e I should? (Honest question, I really
don't know.)
--
Wolfram Hinderer
--
https://mail.python.org/mailman/listinfo/python-list
Am 13.01.2021 um 22:20 schrieb Bischoop:
I want to to display a number or an alphabet which appears mostly
consecutive in a given string or numbers or both
Examples
s= ' aabskaaabad'
output: c
# c appears 4 consecutive times
8bbakebaoa
output: b
#b appears 2 consecutive times
You can
Am 21.05.2018 um 01:16 schrieb bruceg113...@gmail.com:
If I decide I need the parentheses, this works.
"(" + ",".join([str(int(i)) for i in s[1:-1].split(",")]) + ")"
'(128,20,8,255,-1203,1,0,-123)'
Thanks,
Bruce
Creating the tuple seems to be even simpler.
>>> str(tuple(map(int,
Am 10.11.2016 um 03:06 schrieb Paul Rubin:
This can probably be cleaned up some:
from itertools import islice
from collections import deque
def ngram(n, seq):
it = iter(seq)
d = deque(islice(it, n))
if len(d) != n:
return
for s in
Am Samstag, 11. April 2015 09:14:50 UTC+2 schrieb Marko Rauhamaa:
Paul Rubin no.email@nospam.invalid:
This takes about 4 seconds on a Intel(R) Core(TM) i5-3230M CPU @ 2.60GHz
laptop (64 bit linux):
Converted to Python3:
On 3 Feb., 11:47, John O'Hagan resea...@johnohagan.com wrote:
But isn't it equally true if we say that z = t[1], then t[1] += x is
syntactic sugar for z = z.__iadd__(x)? Why should that fail, if z can handle
it?
It's more like syntactic sugar for
y = t; z = y.__getitem__(1); z.__iadd__(x);
On 17 Mai, 20:56, geremy condra debat...@gmail.com wrote:
On Tue, May 17, 2011 at 10:19 AM, Jussi Piitulainen
jpiit...@ling.helsinki.fi wrote:
geremy condra writes:
or O(1):
ö = (1 + sqrt(5)) / 2
def fib(n):
numerator = (ö**n) - (1 - ö)**n
denominator = sqrt(5)
On 27 Okt., 10:27, Arnaud Delobelle arno...@gmail.com wrote:
True. It's far too verbose. I'd go for something like:
f=lambda n:n=0 or n*f(~-n)
I've saved a few precious keystrokes and used the very handy ~- idiom!
You can replace n=0 with n1. Then you can leave out the space
before or
On 6 Aug., 22:07, John Posner jjpos...@optimum.net wrote:
On 8/2/2010 11:00 PM, John Posner wrote:
On 7/31/2010 1:31 PM, John Posner wrote:
Caveat -- there's another description of defaultdict here:
http://docs.python.org/library/collections.html#collections.defaultdict
... and it's
On 8 Jul., 15:10, Ethan Furman et...@stoneleaf.us wrote:
Interesting. I knew when I posted my above comment that I was ignoring
such situations. I cannot comment on the code itself as I am unaware of
the algorithm, and haven't studied numbers extensively (although I do
find them very
On 7 Jul., 19:32, Ethan Furman et...@stoneleaf.us wrote:
Nobody wrote:
On Wed, 07 Jul 2010 15:08:07 +0200, Thomas Jollans wrote:
you should never rely on a floating-point number to have exactly a
certain value.
Never is an overstatement. There are situations where you can rely
upon a
On 1 Jul., 06:04, Stephen Hansen me+list/pyt...@ixokai.io wrote:
The 'reversed' and 'sorted' functions are generators that lazilly
convert an iterable as needed.
'sorted' returns a new list (and is not lazy).
--
http://mail.python.org/mailman/listinfo/python-list
On 8 Mai, 21:46, Steven D'Aprano st...@remove-this-
cybersource.com.au wrote:
On Sat, 08 May 2010 12:15:22 -0700, Wolfram Hinderer wrote:
Returning s[:-1 - len(t)] is faster.
I'm sure it is. Unfortunately, it's also incorrect.
However, s[:-len(t)] should be both faster and correct.
Ouch
On 8 Mai, 20:46, Steven D'Aprano st...@remove-this-
cybersource.com.au wrote:
def get_leading_whitespace(s):
t = s.lstrip()
return s[:len(s)-len(t)]
c = get_leading_whitespace(a)
assert c == leading_whitespace
Unless your strings are very large, this is likely to be faster than
On 17 Feb., 19:10, Andrej Mitrovic andrej.mitrov...@gmail.com wrote:
Hi,
I couldn't figure out a better description for the Subject line, but
anyway, I have the following:
_num_frames = 32
_frames = range(0, _num_frames) # This is a list of actual objects,
I'm just pseudocoding here.
On 19 Jan., 16:30, Gerald Britton gerald.brit...@gmail.com wrote:
Timer(' '.join([x for x in l]), 'l = map(str,range(10))').timeit()
2.9967339038848877
Timer(' '.join(x for x in l), 'l = map(str,range(10))').timeit()
7.2045478820800781
[...]
2. Why should the pure list comprehension be
On 19 Jan., 21:06, Gerald Britton gerald.brit...@gmail.com wrote:
[snip]
Yes, list building from a generator expression *is* expensive. And
join has to do it, because it has to iterate twice over the iterable
passed in: once for calculating the memory needed for the joined
string, and
On 14 Jan., 19:48, MRAB pyt...@mrabarnett.plus.com wrote:
Arnaud Delobelle wrote:
D'Arcy J.M. Cain da...@druid.net writes:
On Thu, 14 Jan 2010 09:07:47 -0800
Chris Rebert c...@rebertia.com wrote:
Even more succinctly:
def ishex(s):
return all(c in string.hexdigits for c in s)
On 1 Jan., 02:47, Steven D'Aprano st...@remove-this-
cybersource.com.au wrote:
On Thu, 31 Dec 2009 11:34:39 -0800, Tom Machinski wrote:
On Wed, Dec 30, 2009 at 4:01 PM, Steven D'Aprano
st...@remove-this-cybersource.com.au wrote:
On Wed, 30 Dec 2009 15:18:11 -0800, Tom Machinski wrote:
On 15 Sep., 23:51, Ross ros...@gmail.com wrote:
If I have a list of tuples:
k=[(a, bob, c), (p, joe, d), (x, mary, z)]
and I want to pull the middle element out of each tuple to make a new
list:
myList = [bob, joe, mary]
if a tuple is OK: zip(*k)[1]
--
On 5 Aug., 21:31, Mensanator mensana...@aol.com wrote:
import turtle
tooter = turtle.Turtle()
tooter.tracer
Traceback (most recent call last):
File pyshell#2, line 1, in module
tooter.tracer
AttributeError: 'Turtle' object has no attribute 'tracer'
tooter.hideturtle()
On 5 Mai, 08:08, Steven D'Aprano
ste...@remove.this.cybersource.com.au wrote:
Self-reflective functions like these are (almost?) unique in Python in
that they require a known name to work correctly. You can rename a class,
instance or module and expect it to continue to work, but not so for
On 10 Feb., 21:28, r0g aioe@technicalbloke.com wrote:
def inet2ip(n, l=[], c=4):
if c==0: return ..join(l)
p = 256**( c-1 )
l.append( str(n/p) )
return inet2ip( n-(n/p)*p, l, c-1 )
The results for 1
iterations of each were as follows...
0.113744974136 seconds for old
On 29 Jul., 01:05, Raymond Hettinger [EMAIL PROTECTED] wrote:
[Ervan Ensis]
I have a list like [108, 58, 68]. I want to return
the sorted indices of these items in the same order
as the original list. So I should return [2, 0, 1]
One solution is to think of the list indexes
being
On 10 Jul., 21:57, r.e.s. [EMAIL PROTECTED] wrote:
Can the following program be shortened? ...
def h(n,m):
E=n,
while (E!=())*m0:n=h(n+1,m-1);E=E[:-1]+(E[-1]0)*(E[-1]-1,)*n
return n
h(9,9)
Some ideas...
# h is your version
def h(n,m):
E=n,
while
On 26 Feb., 14:36, [EMAIL PROTECTED] wrote:
A possible solution to this problem is optional delimiters. What's
the path of less resistance to implement such optional delimiters?
Is to use comments. For example: #} or #: or something similar.
If you use such pairs of symbols in a systematic
On 22 Jan., 23:56, [EMAIL PROTECTED] wrote:
So anyone got an answer to which set of numbers gives the most targets
from 100 onwards say (or from 0 onwards)? IsPythonupto the task?
It's (5, 8, 9, 50, 75, 100): 47561 targets altogether (including
negative ones), 25814 targets = 100.
(BTW, 1226
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