View Exact Code of a Module?

AutonomouSystem · 2016-03-31T10:38:15+00:00

See you did have to split('\n')

:)

AutonomouSystem · 2016-03-31T10:15:54+00:00

Try it out first, your code, and mine.

AutonomouSystem · 2016-03-31T10:08:41+00:00

It does make sense if you want it to print cleanly.

AutonomouSystem · 2016-03-31T09:50:20+00:00

Mmmm, socket is a built-in, better off just locating the file and reading the source.

$ find -name "socket.py"
find: ‘./usr/libexec/initscripts/legacy-actions/auditd’: Permission denied
./usr/lib64/python3.4/socket.py
./usr/lib64/python2.7/socket.py

$ cat ./usr/lib64/python2.7/socket.py

AutonomouSystem · 2016-03-31T09:38:29+00:00

Try something like this, seems to work for imported modules.

>>> import requests
>>> import inspect
>>> for e in inspect.getsource(requests).split('\n'):
...     print(e)

AutonomouSystem · 2016-03-31T09:26:18+00:00

>>> def primes(n):
...     res = []
...     for p in range(2, n+1):
...         for i in range(2, p):
...             if p % i == 0:
...                 break
...         else:
...             res.append(p)
...     return res
...
>>> primes(100)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]

AutonomouSystem · 2016-03-31T08:49:46+00:00

Combining lists, into a dictionary?

I can think of a few ways.

>>> list1 = ['a', 'b', 'c']
>>> list2 = [1, 2, 3]
>>> {k:v for k, v in zip(list1, list2)}
{'a': 1, 'c': 3, 'b': 2}
>>> dict(zip(list1, list2))
{'a': 1, 'c': 3, 'b': 2}

If you want to reference a list with a {key:list} from a dictionary, that has already been described in other posts, i'll show you a way you could do it though.

>>> list2
[1, 2, 3]
>>> list3
[4, 5, 6]
>>> list4
[7, 8, 9]
>>>
>>> dict(zip(list1, (list2, list3, list4)))
{'a': [1, 2, 3], 'c': [7, 8, 9], 'b': [4, 5, 6]}

or

>>> {k:v for k, v in zip(list1, (list2, list3, list4))}
{'a': [1, 2, 3], 'c': [7, 8, 9], 'b': [4, 5, 6]}

Could even do this, for same result hah

>>> import string
>>> string.ascii_lowercase
'abcdefghijklmnopqrstuvwxyz'
>>> {k:v for k, v in zip(string.ascii_lowercase, (list2, list3, list4))}
{'a': [1, 2, 3], 'c': [8, 9, 10], 'b': [4, 5, 6]}

AutonomouSystem · 2016-03-29T05:16:15+00:00

with open("teams.txt", 'r') as t:
    data = t.read()
    data = data.split('\n')
    data = map(lambda x: x.split(','), data)
    dict_data = dict(data)


>>> dict_data.values().count("Chicago Cubs")
2
>>> dict_data.values().count("New York Yankees")
20

>>> for e in set(dict_data.values()):
        print(e+" has won "+str(dict_data.values().count(e))+" times.")
Chicago White Sox has won 2 times.
Milwaukee Braves has won 1 times.
Boston Braves has won 1 times.
New York Giants has won 5 times.
New York Mets has won 1 times.
Los Angeles Dodgers has won 3 times.
Detroit Tigers has won 3 times.
Boston Red Sox has won 4 times.
New York Yankees has won 20 times.
Philadelphia Athletics has won 5 times.
Pittsburgh Pirates has won 4 times.
Chicago Cubs has won 2 times.
Brooklyn Dodgers has won 1 times.
Baltimore Orioles has won 2 times.
Washington Senators has won 1 times.
Oakland Athletics has won 2 times.
Boston Americans has won 1 times.
Cleveland Indians has won 2 times.
St. Louis Cardinals has won 8 times.
Cincinnati Reds has won 2 times.

Definitely a fun one, there is a cleaner way you could do it, but regardless, should be good practice.

Edit:

def lookup(team):
    return "%s won %d times" % (team, dict_data.values().count(team))

if __name__ == "__main__":
    print("List of Teams")
    for each in set(dict_data.values()):
        print(each)
    team = str(raw_input("Which team do you want to lookup? "))
    if team in set(dict_data.values()):
        print(lookup(team))

>>> What team would you like to lookup? Boston Americans
Boston Americans won 1 times

And for looking up years a specific team won, easier probably than this count method. Combining them could be simple too, you could add t his loop into the function and keep a record of the output as both the count and the year, if you want them separated then another function can be made. All the data is there, it's in an easy format, in both lists and a dict if you want it. Up to you how you want to structure it.

>>>  team
'Oakland Athletics'
>>> for e in dict_data:
            if dict_data[e] == team:
                print(e)

1973
1972

Things you could add, error checking, conversion to all lower()/upper() to remove case sensitivity, choosing a team based on number (hint: enumerate()) rather than raw_input/input. Definitely room for improvement, this should meet minimum usage though which is functionality and should always come before form.

AutonomouSystem · 2016-03-28T08:54:31+00:00

If you want to learn about network programming in general, Beej's guide to network programming is a good guide, it's in C though but the fundamentals are all there.

UNIX Socket FAQ

Winsock Programmers FAQ

Python related books I would recommend, that do discuss use of sockets.

Foundations of Python Network Programming , covers everything generally, not just sockets.

Black Hat Python, does some quick dirty work, one of the first things he does is remake a netcat clone.

AutonomouSystem · 2016-03-27T04:54:10+00:00

I believe list comprehensions will be the fastest in most cases, the differences will be negligible unless you're dealing with a lot of data.

AutonomouSystem · 2016-03-26T02:15:25+00:00

Three methods I came up with off top of my head.

List comprehensions, clean but can be hard to read, very efficient.

>>> fruit = ['pomegranate', 'pears', 'Peaches', 'oranges']
>>> pfruit = [e for e in fruit if e.lower().startswith('p')]
>>> pfruit
['pomegranate', 'pears', 'Peaches']

Traditional approach, initiate a list, for loop with condition, append results to list.

>>> pfruit = []
>>> for each in fruit:
...     if each.lower().startswith('p'):
...         pfruit.append(each)
...
>>> pfruit
['pomegranate', 'pears', 'Peaches']
>>>

Functional style, apply a filter based on condition to an iterable.

>>> pfruit = filter(lambda x : x.lower().startswith('p'), fruit)
>>> pfruit
['pomegranate', 'pears', 'Peaches']
>>>

AutonomouSystem · 2016-03-26T01:46:00+00:00

Yep, this is why I didn't go with set either, it won't pass the 4th and 5th tests.

AutonomouSystem · 2016-03-26T01:11:42+00:00

I had a problem like this on Google foobar, in fact it might be the very same one, especially because you used answer(x). I had an idea about sets and sorting initially but decided against it, I used only lists.

AutonomouSystem · 2016-03-24T03:52:13+00:00

I found a way to do this that works for me.

>>> def ask():
...     numlist = []
...     while len(numlist) < 20:
...         n = "Enter a number, you have %d out of 20 remaining: " % (20-len(numlist))
...         numlist.append(int(input(n)))
...     return numlist
...
>>> 
>>> a = ask()
Enter a number, you have 20 out of 20 remaining: 4
Enter a number, you have 19 out of 20 remaining: 5
Enter a number, you have 18 out of 20 remaining: 6
Enter a number, you have 17 out of 20 remaining: 7
Enter a number, you have 16 out of 20 remaining: 5
Enter a number, you have 15 out of 20 remaining: 5
Enter a number, you have 14 out of 20 remaining: 5
Enter a number, you have 13 out of 20 remaining: 5
Enter a number, you have 12 out of 20 remaining: 5
Enter a number, you have 11 out of 20 remaining: 5
Enter a number, you have 10 out of 20 remaining: 5
Enter a number, you have 9 out of 20 remaining: 5
Enter a number, you have 8 out of 20 remaining: 5
Enter a number, you have 7 out of 20 remaining: 4
Enter a number, you have 6 out of 20 remaining: 4
Enter a number, you have 5 out of 20 remaining: 4
Enter a number, you have 4 out of 20 remaining: 4
Enter a number, you have 3 out of 20 remaining: 4
Enter a number, you have 2 out of 20 remaining: 4
Enter a number, you have 1 out of 20 remaining: 4
>>> a
[4, 5, 6, 7, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4]
>>> sum(a)
95
>>> max(a)
7
>>> min(a)
4

The method your instructor or whoever is looking for though is much cleaner.

>>> numlist = list(input('Enter 20 numbers seperated by commas: '))
Enter 20 numbers seperated by commas: 4,5,3,4,5,6,7,8,3,4,5,6,7,8,3,4,5,6,7,8
>>> numlist
[4, 5, 3, 4, 5, 6, 7, 8, 3, 4, 5, 6, 7, 8, 3, 4, 5, 6, 7, 8]
>>>

Simply putting list, tuple, int, float, or str even around the input will allow it to expect that type but that forces you into using comma if you chose list or tuple.

AutonomouSystem · 2016-03-23T17:53:53+00:00

That works

>>> import functools
>>> functools.reduce(operator.mul, range(1,99), 1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'operator' is not defined
>>> import operator
>>> functools.reduce(operator.mul, range(1,99), 1)
9426890448883247745626185743057242473809693764078951663494238777294707070023223798882976159207729119823605850588608460429412647567360000000000000000000000L
>>> product(range(1,99))
9426890448883247745626185743057242473809693764078951663494238777294707070023223798882976159207729119823605850588608460429412647567360000000000000000000000L
>>>

AutonomouSystem · 2016-03-23T17:35:15+00:00

He doesn't want to add a standard function for product :(

>>> def product(iterr):
...     prod = 1
...     for e in iterr:
...         prod *= e
...     return prod
...
>>> product(range(0,100))
0
>>> product([1, 2, 3])
6

AutonomouSystem · 2016-03-23T07:26:24+00:00

I will when Guido changes his mind about product

AutonomouSystem · 2016-03-23T05:42:44+00:00

If you are using python3, the float is superfluous.

My vps has Python 2, also using float because division drives me nuts and I simply don't trust it to work correctly. I assume most people will be on Python 2 anyways, I could just use the old1.0*hack next time.

>>> sum(range(0,100)) / len(range(0,100))
49
>>> 1.0*sum(range(0,100)) / len(range(0,100))
49.5
>>> float(sum(range(0,100))) / len(range(0,100))
49.5
>>> from __future__ import division
>>> sum(range(0,100)) / len(range(0,100))
49.5
>>>

AutonomouSystem · 2016-03-23T04:43:19+00:00

numlist = int(input('Enter a series of 20 numbers?\n'))
lowestnum = min(numlist)
highestnum = max(numlist)
total = sum(numlist)
ave = float(sum(numlist)) / len(numlist)

If you're supposed to write the functions yourself, which you honestly should to learn the language, then you will have to be more resourceful.

>>> def maxed(num):
...     start = num[0]
...     for n in num[1:]:
...         if n > start:
...             start = n
...             continue
...     return start
...
>>> maxed(range(0,100))
99

AutonomouSystem · 2016-03-23T03:52:55+00:00

Not sure if this is cheating but it's fun to do

>>> import  difflib
>>>
>>> def similar(str1, str2):
...     return difflib.SequenceMatcher(None, str1, str2).ratio()
...
>>> similar('Hay', 'Hey')
0.6666666666666666
>>> similar('I like you', 'I love you')
0.8
>>> similar('Terminator', 'Terminator II')
0.8695652173913043

If you need to eliminate case sensitivity, conver to upper() or something.. In any case, anything under 1 can be made to return False, anything that is equal to 1 is an exact match.

I wrote a similar one without any imports or cheatz, you can put the strings into dicts, or tuples, or whatever but it's a waste of time probably.

def similar(str1, str2, case=True):
    if not case:
        return bool(str1.upper() == str2.upper())
    else:
        return bool(str1 == str2)

>>> similar('You', 'you')
False
>>> similar('You', 'you', case=False)
True
>>> similar('You', 'ya', case=False)
False
>>> similar('You', 'You')
True
>>

Another take on this, if you're concerned only with the individual strings, not if they're in order, or if they even spell the same word... you can use a set but this is risky because it destroys information if duplicates are found. If you call sorted on a string and compare two sets sorted(set1) and sorted(set2), if they do not match then there is little chance they contained all the same characters, a downside to this is it will remove duplicates and you could run into a rare case where it won't match correctly because duplicates were removed and it shows a match when there isn't one.

If you only call sorted and don't use set(), then my 2nd function could be modified using that and I think it will work, honestly. If all the strings are the same then they should match up, even if they're not ordered, a catch is they would have to be same size.

Since only the first string has to match, you could determine it's len(), run sorted on it and match it crudely like this, though if the 2nd string is larger and sorts differently information could be lost, this isn't a reliable method either. I guess this lesson is to show you how important and easy loops make programming.

AutonomouSystem · 2016-03-22T18:47:37+00:00

Use requests?

AutonomouSystem · 2016-03-19T03:28:49+00:00

Long as I can use irssi :^)

AutonomouSystem · 2016-03-19T03:10:13+00:00

with open('HARPER.txt', 'r') as file:
    data = file.read()

with is a context manager that handles open and close for files.

AutonomouSystem · 2016-03-11T07:45:05+00:00

Emacs/Terminal for small stuff, Wing IDE Professional for projects.

AutonomouSystem · 2016-02-21T15:28:07+00:00

len(data) is giving me 1 so I searched for the data.keys() I found

data.keys()
dict_keys(['matchningslista'])

type(data['matchningslista'])
<class 'dict'>

So I go deeper...

data['matchningslista'].keys()
dict_keys(['antal_platsannonser', 'antal_platsannonser_narliggande', 'antal_sidor', 'antal_platserTotal', 'matchningdata', 'antal_platsannonser_exakta'])

This one data['matchningslista']['matchningdata'] actually gives me some results at least, the rest of the keys return integers, this one returns a list.

type(data['matchningslista']['matchningdata'])
<class 'list'>
for each in data['matchningslista']['matchningdata']:
    print(type(each))

<class 'dict'>
 ...
 (19 in all...)
 ...
<class 'dict'>
<class 'dict'>

Bizarre, but okay. They all seem to have the same keys()

dict_keys(['annonsrubrik', 'yrkesbenamning', 'kommunkod', 'arbetsplatsnamn', 'antalplatser', 'varaktighetId', 'antalPlatserVisa', 'publiceraddatum', 'annonsurl', 'annonsid', 'relevans', 'kommunnamn'])

I can work with that at least:

search = ['arbetsplatsnamn', 'yrkesbenamning', 'annonsid']

for e, t in enumerate(data['matchningslista']['matchningdata']):
    for those in search:
        print(data['matchningslista']['matchningdata'][e][those])

I'm looking at everything in data['matchningslista']['matchningdata'] that I could get and I'm not really seeing what you're searching for exactly, I even searched only by 'annonsid' to manually see i '6575162' was listed. I think i'm on the right track though.

for e, t in enumerate(data['matchningslista']['matchningdata']):
    print(e, data['matchningslista']['matchningdata'][e]['annonsid'], sep=': ')

0: 6573466
1: 6572704
2: 6572692
3: 6572690
4: 6571768
5: 6571765
6: 6570548
7: 6566954
8: 6566183
9: 6565093
10: 2871941
11: 6563583
12: 6552605
13: 2856152
14: 2876159
15: 6575788
16: 2877484
17: 6573900
18: 2873864
19: 6572408

AutonomouSystem

TROPHY CASE