TryPyPy · Accepted Answer · 2012-03-30 00:45:38Z

2

While in Python, when you assign a list to a new list, the new list is just a pointer rather than a brand new one.

So when you trying to modify the value of "vector", you are actually changing the value of "vectorized". And in your case, vector[i] = 1 is same as vectorized[i] = 1

edited Mar 30, 2012 at 0:45

TryPyPy

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answered Mar 30, 2012 at 0:17

Rocky

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soundslikeneon · Accepted Answer · 2012-03-30 00:20:29Z

1

Your problem is that when you write vector = vectorized it is not creating a copy of the array, rather it is creating a binding between the two.

Assignment statements in Python do not copy objects, they create bindings between a target and an object.

http://docs.python.org/library/copy.html

This should help you get it sorted out.

And here's a little snippet from the python REPL to show you what I mean.

>>> vectorized = [0] * 4  
>>> print vectorized  
[0, 0, 0, 0]  
>>> vector = vectorized  
>>> vector[1] = 1  
>>> print vectorized  
[0, 1, 0, 0]

EDIT: Jeez you guys are fast!

answered Mar 30, 2012 at 0:20

soundslikeneon

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