From deeppunster at gmail.com  Thu Mar  7 11:57:20 2019
From: deeppunster at gmail.com (Travis Risner)
Date: Thu, 07 Mar 2019 11:57:20 -0500
Subject: [CentralOH] [COhPy] DoJo Rumblings: Tonight's Python Challenge
Message-ID: <BC6DFE4C-37C2-40DB-A071-E346509A5924@gmail.com>

Here is a small problem for us to consider.  At tonight?s DoJo meeting 
we will work together to solve this as individuals or small groups.  We 
will also discuss the advantages of each approach.

	Given a array of numbers representing the stock prices of a company in
	chronological order, write a function that calculates the maximum 
profit you
	could have made from buying and selling that stock once. You must buy 
before
	you can sell it.

	For example, given [9, 11, 8, 5, 7, 10], you should return 5, since you
	could buy the stock at 5 dollars and sell it at 10 dollars.

DoJo is at Smokehouse Brewing, 1130 Dublin Road, Columbus, Ohio.  Join 
us at 6:00 or any time after that!

Travis

From neil.ludban at gmail.com  Thu Mar  7 12:15:20 2019
From: neil.ludban at gmail.com (neil ludban)
Date: Thu, 7 Mar 2019 19:15:20 +0200
Subject: [CentralOH] [COhPy] DoJo Rumblings: Tonight's Python Challenge
In-Reply-To: <BC6DFE4C-37C2-40DB-A071-E346509A5924@gmail.com>
References: <BC6DFE4C-37C2-40DB-A071-E346509A5924@gmail.com>
Message-ID: <CAA5SVLOg36WtycimzsY3KA3H=JdQ8EkHyM94qFuD2_xbF6GLTQ@mail.gmail.com>

With a minor pre-processing of the inputs, there's a O(n) solution with
code (in python) on Wikipedia.  :)


On Thu, Mar 7, 2019, 18:57 Travis Risner <deeppunster at gmail.com> wrote:

> Here is a small problem for us to consider.  At tonight?s DoJo meeting
> we will work together to solve this as individuals or small groups.  We
> will also discuss the advantages of each approach.
>
>         Given a array of numbers representing the stock prices of a
> company in
>         chronological order, write a function that calculates the maximum
> profit you
>         could have made from buying and selling that stock once. You must
> buy
> before
>         you can sell it.
>
>         For example, given [9, 11, 8, 5, 7, 10], you should return 5,
> since you
>         could buy the stock at 5 dollars and sell it at 10 dollars.
>
> DoJo is at Smokehouse Brewing, 1130 Dublin Road, Columbus, Ohio.  Join
> us at 6:00 or any time after that!
>
> Travis
> _______________________________________________
> CentralOH mailing list
> CentralOH at python.org
> https://mail.python.org/mailman/listinfo/centraloh
>
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From deeppunster at gmail.com  Thu Mar  7 20:05:22 2019
From: deeppunster at gmail.com (Travis Risner)
Date: Thu, 07 Mar 2019 20:05:22 -0500
Subject: [CentralOH] Anonymous Solution
Message-ID: <89EBFD45-E7C9-49BD-8325-17CECEC82F6B@gmail.com>



#!/usr/bin/env python3

"""
Anonymous solution to the following problem:

Given a array of numbers representing the stock prices of a company in
chronological order, write a function that calculates the maximum profit you
could have made from buying and selling that stock once. You must buy
before you can sell it.

For example, given [9, 11, 8, 5, 7, 10], you should return 5, since you
could buy the stock at 5 dollars and sell it at 10 dollars.

This solution accepts the array on th command line.
"""

import sys


def get_prices_from_arguments(args):
    return list(map(int, args[1:]))


def calculate_maximum_profit(prices):
    maximum_profit = 0
    maximum_price = None

    for price in reversed(prices):
        if maximum_price is None or price > maximum_price:
            maximum_price = price
        profit = maximum_price - price
        if profit > maximum_profit:
            maximum_profit = profit

    return maximum_profit


def main(args):
    prices = get_prices_from_arguments(args)
    maximum_profit = calculate_maximum_profit(prices)
    print(maximum_profit)


if __name__ == '__main__':
    main(sys.argv)
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From kennethpatierno at gmail.com  Thu Mar  7 20:04:09 2019
From: kennethpatierno at gmail.com (Kenneth Patierno)
Date: Thu, 7 Mar 2019 20:04:09 -0500
Subject: [CentralOH] Solution for Max Stock Profit
Message-ID: <CAAZu_HcqEZ7WVG0LKkqtTqyhn40qjwCZvEv7VVijdjyOZ4kUZQ@mail.gmail.com>

Cheers!

KENNY PATIERNO

*p : *614 <614-902-7379>-902-7379
*e : *kennethpatierno at gmail.com
[image: Facebook]  <https://www.facebook.com/kenny.patierno>[image:
Instagram]  <https://www.instagram.com/kennypatierno/>[image: Twitter]
<https://twitter.com/kenny_patierno>

?
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stock_prices = [40,13,70,3,8,43,54,12,7,90,8]

max_profit = -99

for stock_price in stock_prices:
	buy_price = stock_price
	for sell_price in stock_prices:
		net_profit = sell_price - buy_price
		if net_profit > max_profit:
			max_profit = net_profit
	stock_prices.pop(stock_prices.index(stock_price))
print("\nTotal Max Profit Available = $" + str(max_profit))

From jim.prior at accenture.com  Fri Mar  8 14:22:10 2019
From: jim.prior at accenture.com (Prior, Jim)
Date: Fri, 8 Mar 2019 19:22:10 +0000
Subject: [CentralOH] 2019-03-25 Python Meeting
Message-ID: <DM5P114MB066854FBF50548E678880B2BFF4D0@DM5P114MB0668.NAMP114.PROD.OUTLOOK.COM>

The March Monthly Python Meeting _will_ be at the (old) Columbus Forge
on the ground floor of the Smith Brothers Hardware Company Building.

https://www.meetup.com/Central-Ohio-Python-Users-Group/events/259333335/

Although Pillar will be moving in the next month or two,
the March Python meeting will be at the regular place.

________________________________

This message is for the designated recipient only and may contain privileged, proprietary, or otherwise confidential information. If you have received it in error, please notify the sender immediately and delete the original. Any other use of the e-mail by you is prohibited. Where allowed by local law, electronic communications with Accenture and its affiliates, including e-mail and instant messaging (including content), may be scanned by our systems for the purposes of information security and assessment of internal compliance with Accenture policy. Your privacy is important to us. Accenture uses your personal data only in compliance with data protection laws. For further information on how Accenture processes your personal data, please see our privacy statement at https://www.accenture.com/us-en/privacy-policy.
______________________________________________________________________________________

www.accenture.com

From deeptinker at gmail.com  Sat Mar  9 00:06:40 2019
From: deeptinker at gmail.com (Travis Risner)
Date: Sat, 09 Mar 2019 00:06:40 -0500
Subject: [CentralOH] Solution for Max Stock Profit
In-Reply-To: <CAAZu_HcqEZ7WVG0LKkqtTqyhn40qjwCZvEv7VVijdjyOZ4kUZQ@mail.gmail.com>
References: <CAAZu_HcqEZ7WVG0LKkqtTqyhn40qjwCZvEv7VVijdjyOZ4kUZQ@mail.gmail.com>
Message-ID: <B4671F07-6D5D-4DAD-9A54-40319BE4EB74@gmail.com>

Hi Ken,

Thanks for posting this solution to the list!

Travis

On 7 Mar 2019, at 20:04, Kenneth Patierno wrote:

> Cheers!
>
> KENNY PATIERNO
>
> *p : *614 <614-902-7379>-902-7379
> *e : *kennethpatierno at gmail.com
> [image: Facebook]  <https://www.facebook.com/kenny.patierno>[image:
> Instagram]  <https://www.instagram.com/kennypatierno/>[image: Twitter]
> <https://twitter.com/kenny_patierno>
>
> ?


> _______________________________________________
> CentralOH mailing list
> CentralOH at python.org
> https://mail.python.org/mailman/listinfo/centraloh
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From jim.prior at accenture.com  Mon Mar 11 12:22:47 2019
From: jim.prior at accenture.com (Prior, Jim)
Date: Mon, 11 Mar 2019 16:22:47 +0000
Subject: [CentralOH] Kenny's Solution for Max Stock Profit
In-Reply-To: <CAAZu_HcqEZ7WVG0LKkqtTqyhn40qjwCZvEv7VVijdjyOZ4kUZQ@mail.gmail.com>
References: <CAAZu_HcqEZ7WVG0LKkqtTqyhn40qjwCZvEv7VVijdjyOZ4kUZQ@mail.gmail.com>
Message-ID: <DM5P114MB066894748C69BBC679F1A2B0FF480@DM5P114MB0668.NAMP114.PROD.OUTLOOK.COM>

On Thursday, March 7, 2019 8:04:09 PM, Kenny wrote a solution to the challenge.

If an extra value of 140 is appended to the initial stock_prices,
what output does one expect? Does the code output that?

________________________________

This message is for the designated recipient only and may contain privileged, proprietary, or otherwise confidential information. If you have received it in error, please notify the sender immediately and delete the original. Any other use of the e-mail by you is prohibited. Where allowed by local law, electronic communications with Accenture and its affiliates, including e-mail and instant messaging (including content), may be scanned by our systems for the purposes of information security and assessment of internal compliance with Accenture policy. Your privacy is important to us. Accenture uses your personal data only in compliance with data protection laws. For further information on how Accenture processes your personal data, please see our privacy statement at https://www.accenture.com/us-en/privacy-policy.
______________________________________________________________________________________

www.accenture.com

From neil.ludban at gmail.com  Tue Mar 12 18:41:54 2019
From: neil.ludban at gmail.com (neil ludban)
Date: Tue, 12 Mar 2019 18:41:54 -0400
Subject: [CentralOH] [COhPy] DoJo Rumblings: Tonight's Python Challenge
In-Reply-To: <BC6DFE4C-37C2-40DB-A071-E346509A5924@gmail.com>
References: <BC6DFE4C-37C2-40DB-A071-E346509A5924@gmail.com>
Message-ID: <CAA5SVLMC4UyoQ4KL939MH63MsKSt9PtiooDBHB_yK+bVHsr-hA@mail.gmail.com>

#!/usr/bin/env python3.7

def delta(x):
    a = next(x)
    for b in x:
        yield b - a
        a = b

def positive_sums(x):
    s = 0
    for c in x:
        s += c
        if s > 0:
            yield s
        else:
            s = 0

prices = [ 9, 11, 8, 5, 7, 10 ]
print(max(positive_sums(delta(iter(prices)))))


On Thu, Mar 7, 2019 at 11:57 AM Travis Risner <deeppunster at gmail.com> wrote:

> Here is a small problem for us to consider.  At tonight?s DoJo meeting
> we will work together to solve this as individuals or small groups.  We
> will also discuss the advantages of each approach.
>
>         Given a array of numbers representing the stock prices of a
> company in
>         chronological order, write a function that calculates the maximum
> profit you
>         could have made from buying and selling that stock once. You must
> buy
> before
>         you can sell it.
>
>         For example, given [9, 11, 8, 5, 7, 10], you should return 5,
> since you
>         could buy the stock at 5 dollars and sell it at 10 dollars.
>
> DoJo is at Smokehouse Brewing, 1130 Dublin Road, Columbus, Ohio.  Join
> us at 6:00 or any time after that!
>
> Travis
> _______________________________________________
> CentralOH mailing list
> CentralOH at python.org
> https://mail.python.org/mailman/listinfo/centraloh
>
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From pybokeh at gmail.com  Wed Mar 13 00:46:32 2019
From: pybokeh at gmail.com (pybokeh)
Date: Wed, 13 Mar 2019 00:46:32 -0400
Subject: [CentralOH] Anyone uses nteract papermill? Call for opinions on
 using jupyter notebooks in production
Message-ID: <CAOoMBVgKtVNNJsgAE_nJRrpz52dC0Ngf_TCKHZpr=8D6ULjm9Q@mail.gmail.com>

Hello,
At my new role, I started to use papermill (admittedly with reservation and
skepticism) to orchestrate the execution of parameterized jupyter notebooks
used to automate my group's data processing needs.  The result?  I am now
totally blown away and thrilled with papermill (
https://github.com/nteract/papermill).  At my previous role, I had used
jupyter notebook to prototype a data process to automatically create
invoices.  But to productionalize (sadly, didn't know papermill existed at
the time), I had to port my jupyter notebook code to using a data
pipelining framework called Luigi (https://github.com/spotify/luigi).

Now with papermill, I see no need for me to port my jupyter notebooks to
something completely different and thus incur more technical debt from
having to learn yet another framework.  With papermill, I define which
jupyter notebooks run and what they depend on (dependency management).
Furthermore, when a failure occurs, I can fix the bug, then run only the
notebooks that didn't run (failure recovery) instead of re-running
everything.   So with these features afforded to me in papermill, the
beauty is I can keep developing in jupyter notebooks as data pipeline
scripts and use them for production.  Also, with papermill,  those that
don't come from a software engineering or computer science background, can
perform practical "data engineering" without having to learn sophisticated
data engineer tools like Apache Airflow or Spotify Luigi, since papermill
is far easier to learn, use, and get started than those heavyweight
frameworks.  Furthermore, a well annotated jupyter notebook is so much
easier to debug and explain what it is doing to a team member.  If you've
avoided jupyter notebooks because of the tediousness of version controlling
them or not being able to work with them using your favorite text editor or
IDE, there is now jupytext (https://github.com/mwouts/jupytext) which
allows you to edit notebooks as plain text.  I think with all of these
things in place, there is little to not like about using papermill to
execute notebooks as workflows.  However, since I am just a data analyst or
"citizen developer", I would like to hear opinions/thoughts from seasoned
IT professionals or data engineers who have worked with jupyter notebooks.
I know of a couple limitations with papermill, it does not come with a
means for scheduling tasks and it doesn't come with a fancy GUI or
visualization tool.  But would like to see what others come up with.

I was inspired to give papermill a spin from Netflix's blog article:
https://medium.com/netflix-techblog/notebook-innovation-591ee3221233

I figured if papermill/jupyter notebooks were a good fit for Netflix, I
explored if they could be a good fit for my uses at work.  So far, I am
pleased with the initial results.

- Daniel
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From deeppunster at gmail.com  Thu Mar 14 15:22:19 2019
From: deeppunster at gmail.com (Travis Risner)
Date: Thu, 14 Mar 2019 15:22:19 -0400
Subject: [CentralOH] Python Problem for DoJo Tonight
Message-ID: <85828E70-4515-499C-B036-8F061AA4E7F0@gmail.com>

Hi folks,

Here is today?s problem for people who show up at the DoJo tonight.

	Implement a URL shortener with the following methods:

	shorten(URL), which shortens the URL into a six-character alphanumeric
	string, such as zLg6wl.

	restore(short), which expands the shortened string into the
	original URL. If no such shortened string exists, return null.

	Hint: What if we enter the same URL twice?

See you at the DoJo:

	Smokehouse Brewing
	1130 Dublin Road
	Columbus, OH  43215
	(614) 485-0227

Travis
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From mrehner at e-wrench.net  Fri Mar 15 15:01:08 2019
From: mrehner at e-wrench.net (mrehner)
Date: Fri, 15 Mar 2019 19:01:08 +0000
Subject: [CentralOH] Python Problem for DoJo Tonight
In-Reply-To: <85828E70-4515-499C-B036-8F061AA4E7F0@gmail.com>
References: <85828E70-4515-499C-B036-8F061AA4E7F0@gmail.com>
Message-ID: <0F9E1D0DC01BC4489DC8D649BF9F27F501CB3EF026@EXMBX01.prov.ad.hostmanagement.net>

Hi

Since no one posted a solution here's one with a list of python tutorial links.

import random
import string

# a list of urls for testing
url_list = ["https://astro-pi.org/",
            "https://wiki.python.org/moin/BeginnersGuide",
            "http://gvr.sourceforge.net/",
            "https://docs.python-guide.org/",
            "https://docs.python.org/3/tutorial/",
            "https://developers.google.com/edu/python/",
            "https://docs.python.org/3/tutorial/"]

# the data store- a key,value pair
url_dict = {}


# if url is not in the url dictionary
# creates a random 6 character key
# updates url dictionary with key and url
def shorten(url):
    if not search_for_url(url):
        key = make_key()
        # Add new key value pair
        url_dict.update({key: url})


# returns a url value if key is in dictionary, if not returns None (null)
def restore(short):
    return url_dict.get(short, None)


# Create 6 char key using upper case, lower case and numerical digits
def make_key():
    # seed function automatically called with system current time value as the seed value.
    random.seed()
    return ''.join(random.choice(string.ascii_letters + string.digits) for _i in range(6))


# returns true if a url is a value in the dictionary
def search_for_url(url):
    return url in url_dict.values()


if __name__ == '__main__':

    # fill the dictionary
    for item in url_list:
        shorten(item)

    # Verify results
    for k in url_dict:
        print(k, ": ",  url_dict[k])
        print("url for key: ", k, " is ", restore(k), "\n")

    print("Should return None: ", restore("1a2b3c"))


I'm sure it can be improved and some may worry about a collision with a 62^6 random character generator, but that's it for me.

Cheers,

Mike


______________________________________
From: CentralOH [centraloh-bounces+mrehner=e-wrench.net at python.org] on behalf of Travis Risner [deeppunster at gmail.com]
Sent: Thursday, March 14, 2019 3:22 PM
To: COhPy Mailing List
Subject: [CentralOH] Python Problem for DoJo Tonight

Hi folks,

Here is today?s problem for people who show up at the DoJo tonight.

Implement a URL shortener with the following methods:

shorten(URL), which shortens the URL into a six-character alphanumeric
string, such as zLg6wl.

restore(short), which expands the shortened string into the
original URL. If no such shortened string exists, return null.

Hint: What if we enter the same URL twice?


See you at the DoJo:

Smokehouse Brewing
1130 Dublin Road
Columbus, OH  43215
(614) 485-0227


Travis


From deeppunster at gmail.com  Fri Mar 15 16:25:28 2019
From: deeppunster at gmail.com (Travis Risner)
Date: Fri, 15 Mar 2019 16:25:28 -0400
Subject: [CentralOH] Python Problem for DoJo Tonight
In-Reply-To: <85828E70-4515-499C-B036-8F061AA4E7F0@gmail.com>
References: <85828E70-4515-499C-B036-8F061AA4E7F0@gmail.com>
Message-ID: <4707E9E2-8E9D-46B3-8E0B-026E7C3C6777@gmail.com>

Hi folks,

Mike Rehner has published a solution (see his email in this thread).  I 
have created a (perhaps overly complicated) solution and posted it on 
GitHub at https://github.com/deeppunster/ShortLongURL.

If you have a solution you are willing to share, please post it here (or 
a link to it).

Travis

On 14 Mar 2019, at 15:22, Travis Risner wrote:

> Hi folks,
>
> Here is today?s problem for people who show up at the DoJo tonight.
>
> 	Implement a URL shortener with the following methods:
>
> 	shorten(URL), which shortens the URL into a six-character 
> alphanumeric
> 	string, such as zLg6wl.
>
> 	restore(short), which expands the shortened string into the
> 	original URL. If no such shortened string exists, return null.
>
> 	Hint: What if we enter the same URL twice?
>
> See you at the DoJo:
>
> 	Smokehouse Brewing
> 	1130 Dublin Road
> 	Columbus, OH  43215
> 	(614) 485-0227
>
> Travis
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From deeppunster at gmail.com  Thu Mar 21 15:38:49 2019
From: deeppunster at gmail.com (Travis Risner)
Date: Thu, 21 Mar 2019 15:38:49 -0400
Subject: [CentralOH] DoJo Problem for Tonight
Message-ID: <ADBABA23-036D-43C6-8D4F-ACA099F6A38B@gmail.com>

Hi folks,

If you are interested, here is a problem to consider for tonight?s 
meeting.

????????

Given a 2D matrix of characters and a target word, write a function that 
returns whether the word can be found in the matrix by going 
left-to-right, or up-to-down.

For example, given the following matrix:

[['F', 'A', 'C', 'I'],
  ['O', 'B', 'Q', 'P'],
  ['A', 'N', 'O', 'B'],
  ['M', 'A', 'S', 'S']]
and the target word 'FOAM', you should return true, since it's the 
leftmost column. Similarly, given the target word 'MASS', you should 
return true, since it's the last row.

????????

See you tonight!

Travis

From winningham at gmail.com  Thu Mar 21 16:12:53 2019
From: winningham at gmail.com (Thomas Winningham)
Date: Thu, 21 Mar 2019 16:12:53 -0400
Subject: [CentralOH] DoJo Problem for Tonight
In-Reply-To: <ADBABA23-036D-43C6-8D4F-ACA099F6A38B@gmail.com>
References: <ADBABA23-036D-43C6-8D4F-ACA099F6A38B@gmail.com>
Message-ID: <CAJShLf9qRNG1Ne4zqm2qHDaef-0LoE=YXPoqyMt1h-+q59i-zw@mail.gmail.com>

I tried to make a one-liner for this. I can imagine a possibly simpler
NumPy based idea though maybe? Essentially joins the strings by column and
then simply by row, appends both of those, and then tests for membership.
This would not find partial matches.

the_matrix=[['F', 'A', 'C', 'I'],
  ['O', 'B', 'Q', 'P'],
  ['A', 'N', 'O', 'B'],
  ['M', 'A', 'S', 'S']]


test_function = lambda t : t in [
  ''.join([y[z] for y in the_matrix])
    for z in range(len(the_matrix[0]))] + [
  ''.join(y) for y in the_matrix]

>>> test_function('FOAM')
True
>>> test_function('FOA')
False
>>> test_function('FOAMS')
False
>>> test_function('MASS')
True
>>> test_function('MAS')
False

On Thu, Mar 21, 2019 at 3:39 PM Travis Risner <deeppunster at gmail.com> wrote:

> Hi folks,
>
> If you are interested, here is a problem to consider for tonight?s
> meeting.
>
> ????????
>
> Given a 2D matrix of characters and a target word, write a function that
> returns whether the word can be found in the matrix by going
> left-to-right, or up-to-down.
>
> For example, given the following matrix:
>
> [['F', 'A', 'C', 'I'],
>   ['O', 'B', 'Q', 'P'],
>   ['A', 'N', 'O', 'B'],
>   ['M', 'A', 'S', 'S']]
> and the target word 'FOAM', you should return true, since it's the
> leftmost column. Similarly, given the target word 'MASS', you should
> return true, since it's the last row.
>
> ????????
>
> See you tonight!
>
> Travis
> _______________________________________________
> CentralOH mailing list
> CentralOH at python.org
> https://mail.python.org/mailman/listinfo/centraloh
>
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From neil.ludban at gmail.com  Thu Mar 21 21:27:26 2019
From: neil.ludban at gmail.com (neil ludban)
Date: Thu, 21 Mar 2019 21:27:26 -0400
Subject: [CentralOH] DoJo Problem for Tonight
In-Reply-To: <ADBABA23-036D-43C6-8D4F-ACA099F6A38B@gmail.com>
References: <ADBABA23-036D-43C6-8D4F-ACA099F6A38B@gmail.com>
Message-ID: <CAA5SVLNLwReE5C=b6wdsGk0cErJeg3o0FwSDe1aqF1M9RMEszw@mail.gmail.com>

#!/usr/bin/env python3.7

def row_generator(p):
    for row in p:
        yield ''.join(row)

def col_generator(p):
    for col in zip(*p):
        yield ''.join(col)

def word_in_puzzle(w, p):
    for x in row_generator(p):
        if w in x:
            return True
    for x in col_generator(p):
        if w in x:
            return True
    return False

puzzle = [
    [ 'F', 'A', 'C', 'I' ],
    [ 'O', 'B', 'Q', 'P' ],
    [ 'A', 'N', 'O', 'B' ],
    [ 'M', 'A', 'S', 'S' ],
]

print(word_in_puzzle('FOAM', puzzle))
print(word_in_puzzle('MASS', puzzle))


On Thu, Mar 21, 2019 at 3:39 PM Travis Risner <deeppunster at gmail.com> wrote:

> Hi folks,
>
> If you are interested, here is a problem to consider for tonight?s
> meeting.
>
> ????????
>
> Given a 2D matrix of characters and a target word, write a function that
> returns whether the word can be found in the matrix by going
> left-to-right, or up-to-down.
>
> For example, given the following matrix:
>
> [['F', 'A', 'C', 'I'],
>   ['O', 'B', 'Q', 'P'],
>   ['A', 'N', 'O', 'B'],
>   ['M', 'A', 'S', 'S']]
> and the target word 'FOAM', you should return true, since it's the
> leftmost column. Similarly, given the target word 'MASS', you should
> return true, since it's the last row.
>
> ????????
>
> See you tonight!
>
> Travis
> _______________________________________________
> CentralOH mailing list
> CentralOH at python.org
> https://mail.python.org/mailman/listinfo/centraloh
>
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From ducharmemp at gmail.com  Thu Mar 21 23:35:32 2019
From: ducharmemp at gmail.com (Matthew DuCharme)
Date: Thu, 21 Mar 2019 23:35:32 -0400
Subject: [CentralOH] DoJo Problem for Tonight
In-Reply-To: <CAA5SVLNLwReE5C=b6wdsGk0cErJeg3o0FwSDe1aqF1M9RMEszw@mail.gmail.com>
References: <ADBABA23-036D-43C6-8D4F-ACA099F6A38B@gmail.com>
 <CAA5SVLNLwReE5C=b6wdsGk0cErJeg3o0FwSDe1aqF1M9RMEszw@mail.gmail.com>
Message-ID: <CAKS_gnmvCu6UkH92Mrv92BorTmvOneutL3etwY5qF0nxMdT5uQ@mail.gmail.com>

Since everyone has gone for the readable solutions so far, here's mine. I
wish I could figure out some way to avoid allocating the extra strings from
the slices overall.

from itertools import chain, product

def is_word_in_puzzle(word, puzzle):
    rows = len(puzzle)
    columns = len(puzzle[0])
    puzzle = "".join(chain.from_iterable(puzzle))
    return any(
        word in (
            puzzle[i * columns: i * columns + columns], puzzle[j::columns]
        ) for i, j in product(range(rows), range(columns))
    )

puzzle = [
    [ 'F', 'A', 'C', 'I' ],
    [ 'O', 'B', 'Q', 'P' ],
    [ 'A', 'N', 'O', 'B' ],
    [ 'M', 'A', 'S', 'S' ],
]

assert is_word_in_puzzle('FOAM', puzzle) == True
assert is_word_in_puzzle('ABNA', puzzle) == True
assert is_word_in_puzzle('CQOS', puzzle) == True
assert is_word_in_puzzle('ANOBFFF', puzzle) == False
assert is_word_in_puzzle('MASS', puzzle) == True


On Thu, Mar 21, 2019 at 9:27 PM neil ludban <neil.ludban at gmail.com> wrote:

> #!/usr/bin/env python3.7
>
> def row_generator(p):
>     for row in p:
>         yield ''.join(row)
>
> def col_generator(p):
>     for col in zip(*p):
>         yield ''.join(col)
>
> def word_in_puzzle(w, p):
>     for x in row_generator(p):
>         if w in x:
>             return True
>     for x in col_generator(p):
>         if w in x:
>             return True
>     return False
>
> puzzle = [
>     [ 'F', 'A', 'C', 'I' ],
>     [ 'O', 'B', 'Q', 'P' ],
>     [ 'A', 'N', 'O', 'B' ],
>     [ 'M', 'A', 'S', 'S' ],
> ]
>
> print(word_in_puzzle('FOAM', puzzle))
> print(word_in_puzzle('MASS', puzzle))
>
>
> On Thu, Mar 21, 2019 at 3:39 PM Travis Risner <deeppunster at gmail.com>
> wrote:
>
>> Hi folks,
>>
>> If you are interested, here is a problem to consider for tonight?s
>> meeting.
>>
>> ????????
>>
>> Given a 2D matrix of characters and a target word, write a function that
>> returns whether the word can be found in the matrix by going
>> left-to-right, or up-to-down.
>>
>> For example, given the following matrix:
>>
>> [['F', 'A', 'C', 'I'],
>>   ['O', 'B', 'Q', 'P'],
>>   ['A', 'N', 'O', 'B'],
>>   ['M', 'A', 'S', 'S']]
>> and the target word 'FOAM', you should return true, since it's the
>> leftmost column. Similarly, given the target word 'MASS', you should
>> return true, since it's the last row.
>>
>> ????????
>>
>> See you tonight!
>>
>> Travis
>> _______________________________________________
>> CentralOH mailing list
>> CentralOH at python.org
>> https://mail.python.org/mailman/listinfo/centraloh
>>
> _______________________________________________
> CentralOH mailing list
> CentralOH at python.org
> https://mail.python.org/mailman/listinfo/centraloh
>
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From deeppunster at gmail.com  Thu Mar 28 15:28:00 2019
From: deeppunster at gmail.com (Travis Risner)
Date: Thu, 28 Mar 2019 15:28:00 -0400
Subject: [CentralOH] DoJo Mumbilings for March 28, 2019
Message-ID: <1DCBAA90-1492-48C9-A1A9-E5AEDD52384B@gmail.com>

Hi folks,

Here is a problem proposed by one of our own members.  (The problem is 
in the attached PDF.)

Travis

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From winningham at gmail.com  Thu Mar 28 17:41:36 2019
From: winningham at gmail.com (Thomas Winningham)
Date: Thu, 28 Mar 2019 17:41:36 -0400
Subject: [CentralOH] DoJo Mumbilings for March 28, 2019
In-Reply-To: <1DCBAA90-1492-48C9-A1A9-E5AEDD52384B@gmail.com>
References: <1DCBAA90-1492-48C9-A1A9-E5AEDD52384B@gmail.com>
Message-ID: <CAJShLf-ry6P62a_93zLDG1x1mUvnisp5vPQ2Zm2_B1uCqszdsQ@mail.gmail.com>

1. Fixes row 9 to have category 2
2. Creates an EAV "facts" table, sort of like a graph
3. Assumes only one matching value, no aggregate sums, etc
4. Extendable to other similar problems, although report generation is
rather hard coded for this dataset

import uuid

fields = ['id','category','type','value']

data = [
        [1,1,"A",2],
        [2,3,"D",5],
        [3,4,"B",7],
        [4,2,"C",8],
        [5,4,"D",9],
        [6,4,"C",1],
        [7,1,"D",3],
        [8,3,"C",9],
        [9,2,"A",5],
        [10,3,"A",4],
        [11,1,"B",6],
        [12,4,"A",2],
        [13,2,"D",6],
        [14,3,"B",3],
        [15,2,"B",8],
        [16,1,"C",0]
        ]


# Create fact table

facts = []

for x in data:
    this_fact=str(uuid.uuid1())
    for f in range(len(fields)):
        facts.append((this_fact, fields[f], x[f]))

# Create some query functions for EAV structure

E,A,V = 0,1,2

q_e4av = lambda a,v : [x[E] for x in facts if x[A]==a and x[V]==v]
q_v4a  = lambda a   : [x[V] for x in facts if x[A]==a]
q_v4ea = lambda e,a : [x[V] for x in facts if x[E]==e and x[A]==a]

# Generate report

uniq_cats = sorted(list(set(q_v4a('category'))))
uniq_typs = sorted(list(set(q_v4a('type'))))

result=[]
for c in uniq_cats:
    row=[]
    for t in uniq_typs:
       cat_ids = set(q_e4av('category',c))
       typ_ids = set(q_e4av('type',t))
       result_entity = list(cat_ids & typ_ids)
       if len(result_entity) > 0:
           row.append(q_v4ea(result_entity[0],'value')[0])
       else:
           row.append(None)
    result.append(row)

print('\t'+'\t'.join(uniq_typs))
for c in range(len(uniq_cats)):
    print('\t'.join([str(uniq_cats[c])]+[repr(x) for x in result[c]]))

#-----------------------

$ python3 test1.py

        A       B       C       D
1       2       6       0       3
2       5       8       8       6
3       4       3       9       5
4       2       7       1       9




On Thu, Mar 28, 2019 at 3:53 PM Travis Risner <deeppunster at gmail.com> wrote:

> Hi folks,
>
> Here is a problem proposed by one of our own members.  (The problem is
> in the attached PDF.)
>
> Travis
>
> _______________________________________________
> CentralOH mailing list
> CentralOH at python.org
> https://mail.python.org/mailman/listinfo/centraloh
>
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From neil.ludban at gmail.com  Thu Mar 28 22:09:20 2019
From: neil.ludban at gmail.com (neil ludban)
Date: Thu, 28 Mar 2019 22:09:20 -0400
Subject: [CentralOH] DoJo Mumbilings for March 28, 2019
In-Reply-To: <1DCBAA90-1492-48C9-A1A9-E5AEDD52384B@gmail.com>
References: <1DCBAA90-1492-48C9-A1A9-E5AEDD52384B@gmail.com>
Message-ID: <CAA5SVLOKvzzP_UrA9NjJJa=oHnW5YP_0UTe7c_sO=VCPVnCw+w@mail.gmail.com>

#!/usr/bin/env python3.7

import re
from itertools import chain

def load_table(text):
    for row in text.splitlines():
        m = re.match(r'[0-9]+\s+([1-4])\s+([A-D])\s+([0-9])', row)
        if not m:
            continue
        yield m.groups()
    return

def make_report(data):
    sep = '+----------+-----+-----+-----+-----+'
    yield sep
    headers = [ 'A', 'B', 'C', 'D' ]
    it = chain(headers, data)
    sep = sep.replace('-', '=')
    for Category in ( 'Category', '1', '2', '3', '4' ):
        A, B, C, D = next(it), next(it), next(it), next(it)
        yield F'| {Category:>8} |  {A}  |  {B}  |  {C}  |  {D}  |'
        yield sep
        sep = sep.replace('=', '-')
    return

sample_table_1 = '''
===  =========  =====  =====
ID   Category   Type   Value
===  =========  =====  =====
1    1          A      2
2    3          D      5
3    4          B      7
4    2          C      8
5    4          D      9
6    4          C      1
7    1          D      3
8    3          C      9
9    2          A      5  XXX
10   3          A      4
11   1          B      6
12   4          A      2
13   2          D      6
14   3          B      3
15   2          B      8
16   1          C      0
===  =========  =====  =====
'''

sample_report_1 = '''
+----------+-----+-----+-----+-----+
| Category |  A  |  B  |  C  |  D  |
+==========+=====+=====+=====+=====+
|        1 |  2  |  6  |  0  |  3  |
+----------+-----+-----+-----+-----+
|        2 |  5  |  8  |  8  |  6  |
+----------+-----+-----+-----+-----+
|        3 |  4  |  3  |  9  |  5  |
+----------+-----+-----+-----+-----+
|        4 |  2  |  7  |  1  |  9  |
+----------+-----+-----+-----+-----+
'''

rows = [ row[2] for row in sorted(load_table(sample_table_1)) ]
report = '\n'.join(make_report(rows))

print(report)
assert report.strip() == sample_report_1.strip()

#--#


On Thu, Mar 28, 2019 at 3:53 PM Travis Risner <deeppunster at gmail.com> wrote:

> Hi folks,
>
> Here is a problem proposed by one of our own members.  (The problem is
> in the attached PDF.)
>
> Travis
>
> _______________________________________________
> CentralOH mailing list
> CentralOH at python.org
> https://mail.python.org/mailman/listinfo/centraloh
>
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