I have a some code:
first = ['a','b']
second = first
second.append('c')
print('Test results: ',first == second, first is second)
Which returns Test results: True True.
I expected to get False False. I thought that because the second.append('c') by appending the 'c', the two variables stores different objects - meaning first = ['a','b'] and second = ['a','b','c']
Why do I get True True?
Because second = first does not make a copy. It makes second and first two references to the same object.
Actually the issue is in second = first, after execution of this statement, there is no new variable or reference created to some other object instead second is pointing exactly to the same memory location as first. So any changes made in the second would ultimately reflected in the first as well:
first = [1, 2, 3]
second = first
print second
>>> [1, 2, 3]
second.append(4)
print second
>>> [1, 2, 3, 4]
print first
>>> [1, 2, 3, 4]
To avoid such issues you deepcopy to initialize a new list.
from copy import deepcopy
a = [1, 2, 3]
b = deepcopy(a)
b.append(4)
print b
>>> 1, 2, 3, 4
print a
>>> 1, 2, 3
Option # 1 - here references are being compared instead the values
first = ['a','b']
second = first
second.append('c')
print('Test results: ',first == second, first is second)
## >> result : it will always return TRUE
Here I recommend to use first is second for individual array element than on whole array though Python wont stop you from doing that.
Option # 2 - its more appropriate way
print cmp(second, first)
## >> result : it will return 0 i.e. False
Option # 3 - make use of numpy array
print np.array_equal(first, second)
## >> result : it will return False too.
You can also use deepcopy as explained by ZdaR in the answer above.
