Dictionary inside list assignment results in strange output

Question

I've many lists holding many dictionaries inside a list. I simply assign to one dictionary in one of the lists in the outer list. But it leads to assignment to all dictionaries in all lists in the outer list.

The code:

CL=3*[0]
DL=4*[0]
di= {
     'A':[],
     'B':[],
     'C':CL,
     'D':DL
    }
R=[[],[]]
R[0].append(di)
R[1].append(di)

def func(dd):
    dd[0][0]['A'].append("BANANA")
    dd[0][0]['B'].append("ELEPHANT")
    dd[0][0]['C'][0]='BLUE'
    dd[0][0]['D'][3]='ROCK'
    dd[0][0]['D'][2]=1111

print(R[0])
print(R[1])

print("\n")

func(R)

print(R[0])
print(R[1])

The output:

[{'A': [], 'C': [0, 0, 0], 'B': [], 'D': [0, 0, 0, 0]}]                                                                                                    
[{'A': [], 'C': [0, 0, 0], 'B': [], 'D': [0, 0, 0, 0]}]                                                                                                    


[{'A': ['BANANA'], 'C': ['BLUE', 0, 0], 'B': ['ELEPHANT'], 'D': [0, 0, 1111, 'ROCK']}]                                                                     
[{'A': ['BANANA'], 'C': ['BLUE', 0, 0], 'B': ['ELEPHANT'], 'D': [0, 0, 1111, 'ROCK']}]

As you can see, even though I assigned values to the dictionary in the first list in the outer list (func() operates only on dd[0]..), both the lists got assigned.

Is there any mistake in my indexing anywhere? Why does this happen?

R[0] and R[1] hold the same dictionary, not two separate copies of the initial dictionary. see http://lucumr.pocoo.org/2011/7/9/python-and-pola/#pass-by-what-exactly — TheoretiCAL, Jul 27 '17 at 17:28
You may find this article helpful: [Facts and myths about Python names and values](http://nedbatchelder.com/text/names.html), which was written by SO veteran Ned Batchelder. — PM 2Ring, Jul 27 '17 at 17:36

score 3 · Answer 1 · answered Jul 27 '17 at 17:27

di= {
     'A':[],
     'B':[],
     'C':CL,
     'D':DL
    }
R=[[],[]]
R[0].append(di)
R[1].append(di)

di is di is di. Each use of di doesn't create separate copies. There is only one di and so R[0][0] is R[1][[0] and all changes appear in the same dict.

To fix your code:

from copy import deepcopy
R[0].append(deepcopy(di))
R[1].append(deepcopy(di))

score 2 · Answer 2 · answered Jul 27 '17 at 17:27

2

The code

R[0].append(di)
R[1].append(di)

adds di to R by reference. That means that both R[0] and R[1] refer to the same underlying object. Thus, changing one changes the underlying object and therefore changes both.

answered Jul 27 '17 at 17:27

Sumner Evans

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Hi..Thanks for the answer....But could you please explain how can I get around this problem? – nPab Jul 27 '17 at 17:30
@nPab, Alex Hall has a way to fix your code in his answer. I recommend that approach. – Sumner Evans Jul 27 '17 at 17:32

Dictionary inside list assignment results in strange output

2 Answers2