Re: looking for a neat solution to a nested loop problem

2012-08-14 Thread 88888 Dihedral 
Nobody於 2012年8月7日星期二UTC+8下午11時32分55秒寫道：
 On Mon, 06 Aug 2012 21:02:33 -0700, Larry Hudson wrote:
 
 
 
for i in range(N,N+100):
 
for j in range(M,M+100):
 
do_something(i % 100 ,j % 100)
 
 
 
  Emile
 
  
 
  How about...
 
  
 
  for i in range(100):
 
   for j in range(100):
 
   do_something((i + N) % 100, (j + M) % 100)
 
 
 
 Both of these approaches move the modifications to the sequence into the
 
 body of the loop. It may be preferable to iterate over the desired
 
 sequence directly. E.g.
 
 
 
   for i in ((N + ii) % 100 for ii in xrange(100)):
 
   for j in ((M + jj) % 100 for jj in xrange(100)):
 
   do_something(i, j)

I'll contrubute another version by the xrange function in python.

for i in xrange(N, N+100): # not list based
if i100: i2=i else: i2=i-100 # not in the inner most loop
for j in xrange(M, M+100):
if j100: j2=j else: j2=j-100
do_something(i2,j2)



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Re: looking for a neat solution to a nested loop problem

2012-08-09 Thread 88888 Dihedral 
Nobody於 2012年8月7日星期二UTC+8下午11時32分55秒寫道：
 On Mon, 06 Aug 2012 21:02:33 -0700, Larry Hudson wrote:
 
 
 
for i in range(N,N+100):
 
for j in range(M,M+100):
 
do_something(i % 100 ,j % 100)
 
 
 
  Emile
 
  
 
  How about...
 
  
 
  for i in range(100):
 
   for j in range(100):
 
   do_something((i + N) % 100, (j + M) % 100)
 
 
 
 Both of these approaches move the modifications to the sequence into the
 
 body of the loop. It may be preferable to iterate over the desired
 
 sequence directly. E.g.
 
 
 
   for i in ((N + ii) % 100 for ii in xrange(100)):
 
   for j in ((M + jj) % 100 for jj in xrange(100)):
 
   do_something(i, j)

list(range(N,100))+ list(range(0,N))  //good for j

Well, this is the way!
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Re: looking for a neat solution to a nested loop problem

2012-08-09 Thread 88888 Dihedral 
Nobody於 2012年8月7日星期二UTC+8下午11時32分55秒寫道：
 On Mon, 06 Aug 2012 21:02:33 -0700, Larry Hudson wrote:
 
 
 
for i in range(N,N+100):
 
for j in range(M,M+100):
 
do_something(i % 100 ,j % 100)
 
 
 
  Emile
 
  
 
  How about...
 
  
 
  for i in range(100):
 
   for j in range(100):
 
   do_something((i + N) % 100, (j + M) % 100)
 
 
 
 Both of these approaches move the modifications to the sequence into the
 
 body of the loop. It may be preferable to iterate over the desired
 
 sequence directly. E.g.
 
 
 
   for i in ((N + ii) % 100 for ii in xrange(100)):
 
   for j in ((M + jj) % 100 for jj in xrange(100)):
 
   do_something(i, j)



Nobody於 2012年8月7日星期二UTC+8下午11時32分55秒寫道：
 On Mon, 06 Aug 2012 21:02:33 -0700, Larry Hudson wrote:
 
 
 
for i in range(N,N+100):
 
for j in range(M,M+100):
 
do_something(i % 100 ,j % 100)
 
 
 
  Emile
 
  
 
  How about...
 
  
 
  for i in range(100):
 
   for j in range(100):
 
   do_something((i + N) % 100, (j + M) % 100)
 
 
 
 Both of these approaches move the modifications to the sequence into the
 
 body of the loop. It may be preferable to iterate over the desired
 
 sequence directly. E.g.
 
 
 
   for i in ((N + ii) % 100 for ii in xrange(100)):
 
   for j in ((M + jj) % 100 for jj in xrange(100))

list(range(N,100))+list(range(1,N)) //good for  i

I contribute my python code here to avoid any divison.
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Re: looking for a neat solution to a nested loop problem

2012-08-07 Thread Nobody 
On Mon, 06 Aug 2012 21:02:33 -0700, Larry Hudson wrote:

   for i in range(N,N+100):
   for j in range(M,M+100):
   do_something(i % 100 ,j % 100)

 Emile
 
 How about...
 
 for i in range(100):
  for j in range(100):
  do_something((i + N) % 100, (j + M) % 100)

Both of these approaches move the modifications to the sequence into the
body of the loop. It may be preferable to iterate over the desired
sequence directly. E.g.

for i in ((N + ii) % 100 for ii in xrange(100)):
for j in ((M + jj) % 100 for jj in xrange(100)):
do_something(i, j)

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Re: looking for a neat solution to a nested loop problem

2012-08-07 Thread 88888 Dihedral 
Nobody於 2012年8月7日星期二UTC+8下午11時32分55秒寫道：
 On Mon, 06 Aug 2012 21:02:33 -0700, Larry Hudson wrote:
 
 
 
for i in range(N,N+100):
 
for j in range(M,M+100):
 
do_something(i % 100 ,j % 100)
 
 
 
  Emile
 
  
 
  How about...
 
  
 
  for i in range(100):
 
   for j in range(100):
 
   do_something((i + N) % 100, (j + M) % 100)
 
 
 
 Both of these approaches move the modifications to the sequence into the
 
 body of the loop. It may be preferable to iterate over the desired
 
 sequence directly. E.g.
 
 
 
   for i in ((N + ii) % 100 for ii in xrange(100)):
 
   for j in ((M + jj) % 100 for jj in xrange(100)):
 
   do_something(i, j)



Nobody於 2012年8月7日星期二UTC+8下午11時32分55秒寫道：
 On Mon, 06 Aug 2012 21:02:33 -0700, Larry Hudson wrote:
 
 
 
for i in range(N,N+100):
 
for j in range(M,M+100):
 
do_something(i % 100 ,j % 100)
 
 
 
  Emile
 
  
 
  How about...
 
  
 
  for i in range(100):
 
   for j in range(100):
 
   do_something((i + N) % 100, (j + M) % 100)
 
 
 
 Both of these approaches move the modifications to the sequence into the
 
 body of the loop. It may be preferable to iterate over the desired
 
 sequence directly. E.g.
 
 
 
   for i in ((N + ii) % 100 for ii in xrange(100)):
 
   for j in ((M + jj) % 100 for jj in xrange(100)):
 
   do_something(i, j)

This is a good example to be tuned into some example 
 such that  this kind of loops by iterators of parameters in python 
wich do not use any division at all.

But I am getting lazy recently for non-critical parts.
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looking for a neat solution to a nested loop problem

2012-08-06 Thread Tom P 

consider a nested loop algorithm -

for i in range(100):
for j in range(100):
do_something(i,j)

Now, suppose I don't want to use i = 0 and j = 0 as initial values, but 
some other values i = N and j = M, and I want to iterate through all 
10,000 values in sequence - is there a neat python-like way to this? I 
realize I can do things like use a variable for k in range(1): and 
then derive values for i and j from k, but I'm wondering if there's 
something less clunky.

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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Oscar Benjamin 
Are you familiar with the itertools module?

itertools.product is designed for this purpose:
http://docs.python.org/library/itertools#itertools.product

Oscar.

On 6 August 2012 16:52, Tom P werot...@freent.dd wrote:

 consider a nested loop algorithm -

 for i in range(100):
 for j in range(100):
 do_something(i,j)

 Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
 some other values i = N and j = M, and I want to iterate through all 10,000
 values in sequence - is there a neat python-like way to this? I realize I
 can do things like use a variable for k in range(1): and then derive
 values for i and j from k, but I'm wondering if there's something less
 clunky.
 --
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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Oscar Benjamin 
On 6 August 2012 16:52, Tom P werot...@freent.dd wrote:

 consider a nested loop algorithm -

 for i in range(100):
 for j in range(100):
 do_something(i,j)

 Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
 some other values i = N and j = M, and I want to iterate through all 10,000
 values in sequence - is there a neat python-like way to this? I realize I
 can do things like use a variable for k in range(1): and then derive
 values for i and j from k, but I'm wondering if there's something less
 clunky.
 --
 http://mail.python.org/**mailman/listinfo/python-listhttp://mail.python.org/mailman/listinfo/python-list


Are you familiar with the itertools module?

itertools.product is designed for this purpose:
http://docs.python.org/library/itertools#itertools.product
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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Ethan Furman 

Tom P wrote:

consider a nested loop algorithm -

for i in range(100):
for j in range(100):
do_something(i,j)

Now, suppose I don't want to use i = 0 and j = 0 as initial values, but 
some other values i = N and j = M, and I want to iterate through all 
10,000 values in sequence - is there a neat python-like way to this? I 
realize I can do things like use a variable for k in range(1): and 
then derive values for i and j from k, but I'm wondering if there's 
something less clunky.


for i in range(N, N+100):
for j in range(M, M+100):
do_something(i, j)


~Ethan~
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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread John Gordon 
In a8a7hvf8c...@mid.individual.net Tom P werot...@freent.dd writes:

 consider a nested loop algorithm -

 for i in range(100):
  for j in range(100):
  do_something(i,j)

 Now, suppose I don't want to use i = 0 and j = 0 as initial values, but 
 some other values i = N and j = M, and I want to iterate through all 
 10,000 values in sequence - is there a neat python-like way to this? I 
 realize I can do things like use a variable for k in range(1): and 
 then derive values for i and j from k, but I'm wondering if there's 
 something less clunky.

You could define your own generator function that yields values
in whatever order you want:

def my_generator():
yield 9
yield 100
for i in range(200, 250):
yield i
yield 5


-- 
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gor...@panix.com  B is for Basil, assaulted by bears
-- Edward Gorey, The Gashlycrumb Tinies

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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Ian Foote 

The function range can be called with more than one argument. For example:

for i in range(N, N + 10):
for j in range(M, M + 100):
do_something(i, j)

You can also call range with 3 arguments, if want a step size different 
to 1:


for k in range(2, 11, 3):
print(k)

2
5
8

Hope this is clear,
Ian

On 06/08/12 16:52, Tom P wrote:

consider a nested loop algorithm -

for i in range(100):
for j in range(100):
do_something(i,j)

Now, suppose I don't want to use i = 0 and j = 0 as initial values, 
but some other values i = N and j = M, and I want to iterate through 
all 10,000 values in sequence - is there a neat python-like way to 
this? I realize I can do things like use a variable for k in 
range(1): and then derive values for i and j from k, but I'm 
wondering if there's something less clunky.


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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Nobody 
On Mon, 06 Aug 2012 17:52:31 +0200, Tom P wrote:

 consider a nested loop algorithm -
 
 for i in range(100):
  for j in range(100):
  do_something(i,j)
 
 Now, suppose I don't want to use i = 0 and j = 0 as initial values, but 
 some other values i = N and j = M, and I want to iterate through all 
 10,000 values in sequence - is there a neat python-like way to this?

for i in range(N,N+100):
for j in range(M,M+100):
do_something(i,j)

Or did you mean something else?

Alternatively:

import itertools

for i, j in itertools.product(range(N,N+100),range(M,M+100)):
do_something(i,j)

This can be preferable to deeply-nested loops.

Also: in 2.x, use xrange() in preference to range().

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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Tom P 

On 08/06/2012 06:18 PM, Nobody wrote:

On Mon, 06 Aug 2012 17:52:31 +0200, Tom P wrote:


consider a nested loop algorithm -

for i in range(100):
  for j in range(100):
  do_something(i,j)

Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
some other values i = N and j = M, and I want to iterate through all
10,000 values in sequence - is there a neat python-like way to this?


for i in range(N,N+100):
for j in range(M,M+100):
do_something(i,j)

Or did you mean something else?


no, I meant something else ..

  j runs through range(M, 100) and then range(0,M), and i runs through 
range(N,100) and then range(0,N)


.. apologies if I didn't make that clear enough.



Alternatively:

import itertools

for i, j in itertools.product(range(N,N+100),range(M,M+100)):
do_something(i,j)

This can be preferable to deeply-nested loops.

Also: in 2.x, use xrange() in preference to range().



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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Tom P 

On 08/06/2012 06:03 PM, John Gordon wrote:

In a8a7hvf8c...@mid.individual.net Tom P werot...@freent.dd writes:


consider a nested loop algorithm -



for i in range(100):
  for j in range(100):
  do_something(i,j)



Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
some other values i = N and j = M, and I want to iterate through all
10,000 values in sequence - is there a neat python-like way to this? I
realize I can do things like use a variable for k in range(1): and
then derive values for i and j from k, but I'm wondering if there's
something less clunky.


You could define your own generator function that yields values
in whatever order you want:

def my_generator():
 yield 9
 yield 100
 for i in range(200, 250):
 yield i
 yield 5


Thanks, I'll look at that but I think it just moves the clunkiness from 
one place in the code to another.


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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Oscar Benjamin 
On 6 August 2012 18:14, Tom P werot...@freent.dd wrote:

 On 08/06/2012 06:18 PM, Nobody wrote:

 On Mon, 06 Aug 2012 17:52:31 +0200, Tom P wrote:

  consider a nested loop algorithm -

 for i in range(100):
   for j in range(100):
   do_something(i,j)

 Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
 some other values i = N and j = M, and I want to iterate through all
 10,000 values in sequence - is there a neat python-like way to this?


 for i in range(N,N+100):
 for j in range(M,M+100):
 do_something(i,j)

 Or did you mean something else?


 no, I meant something else ..

   j runs through range(M, 100) and then range(0,M), and i runs through
 range(N,100) and then range(0,N)

 .. apologies if I didn't make that clear enough.


How about range(N, 100) + range(0, N)?

Example (Python 2.x):

 range(3, 10)
[3, 4, 5, 6, 7, 8, 9]
 range(0, 3)
[0, 1, 2]
 range(3, 10) + range(0, 3)
[3, 4, 5, 6, 7, 8, 9, 0, 1, 2]

In Python 3.x you'd need to do list(range(...)) + list(range(...)) or use
itertools.chain.

Oscar
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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Emile van Sebille 

On 8/6/2012 10:14 AM Tom P said...

On 08/06/2012 06:18 PM, Nobody wrote:

On Mon, 06 Aug 2012 17:52:31 +0200, Tom P wrote:


consider a nested loop algorithm -

for i in range(100):
  for j in range(100):
  do_something(i,j)

Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
some other values i = N and j = M, and I want to iterate through all
10,000 values in sequence - is there a neat python-like way to this?


for i in range(N,N+100):
for j in range(M,M+100):
do_something(i,j)

Or did you mean something else?


no, I meant something else ..

   j runs through range(M, 100) and then range(0,M), and i runs through
range(N,100) and then range(0,N)

.. apologies if I didn't make that clear enough.



 for i in range(N,N+100):
 for j in range(M,M+100):
 do_something(i % 100 ,j % 100)

Emile



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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread André Malo 
* Tom P wrote:

 consider a nested loop algorithm -
 
 for i in range(100):
  for j in range(100):
  do_something(i,j)

 
 Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
 some other values i = N and j = M, and I want to iterate through all
 10,000 values in sequence - is there a neat python-like way to this? I
 realize I can do things like use a variable for k in range(1): and
 then derive values for i and j from k, but I'm wondering if there's
 something less clunky.

you mean:
do_something((i + N) % 100, (j + M) % 100)

?

I'd define my own range function doing exactly that.

def rrange(count, start=0):
for j in xrange(count):
yield (j + start) % count

(untested)

Or use some itertools magic for that. It might be faster.

nd
-- 
Umfassendes Werk (auch fuer Umsteiger vom Apache 1.3)
  -- aus einer Rezension

http://pub.perlig.de/books.html#apache2
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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Grant Edwards 
On 2012-08-06, Grant Edwards invalid@invalid.invalid wrote:
 On 2012-08-06, Tom P werot...@freent.dd wrote:
 On 08/06/2012 06:18 PM, Nobody wrote:
 On Mon, 06 Aug 2012 17:52:31 +0200, Tom P wrote:

 consider a nested loop algorithm -

 for i in range(100):
   for j in range(100):
   do_something(i,j)

 Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
 some other values i = N and j = M, and I want to iterate through all
 10,000 values in sequence - is there a neat python-like way to this?

 for i in range(N,N+100):
 for j in range(M,M+100):
 do_something(i,j)

 Or did you mean something else?

 no, I meant something else ..

j runs through range(M, 100) and then range(0,M), and i runs through 
 range(N,100) and then range(0,N)

 In 2.x:

 for i in range(M,100)+range(0,M):
 for j in range(N,100)+range(0,N):
 do_something(i,j)

 Dunno if that still works in 3.x.  I doubt it, since I think in 3.x
 range returns an iterator, not?

Indeed it doesn't work in 3.x, but this does:

from itertools import chain

for i in chain(range(M,100),range(0,M)):
for j in chain(range(N,100),range(0,N)):
do_something(i,j)


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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Grant Edwards 
On 2012-08-06, Tom P werot...@freent.dd wrote:
 On 08/06/2012 06:18 PM, Nobody wrote:
 On Mon, 06 Aug 2012 17:52:31 +0200, Tom P wrote:

 consider a nested loop algorithm -

 for i in range(100):
   for j in range(100):
   do_something(i,j)

 Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
 some other values i = N and j = M, and I want to iterate through all
 10,000 values in sequence - is there a neat python-like way to this?

  for i in range(N,N+100):
  for j in range(M,M+100):
  do_something(i,j)

 Or did you mean something else?

 no, I meant something else ..

j runs through range(M, 100) and then range(0,M), and i runs through 
 range(N,100) and then range(0,N)

In 2.x:

for i in range(M,100)+range(0,M):
for j in range(N,100)+range(0,N):
do_something(i,j)

Dunno if that still works in 3.x.  I doubt it, since I think in 3.x
range returns an iterator, not?

-- 
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  at   Cincinnati street corner
  gmail.comholding a clean dog!
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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Tom P 

On 08/06/2012 08:29 PM, Grant Edwards wrote:

On 2012-08-06, Grant Edwards invalid@invalid.invalid wrote:

On 2012-08-06, Tom P werot...@freent.dd wrote:

On 08/06/2012 06:18 PM, Nobody wrote:

On Mon, 06 Aug 2012 17:52:31 +0200, Tom P wrote:


consider a nested loop algorithm -

for i in range(100):
   for j in range(100):
   do_something(i,j)

Now, suppose I don't want to use i = 0 and j = 0 as initial values, but
some other values i = N and j = M, and I want to iterate through all
10,000 values in sequence - is there a neat python-like way to this?


for i in range(N,N+100):
for j in range(M,M+100):
do_something(i,j)

Or did you mean something else?


no, I meant something else ..

j runs through range(M, 100) and then range(0,M), and i runs through
range(N,100) and then range(0,N)


In 2.x:

 for i in range(M,100)+range(0,M):
 for j in range(N,100)+range(0,N):
 do_something(i,j)

Dunno if that still works in 3.x.  I doubt it, since I think in 3.x
range returns an iterator, not?


Indeed it doesn't work in 3.x, but this does:

 from itertools import chain

 for i in chain(range(M,100),range(0,M)):
 for j in chain(range(N,100),range(0,N)):
 do_something(i,j)



 ah, that looks good - I guess it works in 2.x as well?
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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Grant Edwards 
On 2012-08-06, Tom P werot...@freent.dd wrote:

 no, I meant something else ..

 j runs through range(M, 100) and then range(0,M), and i runs through
 range(N,100) and then range(0,N)

 In 2.x:

  for i in range(M,100)+range(0,M):
  for j in range(N,100)+range(0,N):
  do_something(i,j)

 Dunno if that still works in 3.x.  I doubt it, since I think in 3.x
 range returns an iterator, not?

 Indeed it doesn't work in 3.x, but this does:

  from itertools import chain

  for i in chain(range(M,100),range(0,M)):
  for j in chain(range(N,100),range(0,N)):
  do_something(i,j)

   ah, that looks good - I guess it works in 2.x as well?

I don't know.  Let me test that for you...

   $ python
   Python 2.6.8 (unknown, May 18 2012, 11:56:26) 
   [GCC 4.5.3] on linux2
   Type help, copyright, credits or license for more information.
from itertools import chain
for i in chain(range(0,5),range(5,10)):
   ...   print i
   ... 
   0
   1
   2
   3
   4
   5
   6
   7
   8
   9


Yes, it works in 2.x as well.

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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Emile van Sebille 

On 8/6/2012 12:22 PM Grant Edwards said...

On 2012-08-06, Tom P werot...@freent.dd wrote:

snip

   ah, that looks good - I guess it works in 2.x as well?


I don't know.  Let me test that for you...


snip


Yes, it works in 2.x as well.



:)

And from the docs, all the way back to 2.3!

9.7. itertools  Functions creating iterators for efficient looping
New in version 2.3.

Emile

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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Arnaud Delobelle 
On 6 August 2012 16:52, Tom P werot...@freent.dd wrote:
 consider a nested loop algorithm -

 for i in range(100):
 for j in range(100):
 do_something(i,j)

 Now, suppose I don't want to use i = 0 and j = 0 as initial values, but some
 other values i = N and j = M, and I want to iterate through all 10,000
 values in sequence - is there a neat python-like way to this? I realize I
 can do things like use a variable for k in range(1): and then derive
 values for i and j from k, but I'm wondering if there's something less
 clunky.

For example:

for i in range(100):
for j in range(100):
do_something((i + N)%100, (j + N)%100)

Cheers,

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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Steven D'Aprano 
On Mon, 06 Aug 2012 19:16:45 +0200, Tom P wrote:

 def my_generator():
  yield 9
  yield 100
  for i in range(200, 250):
  yield i
  yield 5


 Thanks, I'll look at that but I think it just moves the clunkiness from
 one place in the code to another.

And if there was a built-in command that did exactly what you wanted, it 
too would also move the clunkiness from one place in the code to another.

What you are asking for is clunky:

[quote]
j runs through range(M, 100) and then range(0,M), and 
i runs through range(N,100) and then range(0,N)
[end quote]


There's no magic pixie dust that you can sprinkle on it to make it 
elegant. Assuming M and N are small (under 100), you can do this:

values = range(100)  # or list(range(100)) in Python 3.
for j in (values[M:] + values[:M]):
for i in (values[N:] + values[:N]):
...

which isn't too bad. If you have to deal with much large ranges, you can 
use itertools to chain them together:

import itertools
jvalues = itertools.chain(xrange(M, 100), xrange(M))  # or just range
ivalues = itertools.chain(xrange(N, 250), xrange(N))  # in Python 3
for j in jvalues:
for i in ivalues:
...



-- 
Steven
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Re: looking for a neat solution to a nested loop problem

2012-08-06 Thread Larry Hudson 

On 08/06/2012 11:11 AM, Emile van Sebille wrote:
   snip


  for i in range(N,N+100):
  for j in range(M,M+100):
  do_something(i % 100 ,j % 100)

Emile


How about...

for i in range(100):
for j in range(100):
do_something((i + N) % 100, (j + M) % 100)

 -=- Larry -=-


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Loop problem while generating a new value with random.randint()

2010-02-15 Thread Paulo Repreza 
Greetings,

I'm having problems with a little script that I'm trying to finish, I don't
know if I'm in the right track but I know somebody is going to help me.

The script:

# Import modules random for function randint

import random

# Generating a constant.

var = 65

# Generating a random number.
ranum = random.randint(1,100)

#Creating while loop. Stops until var == ranum
while var != ranum:
if var == ranum:
print var, 'equals to:', ranum
else:
print var, 'does not equal to:', ranum

## End of Script ###


What I'm trying to do is to print the new value that ranum generates if the
condition is not met. So far if you run the script it prints the same value
over and over again, making in an infinite loop. What can I do in order to
print out the new value generated every time the condition is not met?

Thanks!

Paulo Repreza
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Re: Loop problem while generating a new value with random.randint()

2010-02-15 Thread Jean-Michel Pichavant 

Paulo Repreza wrote:

Greetings,

I'm having problems with a little script that I'm trying to finish, I 
don't know if I'm in the right track but I know somebody is going to 
help me.


The script:

# Import modules random for function randint

import random

# Generating a constant.

var = 65

# Generating a random number.
ranum = random.randint(1,100)

#Creating while loop. Stops until var == ranum
while var != ranum:
if var == ranum:
print var, 'equals to:', ranum
else:
print var, 'does not equal to:', ranum

## End of Script ###


What I'm trying to do is to print the new value that ranum generates 
if the condition is not met. So far if you run the script it prints 
the same value over and over again, making in an infinite loop. What 
can I do in order to print out the new value generated every time the 
condition is not met?


Thanks!

Paulo Repreza   
in your script you generate the random number only once, no wonder it 
keep being the same value over  over.


# Initialize ranum with a random number
ranum = random.randint(1,100)

#Creating while loop. Stops until var == ranum
while var != ranum:
   if var == ranum:
   print var, 'equals to:', ranum
   else:
   print var, 'does not equal to:', ranum
   ranum = random.randint(1,100) # generate a new number, that's the 
missing line in your script


JM

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Re: Loop problem while generating a new value with random.randint()

2010-02-15 Thread Bruno Desthuilliers 

Jean-Michel Pichavant a écrit :

Paulo Repreza wrote:

Greetings,

I'm having problems with a little script that I'm trying to finish, I 
don't know if I'm in the right track but I know somebody is going to 
help me.


(snip - problem already addressed by Jean-Michel...)


while var != ranum:
if var == ranum:


Note that this branch will never be executed - the condition in the 
while statement make sure var != ranum in the while block.



print var, 'equals to:', ranum
else:
print var, 'does not equal to:', ranum


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Re: Loop problem while generating a new value with random.randint()

2010-02-15 Thread Arnaud Delobelle 
Jean-Michel Pichavant jeanmic...@sequans.com writes:

 Paulo Repreza wrote:
 Greetings,

 I'm having problems with a little script that I'm trying to finish,
 I don't know if I'm in the right track but I know somebody is going
 to help me.

 The script:

 # Import modules random for function randint

 import random

 # Generating a constant.

 var = 65

 # Generating a random number.
 ranum = random.randint(1,100)

 #Creating while loop. Stops until var == ranum
 while var != ranum:
 if var == ranum:
 print var, 'equals to:', ranum
 else:
 print var, 'does not equal to:', ranum

 ## End of Script ###


 What I'm trying to do is to print the new value that ranum generates
 if the condition is not met. So far if you run the script it prints
 the same value over and over again, making in an infinite loop. What
 can I do in order to print out the new value generated every time
 the condition is not met?

 Thanks!

 Paulo Repreza   
 in your script you generate the random number only once, no wonder it
 keep being the same value over  over.

 # Initialize ranum with a random number
 ranum = random.randint(1,100)

 #Creating while loop. Stops until var == ranum
 while var != ranum:
if var == ranum:
print var, 'equals to:', ranum
else:
print var, 'does not equal to:', ranum
ranum = random.randint(1,100) # generate a new number, that's the
 missing line in your script

 JM

And a more idiomatic way of writing this in Python would be, I guess:

import random

var = 65
while True:
ranum = random.randint(1, 100)
if var == ranum:
print var, 'equals to:', ranum
break
else:
print var, 'does not equal to:', ranum

(following up from a FU as I can't see the OP)

-- 
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Re: Loop problem while generating a new value with random.randint()

2010-02-15 Thread Steve Holden 
Paulo Repreza wrote:
 Greetings,
 
 I'm having problems with a little script that I'm trying to finish, I
 don't know if I'm in the right track but I know somebody is going to
 help me.
 
 The script:
 
 # Import modules random for function randint
 
 import random
 
 # Generating a constant.
 
 var = 65
 
 # Generating a random number.
 ranum = random.randint(1,100)
 
 #Creating while loop. Stops until var == ranum
 while var != ranum:
 if var == ranum:

The if condition can never be true, since if it were then the loop
would have just terminated before starting this iteration!

 print var, 'equals to:', ranum
 else:
 print var, 'does not equal to:', ranum
 
Shouldn't there be something here to set ranum to a new value? Otherwise
the same value will be there the next time around ...

 ## End of Script ###
 
 
 What I'm trying to do is to print the new value that ranum generates if
 the condition is not met. So far if you run the script it prints the
 same value over and over again, making in an infinite loop. What can I
 do in order to print out the new value generated every time the
 condition is not met?
 
 Thanks!

regards
 Steve
-- 
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PyCon is coming! Atlanta, Feb 2010  http://us.pycon.org/
Holden Web LLC http://www.holdenweb.com/
UPCOMING EVENTS:http://holdenweb.eventbrite.com/
d

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Re: Loop problem while generating a new value with random.randint()

2010-02-15 Thread Paulo Repreza 
Thanks for the help!
Using while True helps alot!


Paulo Repreza
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Re: newb loop problem

2008-08-14 Thread Dave 
On Aug 13, 8:46 am, Sion Arrowsmith [EMAIL PROTECTED]
wrote:
 Dave  [EMAIL PROTECTED] wrote:
 hitNum = 0
 stopCnt = 6 + hitNum
 offSet = 5

 for i in range(0,10,1):

 The step argument to range defaults to 1: it's tidier to omit it.
 Similarly, the start argument defaults to 0, so you can drop that too.

 for i in range(10):

 for x in range(hitNum,len(inLst), 1):
 print hitNum, stopCnt

 hitNum and stopCnt are constant in this loop: if you care about
 this print statement, move it into the outer loop and stop yourself
 drowning in output.

 if x == stopCnt: break

 If you want to exit the inner loop when x == stopCnt, why not make
 that condition part of the loop construct?

 for x in range(hitNum, stopCnt):

 That said, if you ever see for i in range(len(lst)) *immediately*
 replace it by for i, x in enumerate(lst), then go through to body
 to see if you really need that i, and if not use for x in lst,
 with slicing if the range is more complex than range(len(lst)). As
 you can do here:

 for x in inLst[hitNum:stopCnt]:
 hitLst.append(x)

 And if all you're doing in a for loop is appending to one list from
 another, that's just what list.extend does:

 hitLst.extend(inLst[hitNum:stopCnt])

 hitNum +=offSet
 stopCnt+=offSet

 Finally, that i in the outer loop isn't being used anywhere. Why
 don't you just loop over hitNum? And noticing that stopCnt is
 increasing in lock-step with hitNum:

 offset = 5

 for hitNum in range(0, 10*offset, offset):
 hitLst.extend(inLst[hitNum:hitNum+6]

 --
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Frankly I have no feelings towards penguins one way or the other
 -- Arthur C. Clarke
her nu becomeþ se bera eadward ofdun hlæddre heafdes bæce bump bump bump

Excellent, thanks for all your help guys.
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Re: newb loop problem

2008-08-13 Thread Dave 
arrrggg, now I feel really dumb..

hitNum = 0
stopCnt = 6 + hitNum
offSet = 5

for i in range(0,10,1):

for x in range(hitNum,len(inLst), 1):
print hitNum, stopCnt
if x == stopCnt: break
hitLst.append(inLst[x])
hitNum +=offSet
stopCnt+=offSet
print hitLst


Beers, Dave


On Aug 13, 12:58 am, Dave [EMAIL PROTECTED] wrote:
 On Aug 13, 12:35 am, Larry Bates [EMAIL PROTECTED] wrote:



  Dave wrote:
   Hey there, having a bit of problem iterating through lists before i go
   on any further, here is
   a snip of the script.
   --
   d = a1 b1 c1 d1 e1 a2 b2 c2 d2 e2 a3 b3 c3 d3 e3 a4 b4 c4 d4 e4 a5 b5
   c5 d5 e5
   inLst = d.split()
   hitLst = []

   hitNum = 0
   stopCnt = 6 + hitNum

   for i in range(hitNum,len(inLst), 1):
  if i == stopCnt: break
  hitLst.append(inLst[i])

   print hitLst
   --
   $ python helper.py
   ['a1', 'b1', 'c1', 'd1', 'e1', 'a2']

   This works fine for my purposes, what I need is an outer loop that
   goes through the original list again and appends my next request.

   ie.

   hitNum = 5

   which will return:

   ['a2', 'b2', 'c2', 'd2', 'e2', 'a3']

   and append it to the previous one.

   ie:

   ['a1', 'b1', 'c1', 'd1', 'e1', 'a2']
   ['a2', 'b2', 'c2', 'd2', 'e2', 'a3']

   not really sure how to do this right now though, been trying several
   methods with no good results.

   btw, just creating lagged values (sort of shift registers) on the
   incoming signal

   Many thanks,

   Dave

  Dave,

  You are going to need to supply us with more info to help.

  1) Are you always going to want to get 6 elements from the list
  2) Are you always going to want to step by 5 elements as your offset each 
  time
  through the outer loop?
  3) Is your requirement just to split inLst into equal length lists and 
  append
  them together into a list?

  Depending on your answers this might be quite easy.

  -Larry

 Hi Larry, well to answer your questions.

 1) Are you always going to want to get 6 elements from the list

 yes for this example, but this will change depending on the length of
 the sig.
 hitNum will be changing depending on what element I need to lag ie:

 if hitNum = 1

 ['b1', 'c1', 'd1', 'e1', 'a2', 'b2']

 so I will be lagging the b elements.

 2) Are you always going to want to step by 5 elements as your offset
 each time
 through the outer loop?

 I think I just answered this

 3) Is your requirement just to split inLst into equal length lists and
 append
 them together into a list?

 yes

 Thanks for all your help,

 Dave

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Re: newb loop problem

2008-08-13 Thread marek . rocki 
Dave napisał(a):
 Hey there, having a bit of problem iterating through lists before i go
 on any further, here is
 a snip of the script.
 --
 d = a1 b1 c1 d1 e1 a2 b2 c2 d2 e2 a3 b3 c3 d3 e3 a4 b4 c4 d4 e4 a5 b5
 c5 d5 e5
 inLst = d.split()
 hitLst = []

 hitNum = 0
 stopCnt = 6 + hitNum

 for i in range(hitNum,len(inLst), 1):
   if i == stopCnt: break
   hitLst.append(inLst[i])

 print hitLst
 --
 $ python helper.py
 ['a1', 'b1', 'c1', 'd1', 'e1', 'a2']


 This works fine for my purposes, what I need is an outer loop that
 goes through the original list again and appends my next request.

 ie.

 hitNum = 5

 which will return:

 ['a2', 'b2', 'c2', 'd2', 'e2', 'a3']

 and append it to the previous one.

 ie:

 ['a1', 'b1', 'c1', 'd1', 'e1', 'a2']
 ['a2', 'b2', 'c2', 'd2', 'e2', 'a3']


 not really sure how to do this right now though, been trying several
 methods with no good results.

 btw, just creating lagged values (sort of shift registers) on the
 incoming signal

 Many thanks,

 Dave

It seems you're going through a lot of trouble just to construct the
list by hand. Why don't you try slicing?

 d = a1 b1 c1 d1 e1 a2 b2 c2 d2 e2 a3 b3 c3 d3 e3 a4 b4 c4 d4 e4 a5 b5 c5 d5 
 e5
 in_lst = d.split()
 hit_lst = in_lst[0:6]
 hit_lst_2 = in_lst[5:11]

Regards,
Marek
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Re: newb loop problem

2008-08-13 Thread Larry Bates 

Dave wrote:

arrrggg, now I feel really dumb..

hitNum = 0
stopCnt = 6 + hitNum
offSet = 5

for i in range(0,10,1):

for x in range(hitNum,len(inLst), 1):
print hitNum, stopCnt
if x == stopCnt: break
hitLst.append(inLst[x])
hitNum +=offSet
stopCnt+=offSet
print hitLst


Beers, Dave


On Aug 13, 12:58 am, Dave [EMAIL PROTECTED] wrote:

On Aug 13, 12:35 am, Larry Bates [EMAIL PROTECTED] wrote:




Dave wrote:

Hey there, having a bit of problem iterating through lists before i go
on any further, here is
a snip of the script.
--
d = a1 b1 c1 d1 e1 a2 b2 c2 d2 e2 a3 b3 c3 d3 e3 a4 b4 c4 d4 e4 a5 b5
c5 d5 e5
inLst = d.split()
hitLst = []
hitNum = 0
stopCnt = 6 + hitNum
for i in range(hitNum,len(inLst), 1):
   if i == stopCnt: break
   hitLst.append(inLst[i])
print hitLst
--
$ python helper.py
['a1', 'b1', 'c1', 'd1', 'e1', 'a2']
This works fine for my purposes, what I need is an outer loop that
goes through the original list again and appends my next request.
ie.
hitNum = 5
which will return:
['a2', 'b2', 'c2', 'd2', 'e2', 'a3']
and append it to the previous one.
ie:
['a1', 'b1', 'c1', 'd1', 'e1', 'a2']
['a2', 'b2', 'c2', 'd2', 'e2', 'a3']
not really sure how to do this right now though, been trying several
methods with no good results.
btw, just creating lagged values (sort of shift registers) on the
incoming signal
Many thanks,
Dave

Dave,
You are going to need to supply us with more info to help.
1) Are you always going to want to get 6 elements from the list
2) Are you always going to want to step by 5 elements as your offset each time
through the outer loop?
3) Is your requirement just to split inLst into equal length lists and append
them together into a list?
Depending on your answers this might be quite easy.
-Larry

Hi Larry, well to answer your questions.

1) Are you always going to want to get 6 elements from the list

yes for this example, but this will change depending on the length of
the sig.
hitNum will be changing depending on what element I need to lag ie:

if hitNum = 1

['b1', 'c1', 'd1', 'e1', 'a2', 'b2']

so I will be lagging the b elements.

2) Are you always going to want to step by 5 elements as your offset
each time
through the outer loop?

I think I just answered this

3) Is your requirement just to split inLst into equal length lists and
append
them together into a list?

yes

Thanks for all your help,

Dave




This can also be written as:

d = a1 b1 c1 d1 e1 a2 b2 c2 d2 e2 a3 b3 c3 d3 e3  \
a4 b4 c4 d4 e4 a5 b5 c5 d5 e5
inLst = d.split()
sliceLen = 6
step = 5
offset = 1
hitLst = list()
for n in xrange(len(inLst)/step):
start = offset + n * step
hitLst.append(inLst[offset:offset + sliceLen])

print hitLst


-Larry
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Re: newb loop problem

2008-08-13 Thread Sion Arrowsmith 
Dave  [EMAIL PROTECTED] wrote:
hitNum = 0
stopCnt = 6 + hitNum
offSet = 5

for i in range(0,10,1):

The step argument to range defaults to 1: it's tidier to omit it.
Similarly, the start argument defaults to 0, so you can drop that too.

for i in range(10):

   for x in range(hitNum,len(inLst), 1):
   print hitNum, stopCnt

hitNum and stopCnt are constant in this loop: if you care about
this print statement, move it into the outer loop and stop yourself
drowning in output.

   if x == stopCnt: break

If you want to exit the inner loop when x == stopCnt, why not make
that condition part of the loop construct?

for x in range(hitNum, stopCnt):

That said, if you ever see for i in range(len(lst)) *immediately*
replace it by for i, x in enumerate(lst), then go through to body
to see if you really need that i, and if not use for x in lst,
with slicing if the range is more complex than range(len(lst)). As
you can do here:

for x in inLst[hitNum:stopCnt]:
hitLst.append(x)

And if all you're doing in a for loop is appending to one list from
another, that's just what list.extend does:

hitLst.extend(inLst[hitNum:stopCnt])

   hitNum +=offSet
   stopCnt+=offSet

Finally, that i in the outer loop isn't being used anywhere. Why
don't you just loop over hitNum? And noticing that stopCnt is
increasing in lock-step with hitNum:

offset = 5

for hitNum in range(0, 10*offset, offset):
hitLst.extend(inLst[hitNum:hitNum+6]

-- 
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   Frankly I have no feelings towards penguins one way or the other
-- Arthur C. Clarke
   her nu becomeþ se bera eadward ofdun hlæddre heafdes bæce bump bump bump
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newb loop problem

2008-08-12 Thread Dave 
Hey there, having a bit of problem iterating through lists before i go
on any further, here is
a snip of the script.
--
d = a1 b1 c1 d1 e1 a2 b2 c2 d2 e2 a3 b3 c3 d3 e3 a4 b4 c4 d4 e4 a5 b5
c5 d5 e5
inLst = d.split()
hitLst = []

hitNum = 0
stopCnt = 6 + hitNum

for i in range(hitNum,len(inLst), 1):
if i == stopCnt: break
hitLst.append(inLst[i])

print hitLst
--
$ python helper.py
['a1', 'b1', 'c1', 'd1', 'e1', 'a2']


This works fine for my purposes, what I need is an outer loop that
goes through the original list again and appends my next request.

ie.

hitNum = 5

which will return:

['a2', 'b2', 'c2', 'd2', 'e2', 'a3']

and append it to the previous one.

ie:

['a1', 'b1', 'c1', 'd1', 'e1', 'a2']
['a2', 'b2', 'c2', 'd2', 'e2', 'a3']


not really sure how to do this right now though, been trying several
methods with no good results.

btw, just creating lagged values (sort of shift registers) on the
incoming signal

Many thanks,

Dave
--
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Re: newb loop problem

2008-08-12 Thread Larry Bates 

Dave wrote:

Hey there, having a bit of problem iterating through lists before i go
on any further, here is
a snip of the script.
--
d = a1 b1 c1 d1 e1 a2 b2 c2 d2 e2 a3 b3 c3 d3 e3 a4 b4 c4 d4 e4 a5 b5
c5 d5 e5
inLst = d.split()
hitLst = []

hitNum = 0
stopCnt = 6 + hitNum

for i in range(hitNum,len(inLst), 1):
if i == stopCnt: break
hitLst.append(inLst[i])

print hitLst
--
$ python helper.py
['a1', 'b1', 'c1', 'd1', 'e1', 'a2']


This works fine for my purposes, what I need is an outer loop that
goes through the original list again and appends my next request.

ie.

hitNum = 5

which will return:

['a2', 'b2', 'c2', 'd2', 'e2', 'a3']

and append it to the previous one.

ie:

['a1', 'b1', 'c1', 'd1', 'e1', 'a2']
['a2', 'b2', 'c2', 'd2', 'e2', 'a3']


not really sure how to do this right now though, been trying several
methods with no good results.

btw, just creating lagged values (sort of shift registers) on the
incoming signal

Many thanks,

Dave


Dave,

You are going to need to supply us with more info to help.

1) Are you always going to want to get 6 elements from the list
2) Are you always going to want to step by 5 elements as your offset each time 
through the outer loop?
3) Is your requirement just to split inLst into equal length lists and append 
them together into a list?


Depending on your answers this might be quite easy.

-Larry
--
http://mail.python.org/mailman/listinfo/python-list

Re: newb loop problem

2008-08-12 Thread Dave 
On Aug 13, 12:35 am, Larry Bates [EMAIL PROTECTED] wrote:
 Dave wrote:
  Hey there, having a bit of problem iterating through lists before i go
  on any further, here is
  a snip of the script.
  --
  d = a1 b1 c1 d1 e1 a2 b2 c2 d2 e2 a3 b3 c3 d3 e3 a4 b4 c4 d4 e4 a5 b5
  c5 d5 e5
  inLst = d.split()
  hitLst = []

  hitNum = 0
  stopCnt = 6 + hitNum

  for i in range(hitNum,len(inLst), 1):
 if i == stopCnt: break
 hitLst.append(inLst[i])

  print hitLst
  --
  $ python helper.py
  ['a1', 'b1', 'c1', 'd1', 'e1', 'a2']

  This works fine for my purposes, what I need is an outer loop that
  goes through the original list again and appends my next request.

  ie.

  hitNum = 5

  which will return:

  ['a2', 'b2', 'c2', 'd2', 'e2', 'a3']

  and append it to the previous one.

  ie:

  ['a1', 'b1', 'c1', 'd1', 'e1', 'a2']
  ['a2', 'b2', 'c2', 'd2', 'e2', 'a3']

  not really sure how to do this right now though, been trying several
  methods with no good results.

  btw, just creating lagged values (sort of shift registers) on the
  incoming signal

  Many thanks,

  Dave

 Dave,

 You are going to need to supply us with more info to help.

 1) Are you always going to want to get 6 elements from the list
 2) Are you always going to want to step by 5 elements as your offset each time
 through the outer loop?
 3) Is your requirement just to split inLst into equal length lists and append
 them together into a list?

 Depending on your answers this might be quite easy.

 -Larry

Hi Larry, well to answer your questions.

1) Are you always going to want to get 6 elements from the list

yes for this example, but this will change depending on the length of
the sig.
hitNum will be changing depending on what element I need to lag ie:

if hitNum = 1

['b1', 'c1', 'd1', 'e1', 'a2', 'b2']

so I will be lagging the b elements.

2) Are you always going to want to step by 5 elements as your offset
each time
through the outer loop?

I think I just answered this

3) Is your requirement just to split inLst into equal length lists and
append
them together into a list?

yes


Thanks for all your help,

Dave




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loop problem

2005-07-13 Thread Shankar Iyer ([EMAIL PROTECTED]) 

Hi, First, I want to thank those who responded to my question about "the plot module" yesterday. I realize now that the question could have been vague. There is a plot module, simply called "plot," that I am using, and I guess it is not a very widely circulated plotting utility. Anyway, I'm a little confused by the output that I get from the following code:
/***/
from Tkinter import *from tkSimpleDialog import *import Pmwimport HP34401Aimport plotimport time
class Application(Frame):
 def __init__(self,master=None): Frame.__init__(self,master) self.pack() self.graph = plot.plot(self) self.beginplotting()
 def beginplotting(self): x = [] y = [] i = 0 fLastOutput = time.time() while 1: fTime = time.time() if((fTime-fLastOutput) = 2): i += 1 x.append(fTime) y.append(2*fTime) self.graph.addSet(x,y)
 fLastOutput = fTime if(i=30):break
app = Application()app.mainloop()
/***/
This is supposed to plot a point every two seconds. When it actually runs, it instead stores all the data points and then plots them after 30 data points have been taken. I'm fairly certain that this is not due to the use of the plot class. If I have the program insteadwrite each time into a Tkinter Text widget, the same type of thing happens. That is, the program stores all 30 times in memory and then dumps them into the Text widget at the end. Since I have written the code so that it plots (or writes to the Text widget) every two seconds, I am confused as to why this is happening.
Shankar

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Re: loop problem

2005-07-13 Thread Paul Boots 
Hi,

First ;-) I don't know Tk or plot, but from looking at your code I see you want 
to plot
a line a minute long, after that, all dots are drawn to the screen. You call

self.graph.addSet(x,y)

My guess is that you have to explicitly draw to the screen using same call from
graph or self? whatever Tkinter wants. Search the tkinter/tk docs for 'draw'
calls.



At 16:18 -0400 13/7/2005, Shankar Iyer ([EMAIL PROTECTED]) wrote:
Content-Language: en
Content-Type: text/html; charset=us-ascii
Content-Disposition: inline
Content-Transfer-Encoding: 7bit

Hi,

 First, I want to thank those who responded to my question about the plot 
 module yesterday.  I realize now that the question could have been vague.  
 There is a plot module, simply called plot, that I am using, and I guess it 
 is not a very widely circulated plotting utility.

Anyway, I'm a little confused by the output that I get from the following 
 code:

/***/

from Tkinter import *
from tkSimpleDialog import *
import Pmw
import HP34401A
import plot
import time

class Application(Frame):

def __init__(self,master=None):
Frame.__init__(self,master)
self.pack()
self.graph = plot.plot(self)
self.beginplotting()

def beginplotting(self):
x = []
y = []
i = 0
fLastOutput = time.time()
while 1:
fTime = time.time()
if((fTime-fLastOutput) = 2):
i += 1
x.append(fTime)
y.append(2*fTime)
self.graph.addSet(x,y)
fLastOutput = fTime
if(i=30):break

app = Application()
app.mainloop()

/***/

This is supposed to plot a point every two seconds.  When it actually runs, it 
instead stores all the data points and then plots them after 30 data points 
have been taken.  I'm fairly certain that this is not due to the use of the 
plot class.  If I have the program instead write each time into a Tkinter Text 
widget, the same type of thing happens.  That is, the program stores all 30 
times in memory and then dumps them into the Text widget at the end.  Since I 
have written the code so that it plots (or writes to the Text widget) every 
two seconds, I am confused as to why this is happening.

Shankar


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-- 
Vriendelijke groet,
Paul
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