On Sat, Mar 31, 2018 at 7:44 PM, Demian Brecht wrote:
> I might be entirely off my face, but figured I'd ask anyways given I
> haven't figured out a clean solution to this problem myself yet:
>
> I'm trying to write a REST API client that supports both async and
> synchronous HTTP transports (init
El miércoles, 28 de agosto de 2013, 21:18:26 (UTC-3), Mohsen Pahlevanzadeh
escribió:
> Dear all,
>
> I'm C++ programmer and unfortunately put semicolon at end of my
> statements in python.
>
> Quesion:
> What's really defferences between putting semicolon and don't put?
>
> Yours,
> Mohsen
We
On 4/1/2018 5:24 PM, David Foster wrote:
My understanding is that the Python interpreter already has enough information
when bytecode-compiling a .py file to determine which names correspond to local
variables in functions. That suggests it has enough information to identify all
valid names in
On 02/04/18 03:24, C W wrote:
Thank you Steven. I am frustrated that I can't enumerate a dictionary by
position index.
Maybe I want to shift by 2 positions, 5 positions...
I want to know/learn how to manipulate dictionary with loop and by its
position location.
Frankly I think you'd be much
Yes, you see right through me!
I was able to conquer it, there's probably better ways:
self.myDict = dict(zip(string.ascii_lowercase + string.ascii_uppercase,
string.ascii_lowercase[shift:26] + string.ascii_lowercase[:shift] +
string.ascii_uppercase[shift:26] + string.ascii_uppercase[:shift]))
Ho
On Sun, 01 Apr 2018 22:24:31 -0400, C W wrote:
> Thank you Steven. I am frustrated that I can't enumerate a dictionary by
> position index.
Why do you care about position index?
> Maybe I want to shift by 2 positions, 5 positions...
Sounds like you are trying to program the Caesar Shift cipher,
Thank you Steven. I am frustrated that I can't enumerate a dictionary by
position index.
Maybe I want to shift by 2 positions, 5 positions...
I want to know/learn how to manipulate dictionary with loop and by its
position location.
On Sun, Apr 1, 2018 at 10:02 PM, Steven D'Aprano <
steve+comp.
On Sun, 01 Apr 2018 20:52:35 -0400, C W wrote:
> Thank you all for the response.
>
> What if I have myDict = {'a': 'B', 'b': 'C',...,'z':'A' }? So now, the
> values are shift by one position.
>
> key:abcdefghijklmnopqrstuvwxyz
> value: BCDEFGHIJKLMNOPQRSTUVWXYZA
>
> Can I fill in a key and
On Sun, 01 Apr 2018 14:24:38 -0700, David Foster wrote:
> My understanding is that the Python interpreter already has enough
> information when bytecode-compiling a .py file to determine which names
> correspond to local variables in functions. That suggests it has enough
> information to identify
I am using Python 3.6. I ran the those lines and got a sorted dictionary by
keys.
On Sun, Apr 1, 2018 at 9:38 PM, Chris Angelico wrote:
> On Mon, Apr 2, 2018 at 11:34 AM, C W wrote:
> > A different but related question:
> >
> > myDict = dict(zip(string.ascii_lowercase + string.ascii_uppercase,
On Mon, Apr 2, 2018 at 11:34 AM, C W wrote:
> A different but related question:
>
> myDict = dict(zip(string.ascii_lowercase + string.ascii_uppercase,
> string.ascii_lowercase + string.ascii_uppercase))
>>myDict
> {'A': 'A', 'B': 'B', 'C': 'C',...,'w': 'w', 'x': 'x', 'y': 'y', 'z': 'z'}
>
> Why ar
A different but related question:
myDict = dict(zip(string.ascii_lowercase + string.ascii_uppercase,
string.ascii_lowercase + string.ascii_uppercase))
>myDict
{'A': 'A', 'B': 'B', 'C': 'C',...,'w': 'w', 'x': 'x', 'y': 'y', 'z': 'z'}
Why are the keys sorted from upper case to lower case? I asked f
On Sun, 01 Apr 2018 19:14:05 +, Arkadiusz Bulski wrote:
> Thanks,
> timeit gives `not any(key)` same performance as `sum(key)==0`.
Have you considered what happens when the key is *not* all zeroes?
key = b'\x11'*100
any(key) bails out on the first byte. sum(key) has to add a million
v
Thank you all for the response.
What if I have myDict = {'a': 'B', 'b': 'C',...,'z':'A' }? So now, the
values are shift by one position.
key:abcdefghijklmnopqrstuvwxyz
value: BCDEFGHIJKLMNOPQRSTUVWXYZA
Can I fill in a key and its corresponding value simultaneously on the fly?
Something in t
David Foster writes:
> My understanding is that the Python interpreter already has enough
> information when bytecode-compiling a .py file to determine which
> names correspond to local variables in functions. That suggests it has
> enough information to identify all valid names in a .py file and
On Mon, Apr 2, 2018 at 8:05 AM, Devin Jeanpierre wrote:
> On Sun, Apr 1, 2018 at 2:38 PM, Chris Angelico wrote:
>> On Mon, Apr 2, 2018 at 7:24 AM, David Foster wrote:
>>> My understanding is that the Python interpreter already has enough
>>> information when bytecode-compiling a .py file to det
On Sun, Apr 1, 2018 at 2:38 PM, Chris Angelico wrote:
> On Mon, Apr 2, 2018 at 7:24 AM, David Foster wrote:
>> My understanding is that the Python interpreter already has enough
>> information when bytecode-compiling a .py file to determine which names
>> correspond to local variables in functi
> But if it is cheap to detect a wide variety of name errors at compile time,
> is there any particular reason it is not done?
>From my perspective, it is done, but by tools that give better output
than Python's parser. :)
Linters (like pylint) are better than syntax errors here, because they
co
On Mon, Apr 2, 2018 at 7:24 AM, David Foster wrote:
> My understanding is that the Python interpreter already has enough
> information when bytecode-compiling a .py file to determine which names
> correspond to local variables in functions. That suggests it has enough
> information to identify
My understanding is that the Python interpreter already has enough information
when bytecode-compiling a .py file to determine which names correspond to local
variables in functions. That suggests it has enough information to identify all
valid names in a .py file and in particular to identify w
Arkadiusz Bulski wrote:
> Thanks,
> timeit gives `not any(key)` same performance as `sum(key)==0`.
Then you did not feed it the "right" data
$ python3 -m timeit -s 'key = b"x" + bytes(10**6)' 'sum(key)'
100 loops, best of 3: 15.7 msec per loop
$ python3 -m timeit -s 'key = b"x" + bytes(10**6)' '
On Mon, Apr 2, 2018 at 5:14 AM, Arkadiusz Bulski wrote:
> Thanks,
> timeit gives `not any(key)` same performance as `sum(key)==0`.
Are you timing these on ten-byte keys, or really large keys? For short
keys, just use whatever looks most elegant, and don't worry about
performance. If the key is ac
2018-04-01 22:03 GMT+03:00 Kirill Balunov :
>
>
> 2018-04-01 20:55 GMT+03:00 Arkadiusz Bulski :
>
>> What would be the most performance efficient way of checking if a bytes is
>> all zeros?
>
>
> Try `not any(key)` ;)
>
>
Sorry, I don't timed it before I posted. In reality, it is far from the
fast
Thanks,
timeit gives `not any(key)` same performance as `sum(key)==0`.
niedz., 1 kwi 2018 o 21:03 użytkownik Kirill Balunov <
kirillbalu...@gmail.com> napisał:
> 2018-04-01 20:55 GMT+03:00 Arkadiusz Bulski :
>
>> What would be the most performance efficient way of checking if a bytes is
>> all z
2018-04-01 20:55 GMT+03:00 Arkadiusz Bulski :
> What would be the most performance efficient way of checking if a bytes is
> all zeros?
Try `not any(key)` ;)
With kind regards,
-gdg
--
https://mail.python.org/mailman/listinfo/python-list
On 01/04/2018 18:55, Arkadiusz Bulski wrote:
What would be the most performance efficient way of checking if a bytes is
all zeros? Currently its `key == b'\x00' * len(key)` however, because its
Python 2/3 compatible:
That doesn't too efficient, if you first have to construct a compatible
objec
Arkadiusz Bulski wrote:
> What would be the most performance efficient way of checking if a bytes is
> all zeros? Currently its `key == b'\x00' * len(key)` however, because its
> Python 2/3 compatible:
>
> sum(key) == 0 is invalid
> key == bytes(len(key)) is invalid
>
> I already considered prec
On 2018-04-01 18:55, Arkadiusz Bulski wrote:
What would be the most performance efficient way of checking if a bytes is
all zeros? Currently its `key == b'\x00' * len(key)` however, because its
Python 2/3 compatible:
sum(key) == 0 is invalid
key == bytes(len(key)) is invalid
I already considere
On 2018-04-01 11:26, Paul Moore wrote:
On 1 April 2018 at 04:15, Mikhail V wrote:
MRAB writes:
> UnicodeEncodeError: 'charmap' codec can't encode character
>
> when it meets a non-ascii char.
>
> e.g. tried this:
> pip search pygame > a.txt
>
Well, _I_ didn't get an error!
One of the lines
Some interesting timeits:
In [7]: timeit('sum(x)==0', 'x=bytes(10)')
Out[7]: 0.30194770699927176
In [11]: timeit('x==bytes(10)', 'x=bytes(10)')
Out[11]: 0.2181608650007547
In [12]: timeit('x==z*10', 'x=bytes(10); z=bytes(1)')
Out[12]: 0.1092393600010837
In [13]: timeit('x==x2', 'x=bytes(10); z=
What would be the most performance efficient way of checking if a bytes is
all zeros? Currently its `key == b'\x00' * len(key)` however, because its
Python 2/3 compatible:
sum(key) == 0 is invalid
key == bytes(len(key)) is invalid
I already considered precomputing the rhs value.
Length of key is
On Mon, Apr 2, 2018 at 3:03 AM, Rustom Mody wrote:
> On Saturday, March 31, 2018 at 4:30:04 PM UTC+5:30, bartc wrote:
>> On 30/03/2018 21:13, C W wrote:
>> > Hello all,
>> >
>> > I want to create a dictionary.
>> >
>> > The keys are 26 lowercase letters. The values are 26 uppercase letters.
>> >
>
On Saturday, March 31, 2018 at 4:30:04 PM UTC+5:30, bartc wrote:
> On 30/03/2018 21:13, C W wrote:
> > Hello all,
> >
> > I want to create a dictionary.
> >
> > The keys are 26 lowercase letters. The values are 26 uppercase letters.
> >
> > The output should look like:
> > {'a': 'A', 'b': 'B',..
On 1 April 2018 at 03:16, MRAB wrote:
> On 2018-04-01 02:50, Mikhail V wrote:
>>
>> Steven D'Aprano writes:
>>
PS: was looking forward to PIP improvements on Windows, on 9.0.3 still
some issues. E.g. trying to redirect output from 'pip search ... >
a.txt' gives a wall of errors
On 1 April 2018 at 04:15, Mikhail V wrote:
> MRAB writes:
>
>
>> > UnicodeEncodeError: 'charmap' codec can't encode character
>> >
>> > when it meets a non-ascii char.
>> >
>> > e.g. tried this:
>> > pip search pygame > a.txt
>> >
>> Well, _I_ didn't get an error!
>>
>> One of the lines is:
>>
>>
35 matches
Mail list logo