CODEWITHSUNDEEP
pythonlist
Justin
Justin

Reputation: 21

why list product duplicates elements after doing append action in python

this is really a puzzle issue, while debugging my code, I found this really strange error (it happened in both python 3.6 and 3.7, have not tested on others)

when I am looping through a simple list, take out the element and assigned in to a dict, creating and new list of dict.

the list.append does not just add on the new element, but it also replaces the previous elements.

simple python code: 

d1 = {}
l1=["A1,1","B2,2"]
l2 =[]
for rows in l1:
    print("----- l2 before append")
    print(l2)
    d1["ID"]=rows
    print("-------dict to append ")
    print(d1)
    l2.append(d1)
    print("----- l2 after append")
    print(l2)

print result: 

----- l2 before append
[]
-------dict to append 
{'ID': 'A1,1'}
----- l2 after append
[{'ID': 'A1,1'}]
----- l2 before append
[{'ID': 'A1,1'}]
-------dict to append 
{'ID': 'B2,2'}
----- l2 after append
[{'ID': 'B2,2'}, {'ID': 'B2,2'}]

I expected the output of l2 to be [{'ID': 'A1,1'}, {'ID': 'B2,2'}] but I get [{'ID': 'B2,2'}, {'ID': 'B2,2'}]

Upvotes: 1

Views: 55

Answers (2)

ComplicatedPhenomenon
ComplicatedPhenomenon

Reputation: 4199

l1=["A1,1","B2,2"]
l2 =[]
for rows in l1:
    d1 = {}
    #print(id(d1)) 
    # you will find it's a different object each time.
    d1["ID"]=rows
    l2.append(d1)
print(l2)

Or you can do it in this way as below

l1=["A1,1","B2,2"]
l2 = [{"ID":i } for i in l1]
print(l2)

Output is 

[{'ID': 'A1,1'}, {'ID': 'B2,2'}]

Upvotes: 2

Sam
Sam

Reputation: 433

A dict can only contain unique keys, so 'ID' is overridden with the new value.

Your list is actually containing the value of the dictd1 if d1 changes, your list will change.

You're using a global dict, and the value changes with the second for loop pass, to to get your intended results, put the d1 = {} inside your for loop instead.

Upvotes: 0

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