On Tue, Feb 14, 2017 at 11:01 PM, Alan Gauld via Tutor wrote:
> To compute it if you don't know x in advance then yes,
> use something like
>
> value = 10**x
>
> But if you know the value in advance you can write it in
> a more compact form as:
>
> value = 1e5 # or 3e7 or whatever...
10**5 is an
As per my offline reply, just send messages to the list and
the whole group will act as a virtual mentor.
One other thing though. Please start new topics with a new message.
Do NOT reply to an existing thread as it messes up the archives.
And especially do not reply to a digest message.
And use a
On 14/02/17 12:03, Aaliyah Ebrahim wrote:
> For the function 1*10^x, is there a specific way of computing *10^x or will
> it just be the following :
>
> 1*10**x
To compute it if you don't know x in advance then yes,
use something like
value = 10**x
But if you know the value in advance you can
Please l will like to have a mentor how can I have it
On Feb 14, 2017 09:53, wrote:
> Send Tutor mailing list submissions to
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>
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> or, via email, send a me
Hi
For the function 1*10^x, is there a specific way of computing *10^x or will
it just be the following :
1*10**x
Thank you.
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On 14/02/17 10:11, Alan Gauld via Tutor wrote:
ts.adfuller(goog['Adj Close'], 1
>
> there seems to be a missing closing paren here?
> Also where does 'ts' come from, its not defined above?
Ah! I see from Peters post that I missed the first line of code.
But I still see a missing ')'...
--
Allan Tanaka via Tutor wrote:
> Hi. Not sure why this code produces the error like this. This error
> appears when i run the code of print "Hurst(GBM): %s" % hurst(gbm):
> Traceback (most recent call last): File "", line 1, in
> print "Hurst(GBM): %s" % hurst(gbm)NameError: name 'hurst'
>
On 14/02/17 01:55, Allan Tanaka via Tutor wrote:
> Hi. Not sure why this code produces the error like this. This error appears
> when i run the code of print "Hurst(GBM): %s" % hurst(gbm):
> Traceback (most recent call last): File "", line 1, in
> print "Hurst(GBM): %s" % hurst(gbm)Name
>I find out that the outcome for using .join() on a dictionary is totally
>different than it using on list or string. for example,
> >>> seq4 = {'hello':1,'good':2,'boy':3,'doiido':4}
> >>>
> boy:good:doiido:hello
>So my question is why the outco
Hi. Not sure why this code produces the error like this. This error appears
when i run the code of print "Hurst(GBM): %s" % hurst(gbm):
Traceback (most recent call last): File "", line 1, in
print "Hurst(GBM): %s" % hurst(gbm)NameError: name 'hurst' is not defined
Here is the full code
Danny Yoo wrote:
> Moreover, most implementations *deliberately* randomize their iteration
> order to avoid a particular kind of hash collision attack out there in
> the wild. See:
In CPython the hash() of a string may change between different runs to fend
off hash collision attacks, but that
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