2

I have a networkx graph. I am adding nodes by adding edges

G.add_edge(route[i-1],route[i]);

Now once the node is created by directly adding edges, I add a list named 

G.node[route[i]]['position'] = list()

and I append positions to it when I get same nodes again and again

G.node[route[i]]['position'].append( i - 3 )

Now when I want to append how do I check whether the list exist? does doing

G.node[route[i]]['position'] = list()

clear the list of already existing elements?

edit----- my earlier question was confusing I want to keep adding to the list but I cant append unless a list exists, right? So I have to do do G.node[route[i]]['position'] = list() in my loop So next time when I want to add to the same list in another loop instance how do I know that a list exists for G.node[route[i]]['position'] and I dont have to create it again.

edit----- I think my list itself is a key here so I did

                    if not 'position' in G.node[route[i]]:

and it works

  • My python is rusty but I think doing `"position" in G.node[route[i]]` to look for the key. (assuming that position does not exist before you create the list()) – Ilion Jan 18 '12 at 09:42
  • for that -- if not G.node[traceroute[i]]['position'] -- I get a invalid syntax –  Jan 18 '12 at 09:47
  • I think you might have misunderstood. I wasn't saying to do an `if not` condition. The "in" is an operator in Python. See this thread: http://bytes.com/topic/python/answers/614846-how-test-if-key-dictionary-exists – Ilion Jan 18 '12 at 09:56

5 Answers5

5

G.node[route[i]]['position'] = list() will leave the slot G.node[route[i]]['position'] holding an empty list, but it will not affect the list that it previously held, which other objects may have a reference to.

Instead, use: del l[:] to empty the list.

If you want to have a list automatically created, use collections.defaultdict to have newly created entries default to a list.

Marcin
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1

Yes, that clears the existing list. You could try

G.node[route[i]].setdefault('position', []).append(...)

whenever you want to append elements.

Johannes Charra
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  • Assigning a new list is not the same as clearing an existing list. – Marcin Jan 18 '12 at 10:17
  • You've got a point there. The existing list is not cleared when one assigns a new list as the value in that dictionary. I focussed on the other aspect of the OP's question, i.e. checking whether `G.node[route[i]]['position']` already exists. – Johannes Charra Jan 18 '12 at 10:32
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Not sure if this is what you mean, but assigning list() should make sure that there is a list to append to. If there's already a list the assignment creates a new one (see answer of Marcin). Test:

>>> a = list()
>>> for i in range(10):
...     a.append(i)
...
>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> b = a
>>> b
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> a = list()
>>> a
[]
>>> b
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
LiMuBei
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0

1) How do I check whether the list exist?

Use isinstance() and do something like:

if not ininstance(G.node[route[i]]['position'], list):
    G.node[route[i]]['position'] = list()
G.node[route[i]]['position'].append(i - 3)

Or use type like:

if not type(G.node[route[i]]['position']) is list

I must say that this kind of checking is rather un-pythonic, usually you should know what G.node[route[i]]['position'] was before becoming a list and check for that.

For example, if it was None you could do (assuming that the key 'position' exists, otherwise just call get('position')):

if G.node[route[i]]['position'] is None:
    G.node[route[i]]['position'] = list()
G.node[route[i]]['position'].append(i - 3)

2) Does doing .. = list() clear the list of already existing elements?

The answer is No.

list() will instantiate a new empty list.

You may want to take a look at this SO question: How to empty a list in Python?.
In short:

G.node[route[i]]['position'][:] = []

will clear your list.

Community
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Rik Poggi
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  • Why put a -1 like that? Just tell me if I misunderstood something. – Rik Poggi Jan 18 '12 at 10:25
  • Don't know who gave the -1. However I think you misunderstood OP question. Problem is G.node[route[i]] does not have 'position' as key at first iteration and all you suggested will fail with a 'KeyError'. – Bogdan Jan 18 '12 at 10:33
  • @Bogdan: I just assumed that he/she initialized that value before the loop, so I didn't thought of using `get` (like you did). I'll clear that now, thank you for your comment! – Rik Poggi Jan 18 '12 at 11:03
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Just use get:

    if G.node[route[i]].get('position') is None:
        G.node[route[i]]['position'] = list()
    else:
        G.node[route[i]]['position'].append(stuff)
Bogdan
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