On Tue, 04 Jan 2005 21:57:46 +0100, Marc 'BlackJack' Rintsch
<[EMAIL PROTECTED]> wrote:
>In <[EMAIL PROTECTED]>, Bulba! wrote:
>
>> I put those dictionaries into the list:
>>
>>oldl=[x for x in orig] # where orig=csv.DictReader(ofile ...
>
>If you don't "do" anything with each `x` you can wr
Bulba! wrote:
[big snip]
Forget the csv-induced dicts for the moment, they're just an artifact
of your first solution attempt. Whether english = csv_row[1], or
english = csv_row_dict["english"], doesn't matter yet. Let's take a few
steps back, and look at what you are trying to do through a telesc
In <[EMAIL PROTECTED]>, Bulba! wrote:
> I put those dictionaries into the list:
>
>oldl=[x for x in orig] # where orig=csv.DictReader(ofile ...
If you don't "do" anything with each `x` you can write this as:
oldl = list(orig)
Ciao,
Marc 'BlackJack' Rintsch
--
http://mail.pyth
Skip Montanaro wrote:
...lotsa great stuff ...
You might want to sort your lists by the 'English' key. I don't know how to
use the new key arg to list.sort(), but you can still do it the
old-fashioned way:
oldl.sort(lambda a,b: cmp(a['English'], b['English']))
newl.sort(lambda a,b: cmp(a['
Bulba> I put those dictionaries into the list:
Bulba>oldl=[x for x in orig] # where orig=csv.DictReader(ofile ...
Bulba> ..and then search for matching source terms in two loops:
Bulba>for o in oldl:
Bulba>for n in newl:
Bulba>if n['English']
Bulba! wrote:
Hello everyone,
I'm reading the rows from a CSV file. csv.DictReader puts
those rows into dictionaries.
The actual files contain old and new translations of software
strings. The dictionary containing the row data looks like this:
o={'TermID':'4', 'English':'System Administration'
Hello everyone,
I'm reading the rows from a CSV file. csv.DictReader puts
those rows into dictionaries.
The actual files contain old and new translations of software
strings. The dictionary containing the row data looks like this:
o={'TermID':'4', 'English':'System Administration',
'Polish':
