Brtdku

I am putting data in a dictionary. I want to classify data per source adress. My data template look like this:

(timestamp, idsource,iddest, counter)

This is an example of my data:

('1547911869', '02141592cc00000002', '02141592cc00000001', '1'), 

('1547911869', '02141592cc00000003', '02141592cc00000004', '2'), 

('1547911869', '02141592cc00000004', '02141592cc00000005', '4'), 

in order to put them in a dictionary, I uuse these lines:

for timestamp, id_src_node ,id_dest_node ,Counter in list_info:

    if init_time == None:

        init_time = float(timestamp)



    if id_src_node not in dict_node_info:

        dict_node_info[id_src_node] = {"timestamp": [float(timestamp) - init_time], 'Counter': [int(1)], 'ID_Dest':[int(id_dest_node)]}

    else:

        dict_node_info[id_src_node]["timestamp"].append(float(timestamp) - init_time)

        dict_node_info[id_src_node]["Counter"].append(int(1))

        dict_node_info[id_src_node]["Counter"].append(int(id_dest_node))



return dict_node_info

But this strategy gives me this error:

dict_node_info[id_src_node] = {"timestamp": [float(timestamp) - init_time], 'Counter': [int(1)], 'ID_Dest':[int(id_dest_node)]}

ValueError: invalid literal for int() with base 10: '01ffff00ffffffffffffffffb00abeff'

If you want to convert your hex inputs to ints you need to tell int that the input is in base 16, as by default it will assume base 10.

for timestamp, id_src_node ,id_dest_node ,Counter in list_info:

    if init_time == None:

        init_time = float(timestamp)



    if id_src_node not in dict_node_info:

        dict_node_info[id_src_node] = {"timestamp": [float(timestamp) - init_time], 'Counter': [int(1)], 'ID_Dest':[int(id_dest_node, 16)]}

    else:

        dict_node_info[id_src_node]["timestamp"].append(float(timestamp) - init_time)

        dict_node_info[id_src_node]["Counter"].append(int(1))

        dict_node_info[id_src_node]["Counter"].append(int(id_dest_node, 16))



return dict_node_info

You got this error because you're trying to convert an hexadecimal string into an int base 10.
Basically your code goese through this part of code:

>> a = '01ffff00ffffffffffffffffb00abeff'

>> int(a)

ValueError: invalid literal for int() with base 10: '01ffff00ffffffffffffffffb00abeff'

So, like you can see you got a ValueError.

So, one way to bypass this problem is to create a handy function that checks if the passed string if it's a valid int or it contains an hexadecimal.

So, you can do something linke this:

def to_int(elm):

    """

    Need to check if there is another exceptions may occure 

    during the conversion of strings into integers

    """

    try:

        return int(elm)

    except ValueError:

        return int(elm, 16)





for timestamp, id_src_node ,id_dest_node ,Counter in list_info:

    if init_time == None:

        init_time = float(timestamp)



    if id_src_node not in dict_node_info:

        dict_node_info[id_src_node] = {"timestamp": [float(timestamp) - init_time], 'Counter': [int(1)], 'ID_Dest':[to_int(id_dest_node)]}

    else:

        dict_node_info[id_src_node]["timestamp"].append(float(timestamp) - init_time)

        dict_node_info[id_src_node]["Counter"].append(int(1))

        dict_node_info[id_src_node]["Counter"].append(to_int(id_dest_node))



return dict_node_info