In the following script, m1() and m2() work fine. I am assuming m2() is
faster although I haven't checked that (loops through the list twice instead
of once).
Now what I am trying to do is something like m3(). As currently written it
does not work, and I have tried different ways, but I haven't managed to
make it work.
Is there a possibility ? Or is m2() the optimum ?
Thanks.
#!/usr/bin/python
l = [ { 'colour': 'black', 'num': 0},
{ 'colour': 'brown', 'num': 1},
{ 'colour': 'red', 'num': 2},
{ 'colour': 'orange', 'num': 3},
{ 'colour': 'yellow', 'num': 4},
{ 'colour': 'green', 'num': 5},
{ 'colour': 'blue', 'num': 6},
{ 'colour': 'violet', 'num': 7},
{ 'colour': 'grey', 'num': 8},
{ 'colour': 'white', 'num': 9}
]
def m1():
colours = [ e['colour'] for e in l ]
nums = [ e['num'] for e in l ]
def m2():
colours = []
nums = []
for e in l:
colours.append(e['colour'])
nums.append(e['num'])
#def m3():
# colours, nums = [ e['colour'], e['num'] for e in l ]
--
Yves.
http://www.SollerS.ca
--
http://mail.python.org/mailman/listinfo/python-list
