Question

我正面临着一个我现在无法自行解决的问题。它关于以下片段：

counter = 0
appendList = []
valueList = [[0], [0]]

for i in range(0,3):

    valueList[1] = counter
    print "Loop " , i  , " valueList: " , valueList
    print "Appending (valueList): " , valueList , " to (appendList): " , appendList
    appendList.append(valueList)
    counter = counter + 1


print "Final appendList: " , appendList

这导致以下输出：

Loop  0  valueList:  [[0], 0]
Appending (valueList):  [[0], 0]  to (appendList):  []
Loop  1  valueList:  [[0], 1]
Appending (valueList):  [[0], 1]  to (appendList):  [[[0], 1]]
Loop  2  valueList:  [[0], 2]
Appending (valueList):  [[0], 2]  to (appendList):  [[[0], 2], [[0], 2]]
Final appendList:  [[[0], 2], [[0], 2], [[0], 2]]

我希望代码段向List-Items添加不同的appendList。最终结果应如下所示：

[[[0], 0], [[0], 1], [[0], 2]]

但正如您所看到的，该代码段使用最高计数器的相同值填充appendList。

有人可以向我解释一下这种行为，还是告诉我，我的错误在哪里？

Answer 1

valueList每次附加时都是相同的对象，因此在一个地方修改它似乎可以在任何地方修改它。

>>> a = [0]
>>> b = a
>>> a[0] = 42
>>> b
[42]

您需要附加一份副本，以便每次都添加新列表。

appendList.append(valueList[:])

Answer 2

您可以尝试：

counter = 0
appendList = []
valueList = [[0], [0]]

for i in range(0,3):

    valueList[1] = counter
    print "Loop " , i  , " valueList: " , valueList
    print "Appending (valueList): " , valueList , " to (appendList): " , appendList
    appendList.append(valueList[:])
    counter = counter + 1


print "Final appendList: " , appendList

输出：

Loop  0  valueList:  [[0], 0]
Appending (valueList):  [[0], 0]  to (appendList):  []
Loop  1  valueList:  [[0], 1]
Appending (valueList):  [[0], 1]  to (appendList):  [[[0], 0]]
Loop  2  valueList:  [[0], 2]
Appending (valueList):  [[0], 2]  to (appendList):  [[[0], 0], [[0], 1]]
Final appendList:  [[[0], 0], [[0], 1], [[0], 2]]

<强>解释

分配valuelist对象时，所有元素都从referance对象分配。因此，b = a代替b = a [:]，因为它每次只能从此对象进行复制。

>>> a = [1, 2, 3]
>>> b = a
>>> a[:] = [4, 5, 6]
>>> b
[4, 5, 6]
>>> a
[4, 5, 6]
>>> b = a [:]
>>> a = [1, 2, 3]
>>> b
[4, 5, 6]
>>> a
[1, 2, 3]
>>>

Answer 3

appendList包含当前值valueList的引用，因此当您修改valueList时，appendList中的副本会发生变化。每次放入valueList时，您需要创建appendList的副本。

Answer 4

valueList仍指向单个对象。您可以通过打印id(valueList)来检查这一点。 id提供了对象的标识。即对象的内存地址。在您的示例中，如果您打印id，它将始终相同。

counter = 0
appendList = []
valueList = [[0], [0]]

for i in range(0,3):
    print "id of valueList before", id(valueList)
    #creates a copy of valueList.
    valueList = valueList[:]
    print "id of valueList after", id(valueList)
    valueList[1] = counter
    print "Loop " , i  , " valueList: " , valueList
    print "Appending (valueList): " , valueList , " to (appendList): ", appendList
    appendList.append(valueList)
    counter = counter + 1


print "Final appendList: " , appendList

Answer 5

counter = 0
appendList = []
valueList = [[0], [0]]

print(valueList[1][0])

for i in range(0,3):

    print(counter)
    valueList[1][0] = counter # this was wrong
    print ("Loop " , i  , " valueList: " , valueList)
    print ("Appending (valueList): " , valueList , " to (appendList): " , appendList)
    appendList.append(valueList)
    valueList = [[0], [0]] # and this was absent
    counter += 1

print(appendList)

[[[0], [0]], [[0], [1]], [[0], [2]]]

Answer 6

一个好习惯也是检查代码的效率：你正在定义多个变量，而对我来说，你对appendList感兴趣。

appendList = [ [[0], x] for x in range(3) ] # doStuff for index in range of n print appendList 使用列表理解，您可以将其简化为必要的计算。

在Python中将列表添加到列表中

6 个答案: