Re: [Tutor] Messy - Very Messy string manipulation.
Thanks Meenu.
str.translate.
Worked like a charm for python 3.5.
And thanks Alan Gauld for the 2.7 version.
Appreciate the help guys. You guys are awesome.
Regards
Vusa
On Wed, Oct 28, 2015 at 6:23 PM, meenu ravi wrote:
> Hi Vusa,
>
> I was not able to reply through mail list due to some issue. So just
> replying through email.
>
> We can make use of string translate method for more pythonic way. If you
> are using python 2.x, the following itself should work for you:
>
> ***
> import string
> def anti_vowel(str):
> word = str.translate(None, 'aeiouAEIOU')
> return word
>
> print anti_vowel('The cow moos louder than the frog')
> ***
>
> And if you are using python 3.x, "None" inside the str.translate method
> doesn't work. So instead, you can use in this way:
>
> ***
> import string
> def anti_vowel(str):
> word = str.translate(str.maketrans("","","aeiouAEIOU"))
> return(word)
>
> print(anti_vowel("The cow moos louder than the frog"))
>
>
> ***
>
> The above code should work with python 2.x as well with python 2 syntax as
> follows:
>
> import string
> def anti_vowel(str):
> word = str.translate(string.maketrans('', ''), 'aeiouAEIOU')
> return word
>
> print anti_vowel('The cow moos louder than the frog')
>
>
> If you want to know more about translate method, please follow the link,
> https://docs.python.org/2/library/string.html#string-functions
>
> I hope you will get much more options through mailing list.
>
> Happy python:)
>
> Thanks,
> Meenakshi
>
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Re: [Tutor] Messy - Very Messy string manipulation.
On 28/10/2015 18:06, Alan Gauld wrote:
On 28/10/15 17:35, Peter Otten wrote:
Alan Gauld wrote:
On 28/10/15 16:37, Peter Otten wrote:
'The cow moos louder than the frog'.translate(str.maketrans("", "",
"aeiouAEIOU"))
'Th cw ms ldr thn th frg'
Even easier, forget the maketrans stuff and just use
'The cow moos louder than the frog'.translate(None,'aeiouAEIOU')
This only works for byte strings, not unicode.
Aha, I tried it in Python 2.7 which worked, but I didn't
think about v3...
Seems like as good a place as any to point out that in Python 3 all of
the following also exist.
static bytes.maketrans(from, to)
bytes.translate(table[, delete])
static bytearray.maketrans(from, to)
bytearray.translate(table[, delete])
--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.
Mark Lawrence
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[Tutor] counter not working in Quick Sort script
Mac OS 10.10
Python 3.4.3
I self-study Python and am using it for a Coursera algorithm class.
Hi! My script sorts correctly on all my test arrays. My accumulator variable
to count the number of comparisons returns nonsense. I'm counting the (length
- one) of each sublist that will be sorted in place, not the individual
comparisons. I put in a print statement for the count variable, and the
expected values print, but the function seems to be returning the first
increment of the count variable instead of the final value. I reread on the
difference between print and return, but I haven't figured this out yet. I
think I'm doing something language-specific wrong would appreciate another set
of eyes.
def partition(A, start, stop):
p = A[start] # make pivot first element in partition
i = start + 1
for j in range(i, stop):
# swap elements if A[j] bigger than pivot
if A[j] < p:
A[j], A[i] = A[i], A[j]
i += 1
# swap pivot into sorted index
A[start], A[i-1] = A[i-1], A[start]
return i
def quick_sort(A, start, stop, count):
if start < stop:
# increment count by length of partition less the pivot
count += (stop - start - 1)
print(count)
split = partition(A, start, stop)
# recursively sort lower partition
quick_sort(A, start, split-1, count)
# recursively count upper partition
quick_sort(A, split, stop, count)
return count
def main():
unsorted = [ 1, 2, 3, 4, 5, 6, 7, 8 ]
count = quick_sort(unsorted, 0, len(unsorted), 0)
print(unsorted)
print("There are {} comparisons.".format(count))
main()
This code is giving me this output:
➜ algorithms python3 quick_sort_first.py
7
13
18
22
25
27
28
28
[1, 2, 3, 4, 5, 6, 7, 8]
There are 7 comparisons.
Thank you,
Patricia
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Re: [Tutor] counter not working in Quick Sort script
On 29/10/15 19:11, Patti Scott via Tutor wrote:
Caveat: I didn't check the algorithms for correctness,
I'll just take your word for that.
My accumulator variable to count the number of comparisons returns nonsense.
def quick_sort(A, start, stop, count):
if start < stop:
# increment count by length of partition less the pivot
count += (stop - start - 1)
print(count)
split = partition(A, start, stop)
# recursively sort lower partition
quick_sort(A, start, split-1, count)
# recursively count upper partition
quick_sort(A, split, stop, count)
return count
Notice that you only set count once at the top of the function.
What the recursive instances of the function do is irrelevant
because you don't use their return values. So the return value
of this function is always (count + stop - start -1) for the
initial invocation value of count.
I suspect you really want to do something to count based on
the returns from the recursive calls too.
def main():
unsorted = [ 1, 2, 3, 4, 5, 6, 7, 8 ]
This looks very sorted to me? Is that correct?
count = quick_sort(unsorted, 0, len(unsorted), 0)
count should return 0+len()-0-1 -> len-1 = 7
print(unsorted)
print("There are {} comparisons.".format(count))
main()
This code is giving me this output:
➜ algorithms python3 quick_sort_first.py
7
This is the outer functions count
13
18
22
25
27
28
28
These are the recursive values of count which are
invisible to the outer invocation
[1, 2, 3, 4, 5, 6, 7, 8]
This is the sorted result
There are 7 comparisons.
And this reflects the outer value of count again.
Your code does exactly what you told it to do.
Your problem is that you are not using the returned
counts from the recursive calls.
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos
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