"peter hodgson" <[EMAIL PROTECTED]> wrote
for key in b:
keylist = key.split(',')
if keylist < a:
i = a.index(keylist[0])
print a[:i] + [b[key]] + a[i+len(keylist):]
now, i get it, mostly; but keylist being shorter than the list "a"
keylist < a
True
This is not testing the length it is testing the contents.
a = ['a','c','e']
b = ['b','d','f']
a<b
True
b<a
False
It sees that 'a' < 'b' so returns True for a<b but
since 'b' is not less than 'a' it returns false for b<a
Now consider more closely aligned examples:
c = ['a','b','c']
c<a
True
Here the a's are equal so it checks the second element
and finds b < c so c<a is true
Finally for unequal length strings
d = ['a','c']
d<a
True
d<c
False
So d winds over a since the third element not existing
is taken to be lower valued than it existing. But the
character values in d<c a still win out regardless of length.
Does that make sense?
So in the original program
for key in b:
keylist = key.split(',')
if keylist < a:
i = a.index(keylist[0])
print a[:i] + [b[key]] + a[i+len(keylist):]
It is checking that it will find a valid index in the i assignment.
Having said that I'm not sure the check is fully effective since
I have a feeling it may fail for duplicated instances of the first
letter.... But to answer your question the list comparison is
more complex than just length.
And finally the a{:i] part is to cover the case where the
index i is greater than 0. So we retain the initial values up
to the matching key.
HTH,
--
Alan Gauld
Author of the Learn to Program web site
http://www.freenetpages.co.uk/hp/alan.gauld
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