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trouble writing results to files

I have two semi related questions...

First, I am trying to output a list of strings to a csv file using the
csv module. The output file separates each letter of the string with a
comma and then puts each string on a separate line. So the code is:

import csv
output = csv.writer(open ('/Python25/working/output.csv', 'a'))
a = ["apple", "cranberry" , "tart"]
for elem in range(len(a)):
output.writerow (a[elem])
.... and it would write to the file:
a,p,p,l,e
c,r,a,n,b,e,r,r ,y
t,a,r,t

How do I get it to write "apple", "cranberry" , "tart" ?

Second, there is a significant delay (5-10 minutes) between when the
program finishes running and when the text actually appears in the
file. Any ideas for why this happens? It is the same for writing with
the csv module or the standard way.

thanks!
Lisa

Nov 29 '06 #1
16 1820
li**********@gm ail.com wrote:
import csv
output = csv.writer(open ('/Python25/working/output.csv', 'a'))
a = ["apple", "cranberry" , "tart"]
for elem in range(len(a)):
output.writerow (a[elem])
output.writerow expects a sequence as an argument. You are passing a
string, which is a sequence of characters. By the way, what output are you
expecting to get? Do you want a file with only one line (apple,
cranberry, tart), or each fruit in a different line?

BTW, iterating over range(len(a)) is an anti-pattern in Python. You should
do it like this:

for item in a:
output.writerow ([item])
Second, there is a significant delay (5-10 minutes) between when the
program finishes running and when the text actually appears in the
file.
Try closing the file explicitly.
Cheers,
--
Roberto Bonvallet
Nov 29 '06 #2
On 2006-11-29, Roberto Bonvallet <Ro************ ***@cern.chwrot e:
BTW, iterating over range(len(a)) is an anti-pattern in Python.
Unless you're modifying elements of a, surely?

--
Neil Cerutti
You can't give him that cutback lane. He's so fast, and he sees it so well. He
can also run away from you if he gets a little bit of crack. --Dick Lebeau
Nov 29 '06 #3
Neil Cerutti wrote:
On 2006-11-29, Roberto Bonvallet <Ro************ ***@cern.chwrot e:
>BTW, iterating over range(len(a)) is an anti-pattern in Python.

Unless you're modifying elements of a, surely?
enumerate is your friend :)

for n, item in enumerate(a):
if f(item):
a[n] = whatever

--
Roberto Bonvallet
Nov 29 '06 #4
On 2006-11-29, Roberto Bonvallet <Ro************ ***@cern.chwrot e:
Neil Cerutti wrote:
>On 2006-11-29, Roberto Bonvallet <Ro************ ***@cern.chwrot e:
>>BTW, iterating over range(len(a)) is an anti-pattern in Python.

Unless you're modifying elements of a, surely?

enumerate is your friend :)

for n, item in enumerate(a):
if f(item):
a[n] = whatever
I was going to bring it up but I had a brainfart about the order
of (n, item) in the tuple and was too lazy to look it up. ;-)

--
Neil Cerutti
Nov 29 '06 #5
Neil Cerutti wrote:
>BTW, iterating over range(len(a)) is an anti-pattern in Python.

Unless you're modifying elements of a, surely?
and needs to run on a Python version that doesn't support enumerate.

</F>

Nov 29 '06 #6

Roberto Bonvallet wrote:
li**********@gm ail.com wrote:
import csv
output = csv.writer(open ('/Python25/working/output.csv', 'a'))
a = ["apple", "cranberry" , "tart"]
for elem in range(len(a)):
output.writerow (a[elem])

output.writerow expects a sequence as an argument. You are passing a
string, which is a sequence of characters. By the way, what output are you
expecting to get? Do you want a file with only one line (apple,
cranberry, tart), or each fruit in a different line?
I want it to print everything on one line and then create a new line
where it will print some more stuff. In my real program I am iterating
and it will eventually print the list a couple hundred times. But it
would be useful to understand how to tell it to do either.
BTW, iterating over range(len(a)) is an anti-pattern in Python. You should
do it like this:

for item in a:
output.writerow ([item])
I can try that. Is using range(len(a)) a bad solution in the sense
that its likely to create an unexpected error? Or because there is a
more efficient way to accomplish the same thing?

thanks!
Lisa

Nov 29 '06 #7
li**********@gm ail.com wrote:
I can try that. Is using range(len(a)) a bad solution in the sense
that its likely to create an unexpected error? Or because there is a
more efficient way to accomplish the same thing?
for-in uses an internal index counter to fetch items from the sequence, so

for item in seq:
function(item)

is simply a shorter and more efficient way to write

for item in range(len(seq)) :
function(seq[item])

also see this article:

http://online.effbot.org/2006_11_01_archive.htm#for

</F>

Nov 29 '06 #8
On Wed, 29 Nov 2006 17:00:30 +0100, Fredrik Lundh wrote:
Neil Cerutti wrote:
>>BTW, iterating over range(len(a)) is an anti-pattern in Python.

Unless you're modifying elements of a, surely?

and needs to run on a Python version that doesn't support enumerate.
This isn't meant as an argument against using enumerate in the common
case, but there are circumstances where iterating with an index variable
is the right thing to do. "Anti-pattern" tends to imply that it is always
wrong.

The advantage of enumerate disappears if you only need to do something to
certain items, not all of them:
>>alist = list("abcdefghi j")
for i in xrange(3, 7, 2):
.... print i, alist[i]
....
3 d
5 f

Nice and clear. But this is just ugly and wasteful:
>>for i,c in enumerate(alist ):
.... if i in xrange(3, 7, 2):
.... print i, c
....
3 d
5 f

although better than the naive alternative using slicing, which is just
wrong:
>>for i,c in enumerate(alist[3:7:2]):
.... print i, c
....
0 d
1 f

The indexes point to the wrong place in the original, non-sliced list, so
if you need to modify the original, you have to adjust the indexes by hand:
>>for i,c in enumerate(alist[3:7:2]):
.... print 2*i+3, c
....
3 d
5 f

And remember that if alist is truly huge, you may take a performance hit
due to duplicating all those megabytes of data when you slice it. If you
are modifying the original, better to skip making a slice.

I wrote a piece of code the other day that had to walk along a list,
swapping adjacent elements like this:

for i in xrange(0, len(alist)-1, 2):
alist[i], alist[i+1] = alist[i+1], alist[i]
The version using enumerate is no improvement:

for i, x in enumerate(alist[0:len(alist)-1:2]):
alist[i*2], alist[i*2+1] = alist[i*2+1], x
In my opinion, it actually is harder to understand what it is doing.
Swapping two items using "a,b = b,a" is a well known and easily recognised
idiom. Swapping two items using "a,b = b,c" is not.
--
Steven.

Nov 30 '06 #9
Steven D'Aprano wrote:
On Wed, 29 Nov 2006 17:00:30 +0100, Fredrik Lundh wrote:
>Neil Cerutti wrote:
>>>BTW, iterating over range(len(a)) is an anti-pattern in Python.

Unless you're modifying elements of a, surely?

and needs to run on a Python version that doesn't support enumerate.

This isn't meant as an argument against using enumerate in the common
case, but there are circumstances where iterating with an index variable
is the right thing to do. "Anti-pattern" tends to imply that it is always
wrong.
Right, I should have said: "iterating over range(len(a)) just to obtain the
elements of a is not the pythonic way to do it".

Cheers,
--
Roberto Bonvallet
Nov 30 '06 #10

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