pythonic way to sort

[micklee74@hotmail.com]

hi
I have a file with columns delimited by '~' like this:

1SOME STRING ~ABC~1231123243 2D~20060401~000 00000
2SOME STRING ~DEF~1353453454 3C~20060401~000 00000
3SOME STRING ~ACD~1435345355 4G~20060401~000 00000

......

What is the pythonic way to sort this type of structured text file?
Say i want to sort by 2nd column , ie ABC, ACD,DEF ? so that it becomes

1SOME STRING ~ABC~1231123243 2D~20060401~000 00000
3SOME STRING ~ACD~1435345355 4G~20060401~000 00000
2SOME STRING ~DEF~1353453454 3C~20060401~000 00000
?
I know for a start, that i have to split on '~', then append all the
second columns into a list, then sort the list using sort(), but i am
stuck with how to get the rest of the corresponding columns after the
sort....

thanks...

[Robert Kern]

Re: pythonic way to sort

micklee74@hotma il.com wrote:[color=blue]
> hi
> I have a file with columns delimited by '~' like this:
>
> 1SOME STRING ~ABC~1231123243 2D~20060401~000 00000
> 2SOME STRING ~DEF~1353453454 3C~20060401~000 00000
> 3SOME STRING ~ACD~1435345355 4G~20060401~000 00000
>
> .....
>
> What is the pythonic way to sort this type of structured text file?
> Say i want to sort by 2nd column , ie ABC, ACD,DEF ? so that it becomes
>
> 1SOME STRING ~ABC~1231123243 2D~20060401~000 00000
> 3SOME STRING ~ACD~1435345355 4G~20060401~000 00000
> 2SOME STRING ~DEF~1353453454 3C~20060401~000 00000
> ?
> I know for a start, that i have to split on '~', then append all the
> second columns into a list, then sort the list using sort(), but i am
> stuck with how to get the rest of the corresponding columns after the
> sort....[/color]

In Python 2.4 and up, you can use the key= keyword to list.sort(). E.g.

In [2]: text = """1SOME STRING ~ABC~1231123243 2D~20060401~000 00000
...: 2SOME STRING ~DEF~1353453454 3C~20060401~000 00000
...: 3SOME STRING ~ACD~1435345355 4G~20060401~000 00000"""

In [3]: lines = text.split('\n' )

In [4]: lines
Out[4]:
['1SOME STRING ~ABC~1231123243 2D~20060401~000 00000',
'2SOME STRING ~DEF~1353453454 3C~20060401~000 00000',
'3SOME STRING ~ACD~1435345355 4G~20060401~000 00000']

In [5]: lines.sort(key= lambda x: x.split('~')[1])

In [6]: lines
Out[6]:
['1SOME STRING ~ABC~1231123243 2D~20060401~000 00000',
'3SOME STRING ~ACD~1435345355 4G~20060401~000 00000',
'2SOME STRING ~DEF~1353453454 3C~20060401~000 00000']

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

[Jay Parlar]

Re: pythonic way to sort


On May 4, 2006, at 12:12 AM, micklee74@hotma il.com wrote:
[color=blue]
> hi
> I have a file with columns delimited by '~' like this:
>
> 1SOME STRING ~ABC~1231123243 2D~20060401~000 00000
> 2SOME STRING ~DEF~1353453454 3C~20060401~000 00000
> 3SOME STRING ~ACD~1435345355 4G~20060401~000 00000
>
> .....
>
> What is the pythonic way to sort this type of structured text file?
> Say i want to sort by 2nd column , ie ABC, ACD,DEF ? so that it becomes
>
> 1SOME STRING ~ABC~1231123243 2D~20060401~000 00000
> 3SOME STRING ~ACD~1435345355 4G~20060401~000 00000
> 2SOME STRING ~DEF~1353453454 3C~20060401~000 00000
> ?
> I know for a start, that i have to split on '~', then append all the
> second columns into a list, then sort the list using sort(), but i am
> stuck with how to get the rest of the corresponding columns after the
> sort....
>
> thanks...
>[/color]

A couple ways. Assume that you have the lines in a list called 'lines',
as follows:

lines = [
"1SOME STRING ~ABC~1231123243 2D~20060401~000 00000",
"3SOME STRING ~ACD~1435345355 4G~20060401~000 00000",
"2SOME STRING ~DEF~1353453454 3C~20060401~000 00000"]


The more traditional way would be to define your own comparison
function:

def my_cmp(x,y):
return cmp( x.split("~")[1], y.split("~")[1])

lines.sort(cmp= my_cmp)


The newer, faster way, would be to define your own key function:

def my_key(x):
return x.split("~")[1]

lines.sort(key= my_key)


The key function is faster because you only have to do the
split("~")[1] once for each line, whereas it will be done many times
for each line if you use a comparison function.

Jay P.

[Boris Borcic]

Re: pythonic way to sort

Jay Parlar wrote:[color=blue]
>
> On May 4, 2006, at 12:12 AM, micklee74@hotma il.com wrote:
> [...][/color]
[color=blue]
> Assume that you have the lines in a list called 'lines',
> as follows:
>
> lines = [
> "1SOME STRING ~ABC~1231123243 2D~20060401~000 00000",
> "3SOME STRING ~ACD~1435345355 4G~20060401~000 00000",
> "2SOME STRING ~DEF~1353453454 3C~20060401~000 00000"]
>
>
> The more traditional way would be to define your own comparison function:
>
> def my_cmp(x,y):
> return cmp( x.split("~")[1], y.split("~")[1])
>
> lines.sort(cmp= my_cmp)
>
>
> The newer, faster way, would be to define your own key function:
>
> def my_key(x):
> return x.split("~")[1]
>
> lines.sort(key= my_key)[/color]

and if the data is in a file rather than a list, you may write eg

lines = sorted(file("/path/tofile"),key=mi ke)

to create it sorted.