Python列表（[]）和[] [英] Python list([]) and []

问题描述

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)

有人告诉我，列表（[]）== [] 。那么，为什么这个code做的结果
oldChain =列表（链）而不是 oldChain =链

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

这是同样的事情，所以也无所谓任何一种方式来做到这一点？

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

推荐答案

列表（链）返回链，它相当于链条[：]

如果您想要列表的浅备份，然后使用的list（），它也被用来有时从一个迭代器获取所有的值。

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

区别 Y =列表（x）和 Y = X ：

浅表复制：

>>> 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的：

请注意，如果 X 包含可变对象然后就的list（）或 [： ] 是不够的：

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

但内心对象仍然在X到对象的引用：

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]])

由于外列表不同，那么修改x将不会影响y和反之亦然

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]])  

要处理这种使用 copy.deepcopy 。

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