Python list([]) and []

Question

from cs1graphics import *
from math import sqrt

numLinks = 50
restingLength = 20.0
totalSeparation = 630.0
elasticityConstant = 0.005
gravityConstant = 0.110
epsilon     = 0.001

def combine(A,B,C=(0,0)):
    return (A[0] + B[0] + C[0], A[1] + B[1] + C[1])

def calcForce(A,B):
    dX = (B[0] - A[0])
    dY = (B[1] - A[1])
    distance = sqrt(dX*dX+dY*dY)
    if distance > restingLength:
        stretch = distance - restingLength
        forceFactor = stretch * elasticityConstant
    else:
        forceFactor = 0
    return (forceFactor * dX, forceFactor * dY)                 #return a tuple


def drawChain(chainData, chainPath, theCanvas):
    for k in range(len(chainData)):
        chainPath.setPoint(Point(chainData[k][0], chainData[k][1]),k)
    theCanvas.refresh()                             #refresh canvas

chain = []                                                             #chain here
for k in range(numLinks + 1):
    X = totalSeparation * k / numLinks
    chain.append( (X,0.0) )

paper = Canvas(totalSeparation, totalSeparation)
paper.setAutoRefresh(False)
curve = Path()
for p in chain:
    curve.addPoint(Point(p[0], p[1]))
paper.add(curve)
graphicsCounter = 100

somethingMoved = True
while somethingMoved:
    somethingMoved = False
    oldChain = list(chain)                                             #oldChain here
    for k in range(1, numLinks):
        gravForce = (0, gravityConstant)
        leftForce = calcForce(oldChain[k], oldChain[k-1])
        rightForce = calcForce(oldChain[k], oldChain[k+1])
        adjust = combine(gravForce, leftForce, rightForce)
        if abs(adjust[0]) > epsilon or abs(adjust[1]) > epsilon:
            somethingMoved = True
        chain[k] = combine(oldChain[k], adjust)
    graphicsCounter -= 1
    if graphicsCounter == 0:
        drawChain(chain, curve, paper)
        graphicsCounter = 100

curve.setBorderWidth(2)
drawChain(chain, curve, paper)

I was told that list([]) == []. So why is this code doing
oldChain = list(chain) instead of oldChain = chain

it's the same thing so it does not matter either way to do it?

Answer 1

list(chain) returns a shallow copy of chain, it is equivalent to chain[:].

If you want a shallow copy of the list then use list(), it also used sometimes to get all the values from an iterator.

Difference between y = list(x) and y = x:

---

Shallow copy:

>>> x = [1,2,3]
>>> y = x         #this simply creates a new referece to the same list object
>>> y is x
True
>>> y.append(4)  # appending to y, will affect x as well
>>> x,y
([1, 2, 3, 4], [1, 2, 3, 4])   #both are changed

#shallow copy   
>>> x = [1,2,3] 
>>> y = list(x)                #y is a shallow copy of x
>>> x is y     
False
>>> y.append(4)                #appending to y won't affect x and vice-versa
>>> x,y
([1, 2, 3], [1, 2, 3, 4])      #x is still same 

---

Deepcopy:

Note that if x contains mutable objects then just list() or [:] are not enough:

>>> x = [[1,2],[3,4]]
>>> y = list(x)         #outer list is different
>>> x is y          
False

But inner objects are still references to the objects in x:

>>> x[0] is y[0], x[1] is y[1]  
(True, True)
>>> y[0].append('foo')     #modify an inner list
>>> x,y                    #changes can be seen in both lists
([[1, 2, 'foo'], [3, 4]], [[1, 2, 'foo'], [3, 4]])

As the outer lists are different then modifying x will not affect y and vice-versa

>>> x.append('bar')
>>> x,y
([[1, 2, 'foo'], [3, 4], 'bar'], [[1, 2, 'foo'], [3, 4]])  

To handle this use copy.deepcopy.

Answer 2

It is true that list([]) is functionally equivalent to [], both creating a new empty list.

But x = list(y) is not the same as x = y. The formers makes a shallow copy, and the latter creates a new reference to the existing list.

Note that list([]) is inefficient -- it creates a new empty list (by doing []), then copies it, resulting with another empty list (by doing list(...)), then deallocates the original, unreferenced, list.