converting to an integer

[Patrick C]

hey everyone, i'm having a mental block can anyone help me out

if i have a list like this
num = ['1;3;2;3;2;1;4' , '\n', '30;2;13;2;1;4; 2', '2;3;2;1;4;2;4' , '3;2;1;4;2;4;1' , '2;1;4;2;4;1;4' , '1;4;2;4;1;4;2']

how might i get python to recognize each elemant as an integer so that i can sum each list

that is have it print
16
\n
54
18

etc

thanks

[bvdet]

[code=Python]>>> num = ['1;3;2;3;2;1;4' , '\n', '30;2;13;2;1;4; 2', '2;3;2;1;4;2;4' , '3;2;1;4;2;4;1' , '2;1;4;2;4;1;4' , '1;4;2;4;1;4;2']
>>> s = '\n'.join([str(result) for result in [sum([int(i) for i in item.split(';')]) for item in num if item != '\n']])
>>> s
'16\n54\n18\n17 \n18\n18'
>>> print s
16
54
18
17
18
18
>>> [/code]

[Patrick C]

that works well, but i'm trying to keep a blank line if the '\n' was there to begin with, how can i tweak that so the solution would have been

'16\n\n54\n18\n 17\n18\n18'

thanks

[bvdet]

[code=Python]results = []
for item in num:
if item == '\n':
results.append( '')
else:
results.append( str(sum([int(i) for i in item.split(';')])))
print '\n'.join(resul ts)[/code]

>>> 16

54
18
17
18
18
>>> print repr('\n'.join(results))
'16\n\n54\n18\n17\n18\n18'
>>>

I got carried away with the list comprehension in my previous post. s could be:[code=Python]s = '\n'.join([str(sum([int(i) for i in item.split(';')])) for item in num if item != '\n'])[/code]

[Patrick C]

hey thanks for breaking it up onto many lines, i'm so fresh at this that when it's all squeezed into one line my mind breaks