o=[]
o1=[1,2,3]
o.append(o1)
o1.append(9)
o.append(o1)
print(o)
for the above code I've been getting the output [[1, 2, 3, 9], [1, 2, 3, 9]] but i want to get the output as [[1, 2, 3], [1, 2, 3, 9]]. So what do i do ?
o=[]
o1=[1,2,3]
o.append(o1)
o1.append(9)
o.append(o1)
print(o)
for the above code I've been getting the output [[1, 2, 3, 9], [1, 2, 3, 9]] but i want to get the output as [[1, 2, 3], [1, 2, 3, 9]]. So what do i do ?
Your code
o=[]
o1=[1,2,3]
o.append(o1)
o1.append(9)
o.append(o1)
print(o)
creates o
holding your o1
list
twice (if you want to check it yourself, print id(o[0])
and id(o[1])
and compare them), you need to append copies i.e. do:
import copy
o=[]
o1=[1,2,3]
o.append(copy.deepcopy(o1))
o1.append(9)
o.append(copy.deepcopy(o1))
print(o)
output:
[[1, 2, 3], [1, 2, 3, 9]]