CODEWITHSUNDEEP
python
Victor Mota
Victor Mota

Reputation: 1269

Method returns variable into its original parameter variable

I am encountering a problem with my program, where a method is saving everything that it does to the parameter variable used in it.

Here's an example: method that moves a list's element to the left by given spaces. 

def moveLeft (perm, pos, spaces):
   permTemp = perm
   for n in range(spaces):
      charMoved = permTemp[pos-n]
      permTemp [pos-n] = permTemp[pos-n-1]
      permTemp[pos-n-1] = charMoved
   return permTemp

permO = [0,1,2,3] #original perm  
print moveLeft(permO, -1, 2)
print moveLeft(permO, -1, 2)
print permO

The expect output would be: the first two to be the same (since it is printing the same returned values of the method) and the last output to be the original list ( [0,1,2,3] ). Instead I get:

    >>> 
[0, 3, 1, 2]
[0, 2, 3, 1]
[0, 2, 3, 1]

Upvotes: 1

Views: 105

Answers (2)

Sven Marnach
Sven Marnach

Reputation: 601769

Use

permTemp = perm[:]

to actually copy the list, instead of just assigning a new name to the same object.

Python assignment does not create new objects, it just names existing ones. That's why you are modifying the original list.

Upvotes: 2

tokland
tokland

Reputation: 67870

Use Sven's list copy, but note also that you don't need a temporal variable to swap values:

def moveLeft(perm, pos, spaces):
   permTemp = perm[:]
   for n in range(spaces):
      permTemp[pos-n], permTemp[pos-n-1] = permTemp[pos-n-1], permTemp[pos-n]
   return permTemp

Upvotes: 1

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