Question

I was writing a Python Code that creates dictionaries dynamically,initializes it to a reference dictionary, and modifying a particular value in the dictionary. But,I found that not only I am getting unexpected results,but the reference dictionary is also getting modified.
My Code:

tdict={'a':'1','b':'2','c':'3'}
newdict={}
for i in range(5):
  newdict['name'+str(i)]=tdict
  newdict['name'+str(i)]['a']='value'+str(i)
  print 'tdict:  ',tdict
print 'newdict:  ',newdict

And the result:

tdict:   {'a': 'value0', 'c': '3', 'b': '2'}
tdict:   {'a': 'value1', 'c': '3', 'b': '2'}
tdict:   {'a': 'value2', 'c': '3', 'b': '2'}
tdict:   {'a': 'value3', 'c': '3', 'b': '2'}
tdict:   {'a': 'value4', 'c': '3', 'b': '2'}
newdict:   {'name4': {'a': 'value4', 'c': '3', 'b': '2'}, 'name2': {'a': 'value4', 'c': '3', 'b': '2'}, 'name3': {'a': 'value4', 'c': '3', 'b': '2'}, 'name0': {'a': 'value4', 'c': '3', 'b': '2'}, 'name1': {'a': 'value4', 'c': '3', 'b': '2'}}

whereas I expected my ‘newdict’ to be like:

newdict:   {'name4': {'a': 'value4', 'c': '3', 'b': '2'}, 'name2': {'a': 'value2', 'c': '3', 'b': '2'}, 'name3': {'a': 'value3', 'c': '3', 'b': '2'}, 'name0': {'a': 'value0', 'c': '3', 'b': '2'}, 'name1': {'a': 'value1', 'c': '3', 'b': '2'}}

Can anyone please help me figuring out why this is happening? Also, why is the reference dictionary ‘tdict’ getting changed when I am not assigning any any value to it?

Thanks in advance

Answer 1

You are storing a reference to tdict in every value of your newdict dictionary:

newdict['name'+str(i)]=tdict

You are then modifying the key 'a' of tdict by doing

# newdict['name'+str(i)] is a reference to tdict
newdict['name'+str(i)]['a']='value'+str(i)
# this is equivalent to doing
tdict['a']='value'+str(i)

What you maybe want is storing a copy of tdict in your newdict dictionary:

newdict['name'+str(i)]=dict(tdict)

Creating a new dictionary by using an existing dictionary as constructor argument creates a shallow copy where you can assign new values to existing keys. What you cannot (or what you don’t want) is modifying mutable values in this dictionary. Example:

>>> a={'a': 1, 'b': 2, 'c': [1,2,3]}
>>> b=dict(a)
>>> b['a']=9
>>> a
{'a': 1, 'c': [1, 2, 3], 'b': 2}
>>> b
{'a': 9, 'c': [1, 2, 3], 'b': 2}
>>> b['c'].append(99)
>>> a
{'a': 1, 'c': [1, 2, 3, 99], 'b': 2}
>>> b
{'a': 9, 'c': [1, 2, 3, 99], 'b': 2}

If you want to modify mutable values in a dictionary you need to create a deep copy:

>>> import copy
>>> a={'a': 1, 'b': 2, 'c': [1,2,3]}
>>> b=copy.deepcopy(a)
>>> b['a']=9
>>> b['c'].append(99)
>>> a
{'a': 1, 'c': [1, 2, 3], 'b': 2}
>>> b
{'a': 9, 'c': [1, 2, 3, 99], 'b': 2}