Add List to List in Python

Question

I am facing an issue that i cannot solve on my own right now. Its about the following snippet:

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

This results in the following output:

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]]

I wanted the Snippet to add different List-Items to the appendList. The final result should have looked like this:

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

But as you can see, the snippet fills the appendList with the same values of the highest counter.

Can someone please explain this behavior to me or tell me, where my mistake is?

Assign a copy every time, like this appendList.append(valueList[:]) — thefourtheye
– thefourtheye, Commented Oct 11, 2015 at 16:06

Ignacio Vazquez-Abrams · Accepted Answer · 2015-10-11 16:11:38Z

2

valueList is the same object each time you append it, so modifying it in one place appears to modify it everywhere.

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

You need to append a copy of it in order to add a new list each time.

appendList.append(valueList[:])

answered Oct 11, 2015 at 16:11

Ignacio Vazquez-Abrams

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Comments

Sakib Ahammed · Accepted Answer · 2015-10-11 16:28:21Z

You can try:

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

Output:

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]]

Explanation:

When you assign valuelist object, all elements are assign from referance object. So, instead b = a, you can do b = a [:] because it's only copy each time from this object.

>>> 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]
>>>

Although it solves the problem, this doesn't explain why and is an exact copy of @thefourtheye's comment.

Perry · Accepted Answer · 2015-10-11 16:12:01Z

0

appendList contains a reference to the current value of valueList, so when you modify valueList, the copies in appendList change. You need to create a copy of valueList each time you put in it appendList.

answered Oct 11, 2015 at 16:12

Perry

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Comments

LetzerWille · Accepted Answer · 2015-10-11 16:18:25Z

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]]]

kmad1729 · Accepted Answer · 2015-10-11 16:21:08Z

valueList still points to a single object. You can check this by printing out the id(valueList). id gives identity of the object. i.e. the memory address of the object. In your example, if you print the id, it will be always same.

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

Ranj · Accepted Answer · 2015-10-11 17:30:20Z

0

A good habit is also to check the efficiency of your code afterwards: you are defining multiple variables while (to me) you are interested in appendList.

appendList = [ [[0], x] for x in range(3) ] # doStuff for index in range of n print appendList With list comprehension you can reduce it to only the necessary computation.

answered Oct 11, 2015 at 17:30

Ranj

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