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 list([]) == []
。那么为什么这段代码在做
oldChain = list(chain)
而不是 oldChain = chain
这是同一件事,所以无论哪种方式都没有关系吗?
最佳答案
list(chain)
返回chain
的浅拷贝,等价于chain[:]
。
如果您想要列表的浅拷贝,请使用 list()
,它有时也用于从迭代器中获取所有值。
y = list(x)
和 y = x
的区别:
浅拷贝:
>>> 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
深拷贝:
请注意,如果 x
包含可变对象,那么仅 list()
或 [:]
是不够的:
>>> x = [[1,2],[3,4]]
>>> y = list(x) #outer list is different
>>> x is y
False
但内部对象仍然是对 x 中对象的引用:
>>> 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]])
由于外部列表不同,因此修改 x 不会影响 y,反之亦然
>>> x.append('bar')
>>> x,y
([[1, 2, 'foo'], [3, 4], 'bar'], [[1, 2, 'foo'], [3, 4]])
要处理此问题,请使用 copy.deepcopy
。
关于Python 列表([])和 [],我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17599175/