Python字典奇怪的行为

Question

1. I can't understand why, when c=2 and c=3, the dict2['A']=1 and dict2['A']=4 respectively and not 0, even though I make dict2 to be equal to dict1, where dict1['A']=0? Why does dict1['A'] change from 0? I don't change any of the dict1 variables!
2. Why does dict3 show correct values within the loop, but only shows values from the last iteration 3 after the loop is finished.

Thank you very much.

import collections

def main():


    dict1 = collections.OrderedDict()
    dict2 = collections.OrderedDict()
    dict3 = collections.OrderedDict()

    dict1['A'] = 0
    dict1['B'] = 0
    dict1['C'] = 0

    for c in [1, 2, 3]:
        print('c=' + str(c))
        dict2 = dict1
        print('dict1A=' + str(dict1['A']))
        print('dict2A=' + str(dict2['A']))
        if c == 1:
            dict2['A'] = 1
            dict2['B'] = 2
            dict2['C'] = 3
        elif c ==2:
            dict2['A'] = 4
            dict2['B'] = 5
            dict2['C'] = 6
        elif c ==3:
            dict2['A'] = 7
            dict2['B'] = 8
            dict2['C'] = 9
        dict3['c' + str(c)] = dict2
        print('dict2A=' + str(dict2['A']))
        print('dict' + str(c) + 'A=' + str(dict3['c' + str(c)]['A']))
        print('dict' + str(c) + 'B=' + str(dict3['c' + str(c)]['B']))
        print('dict' + str(c) + 'C=' + str(dict3['c' + str(c)]['C']))

    print('dict3-c1A='+ str(dict3['c1']['A']))
    print('dict3-c2B=' + str(dict3['c2']['B']))
    print('dict3-c3C=' + str(dict3['c3']['C']))

if __name__ == '__main__':
    main()

Output:

c=1
dict1A=0
dict2A=0
dict2A=1
dict1A=1
dict1B=2
dict1C=3
c=2
dict1A=1
dict2A=1
dict2A=4
dict2A=4
dict2B=5
dict2C=6
c=3
dict1A=4
dict2A=4
dict2A=7
dict3A=7
dict3B=8
dict3C=9
dict3-c1A=7
dict3-c2B=8
dict3-c3C=9

* EDIT * Thank you very much for the answers. I didn't know the '=' operation for dictionaries was not the same as for variables. I found out and as was suggested by gddc, that the copy() is what I wanted:

        dict2 = dict1.copy()

Answer 1

1. This is how the Python object model works. by saying dict2 = dict1 , what in essence happens during the first loop iteration is that the original dict2 object is discarded, and dict2 now refers to the same underlying object as dict 1 . During subsequent loop iterations, the dict2 = dict1 statement has no effect since dict1 and dict2 are already both pointing at the same underlying object.

2. Your print statements for dict3 are one level of indentation further out than your loop body; they are not part of the loop, and therefore only execute once the loop is finished.

Answer 2

dict2 = dict1

This doesn't erase the contents of dict2 and fill it with the contents of dict1 . This makes the names dict1 and dict2 refer to the same object. Anything you do to dict2 affects dict1 , because they're the same thing.

If you want a new dict, make a new dict:

dict2 = collections.OrderedDict()

If you want to clear the old dict, clear it, but that's probably not what you want:

dict2.clear()

Answer 3

When you assign dict2 = dict1 you replace the name which previously existed as an empty OrderedDict and tell the interpreter instead to use dict2 to refer to the same Object that exists in the name dict1 . You can do a couple of things to work around this:

# copy
dict2 = dict1.copy()
# update dict 2
dict2.update(dict1.iteritems())