On 11/10/2010 19.23, Alan Gauld wrote:
...
HTH,
Sure it did! Very enlightening, Alan. THANK YOU!
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"Francesco Loffredo" wrote
lst = []
for n in range(3):
obj = {}
I didn't know that this creates a new obj if obj already exists, I
thought it would just update it. That's my mistake.
Yes you have to remember that in Python, unlike C etc,
names are not aliases for memory locations. They are
Thank you, Alan and Dave, for your spotting this weak point in my
understanding of Python!
On 11/10/2010 2.11, Dave Angel wrote:
On 2:59 PM, Alan Gauld wrote:
"Francesco Loffredo" wrote
did, Roelof's code would work perfectly, and you could store in a list
all the subsequent changes of a d
On 2:59 PM, Alan Gauld wrote:
"Francesco Loffredo" wrote
did, Roelof's code would work perfectly, and you could store in a list
all the subsequent changes of a dictionary without calling them with
different names.
You don;'t need dfifferent names. Provided the name creates a
new object ins
"Francesco Loffredo" wrote
did, Roelof's code would work perfectly, and you could store in a
list
all the subsequent changes of a dictionary without calling them with
different names.
You don;'t need dfifferent names. Provided the name creates a
new object inside the loop you can reuse the
On 09/10/2010 10.25, Alan Gauld wrote:
"Francesco Loffredo" wrote
> On the next iteration you overwrite those two dictionaries
> with new values then append them to the list again.
> So you wind up with 2 copies of the updated dictionaries.
> ...
This is difficult for me too: why does this hap
"Francesco Loffredo" wrote
> On the next iteration you overwrite those two dictionaries
> with new values then append them to the list again.
> So you wind up with 2 copies of the updated dictionaries.
> ...
This is difficult for me too: why does this happen? Or, more
correctly,
why should th
On 09/10/2010 9.37, Steven D'Aprano wrote:
On Sat, 9 Oct 2010 06:05:57 pm Francesco Loffredo wrote:
Alan's answer to Roelof made me think...
I'm sorry, I don't know what your question is. You seem to have quoted
various bits and pieces of text from earlier emails (text beginning
with> signs).
On Sat, 9 Oct 2010 06:05:57 pm Francesco Loffredo wrote:
> Alan's answer to Roelof made me think...
I'm sorry, I don't know what your question is. You seem to have quoted
various bits and pieces of text from earlier emails (text beginning
with > signs). Apart from the sentence beginning with "Al
Alan's answer to Roelof made me think...
On 08/10/2010 13.40, Francesco Loffredo wrote:
Il 08/10/2010 10.02, Alan Gauld ha scritto:
"Roelof Wobben" wrote
I have this programm :
tournooi = [{'thuis': 'A','uit': "B",'thuisscore': 20, 'uitscore':
...
This was your answer to Roelof:
On the
Il 08/10/2010 10.02, Alan Gauld ha scritto:
"Roelof Wobben" wrote
I have this programm :
tournooi = [{'thuis': 'A','uit': "B",'thuisscore': 20, 'uitscore':
...
for wedstrijd in tournooi :
if wedstrijd['thuis'] in stand :
print "True"
stand is a list of dictionaries so this will never be
> To: tutor@python.org
> From: alan.ga...@btinternet.com
> Date: Fri, 8 Oct 2010 09:02:05 +0100
> Subject: Re: [Tutor] list of dict question
>
>
> "Roelof Wobben" wrote
>
>> I have this programm :
>>
>&
"Roelof Wobben" wrote
I have this programm :
tournooi = [{'thuis': 'A','uit': "B",'thuisscore': 20, 'uitscore':
15},{'thuis': 'C','uit': "D",'thuisscore': 80, 'uitscore': 40}]
stand = []
tussen_thuis = {}
tussen_uit = {}
Here you create your dictionary objects.
You never create any more d
Hello,
I have this programm :
tournooi = [{'thuis': 'A','uit': "B",'thuisscore': 20, 'uitscore':
15},{'thuis': 'C','uit': "D",'thuisscore': 80, 'uitscore': 40}]
stand = []
tussen_thuis = {}
tussen_uit = {}
for wedstrijd in tournooi :
if wedstrijd['thuis'] in stand :
print "True"
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