Python referencing array entry

Question

I used to think that python used to work with references or respectively with copies of references when values are passed as function arguments.

Now I tried the following example and do not understand anything anymore even after reading a bit up on the topic.

import numpy as np
import networkx as nx

graph = nx.DiGraph()
test = np.array([1, 1, 1], dtype=np.bool)
graph.add_edge(1, 2, data=True)
print graph[1][2]['data'] # shows True as expected
graph[1][2]['data'] = test[0]
print graph[1][2]['data'] # shows True as expected. Still fine

test[0] = False
print graph[1][2]['data'] # shows True instead of False

Shouldn't it then print False? I thought that assignement would make the graph[1][2]['data'] point to test[0]. However it seems that it actually uses references to True and False and I seem to don't really understand pythonic assignment.

Is there a way how to make it point to the specific entry of the array or is this impossible in python? And not to the content of the array entry?

DorElias · Accepted Answer · 2017-01-27 18:51:02Z

when you did graph[1][2]['data'] = test[0] you took the current value of test[0] and put it in the graph. when you did test[0] = False you changed the value of test[0] to point to a diffrent value, but the graph point to the old value.

a simplified example of your case can be:

x = 1
y = x
x = 2
print y # will print 1

Since you changed x's its value, but y referenced the old value. in python when you assign (use =) you take the value of the right side and place it on the left side, but in your case when you did test[0] = False you didnt change the value, you overriden it with a diffrent value.

if you want to change something and have it shared, you need to manipulate the same instance, for example:

class A(object):
    def __init__(self):
        self.x = 1
y = A()
lst = []
lst.append(y)
y.x += 1
print lst[0].x # will print 2

here the instance of A was shared so the value change was visible by accessing it with the list

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